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Ray Optics and Optical Instruments

From the laws of reflection to the telescope — image formation by mirrors, lenses and prisms

🎯 Overview Chapter hero + roadmap

🔬 Interactive 3D · A single ray meets a mirror, a glass surface and a prism — watch reflection, refraction and dispersion emerge from the same two laws.

Light is the narrow band of electromagnetic radiation, wavelengths of roughly $400$ nm to $750$ nm, that our eyes detect and through which we know and interpret the world around us. Two everyday facts anchor this chapter. First, light travels at an enormous but finite speed, $c = 3 \times 10^8$ m s$^{-1}$ in vacuum, the highest speed attainable in nature. Second, it travels in straight lines. Because the wavelength of light is tiny compared with ordinary objects, a light wave can be treated as travelling along a straight path called a ray, and a bundle of such rays forms a beam. This ray picture is the single idea from which every mirror, lens, prism and telescope in the chapter is built. 🔉⇢

This chapter is best seen as one connected journey. We begin with the two laws of reflection and refraction, fix a sign convention so the algebra stays honest, and derive the mirror equation for curved reflectors. We then follow refraction from a flat interface, through total internal reflection and optical fibres, to a single spherical surface, and combine two such surfaces into the thin lens. From there come the lens maker's formula, the power and combination of lenses, and the prism with its minimum deviation. The payoff is the design of real instruments, the microscope and the telescope, each just a clever arrangement of the same few equations you will have already met. 🔉⇢

Everything that follows rests on just two laws. The law of reflection says the angle of reflection equals the angle of incidence, and the incident ray, reflected ray and the normal to the surface at the point of incidence all lie in one plane. The law of refraction, Snell's law, states that $\frac{\sin i}{\sin r} = n_{21}$, where $n_{21}$ is the refractive index of the second medium with respect to the first. When $n_{21} \gt 1$ the ray bends towards the normal and the second medium is optically denser; when $n_{21} \lt 1$ it bends away. These laws hold at every point of any surface, plane or curved, and the whole chapter is an exercise in applying them to spherical surfaces. 🔉⇢

To turn geometry into formulae that work for every case, we adopt one bookkeeping rule: the Cartesian sign convention. All distances are measured from the pole of a mirror or the optical centre of a lens along the principal axis. Distances measured in the direction of the incident light are taken positive; those measured against it are negative. Heights above the principal axis are positive, heights below are negative. This single convention is why one mirror equation and one lens equation can handle concave and convex surfaces, and real and virtual images, alike. Master it early, because a slip in signs is the most common way marks are quietly lost throughout this chapter and in the examination. 🔉⇢

For spherical mirrors we work with paraxial rays, those incident close to the pole and making small angles with the axis. A parallel beam reflected from a concave mirror converges to the principal focus $F$; from a convex mirror the reflected rays appear to diverge from $F$. A short piece of small-angle geometry then shows that the focal length is half the radius of curvature, $f = R/2$. With the sign convention, $f$ is negative for a concave mirror and positive for a convex mirror. This one relation lets you convert a stated radius of curvature straight into the focal length needed by the mirror equation, so that a single number describes the mirror completely. 🔉⇢

Choosing two convenient rays, one parallel to the axis and one through the centre of curvature, locates the image of any point. Comparing similar triangles and applying the sign convention yields the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, together with the linear magnification $m = \frac{h'}{h} = -\frac{v}{u}$. These two formulae describe every case: a concave mirror can form a real inverted image or, for objects inside the focus, a virtual erect one, while a convex mirror always gives a diminished virtual image. The side-view mirror example, where an approaching jogger's image seems to speed up, shows how much physical intuition these compact relations carry once the signs are handled correctly. 🔉⇢

Turning to refraction, when light crosses into another transparent medium part is reflected and part bends at the interface. The refractive index $n_{21}$ is a characteristic of the pair of media and of the wavelength, but not of the angle of incidence, and it obeys $n_{12} = \frac{1}{n_{21}}$. Simple but important consequences follow directly. A ray passing through a parallel-sided slab emerges parallel to itself, undeviated but laterally shifted. The bottom of a water tank looks raised because, viewed near the normal, the apparent depth equals the real depth divided by the refractive index. Optical density, note carefully, is about the speed of light in the medium, not mass per unit volume. 🔉⇢

When light travels from a denser to a rarer medium it bends away from the normal, and beyond a certain angle refraction becomes impossible. That threshold is the critical angle $i_c$, defined by $\sin i_c = n_{21}$, and for angles greater than $i_c$ the light is totally internally reflected with no transmitted loss. This one phenomenon powers a surprising range of technology. Totally reflecting prisms bend or invert images by $90^\circ$ or $180^\circ$, exploiting the fact that crown and dense flint glass both have critical angles below $45^\circ$. Optical fibres, a core of higher index inside a cladding of lower index, trap a signal by repeated total internal reflection and carry it over kilometres with little loss, even around bends. 🔉⇢

Before tackling lenses we handle a single spherical refracting surface. Treating an infinitesimal patch as planar, applying Snell's law in the small-angle form $n_1 i = n_2 r$, and using exterior-angle geometry gives $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. This relation connects the object and image distances to the two refractive indices and the radius of curvature, and it holds for any spherical surface. It is the workhorse from which the lens formulae are built, because a thin lens is simply two such surfaces in quick succession, the image from the first surface acting as a virtual object for the second. 🔉⇢

Applying that surface relation twice and adding gives the lens maker's formula, $\frac{1}{f} = (n_{21} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$, which lets a designer choose radii of curvature for a desired focal length in a given medium. Setting the object at infinity defines the focus and focal length, and eliminating the refractive index leaves the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, with magnification $m = \frac{h'}{h} = \frac{v}{u}$. A single sign convention again makes both formulae valid for converging and diverging lenses and for real and virtual images. The magician who makes a lens vanish in a matched liquid, where $n_1 = n_2$ so that $f \to \infty$, shows the formula's reach. 🔉⇢

The bending strength of a lens is captured by its power $P = \frac{1}{f}$, measured in dioptres, with $1$ D $= 1$ m$^{-1}$; power is positive for a converging lens and negative for a diverging one, so a prescription of $+2.5$ D means a convex lens of focal length $40$ cm. When thin lenses are placed in contact, their powers simply add, $P = P_1 + P_2 + P_3 + \dots$, equivalently $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \dots$, while the net magnification is the product of the individual magnifications. This additivity is the design principle behind the multi-element lenses of cameras, microscopes and telescopes, where several simple lenses cooperate. 🔉⇢

A prism deviates a ray by an angle $d = i + e - A$, where $A$ is the refracting angle, and its internal angles satisfy $r_1 + r_2 = A$. As the angle of incidence varies, the deviation passes through a minimum $D_m$, where the ray inside runs parallel to the base and $i = e$ with $r_1 = r_2$. At that point the refractive index follows from $n_{21} = \frac{\sin[(A + D_m)/2]}{\sin(A/2)}$, giving a clean experimental method to measure it. For a thin prism this reduces to $D_m = (n_{21} - 1)A$, so thin prisms deviate light only slightly. Because the index depends on wavelength, a prism also disperses white light into its constituent colours. 🔉⇢

Finally the chapter assembles these pieces into instruments. A simple microscope, one short-focus converging lens, gives angular magnification $m = 1 + \frac{D}{f}$ at the near point, or $m = \frac{D}{f}$ for an image at infinity. A compound microscope compounds two lenses to reach $m = \frac{L}{f_o}\cdot\frac{D}{f_e}$, while a telescope, with a long-focus objective and short-focus eyepiece, delivers $m = \frac{f_o}{f_e}$ with tube length $f_o + f_e$. Ray optics is consistently among the highest-weighted topics for the entrance examination, and its questions reward exactly the connected thinking built here: the same two laws, the same sign convention and the same lens and mirror equations reappear from the plane mirror all the way to the reflecting telescope. 🔉⇢

What you will master 🔉⇢

🧠 Concepts Map of the deep dives

This chapter breaks into the concepts below. Each has its own tab with interactive 3D, a full derivation, and a JEE worked example. The Contents tab has the complete prose.

The Cartesian Sign CReflection by SpheriThe Mirror Equation Refraction & Snell's▶Total Internal Refle▶Refraction at a SpheThe Lens Maker's ForThe Thin Lens Formul▶Power of a Lens & CoRefraction Through a▶The Simple MicroscopThe Compound MicroscThe Refracting & Ref
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What you are looking at

A map of the whole chapter. Each box is one concept with its governing relation, and the arrows are the recommended order to learn them in.

Click any box to jump straight to that concept’s tab.

🗺️ Concept map — click any concept to open its tab. Each box shows the concept and its governing formula; the path is the recommended learning order.

The Cartesian Sign Convention 🔉⇢

A single set of rules for measuring distances and heights in reflection and refraction, so that one mirror formula and one lens formula cover every case.

Reflection by Spherical Mirrors & Focal Length 🔉⇢

The laws of reflection applied to concave and convex mirrors, with the focal length shown to be half the radius of curvature (f = R/2) for paraxial rays.

The Mirror Equation & Magnification 🔉⇢

The relation 1/v + 1/u = 1/f between object distance, image distance and focal length, together with the linear magnification m = -v/u.

Refraction & Snell's Law 🔉⇢

The bending of light at an interface between two media, governed by Snell's law sin i / sin r = n21, with the refracted ray bending toward or away from the normal depending on relative optical density.

Total Internal Reflection & Optical Fibres 🔉⇢

When light travelling in a denser medium meets a rarer medium beyond the critical angle, it is reflected entirely back — the principle behind optical fibres and totally reflecting prisms.

Refraction at a Spherical Surface 🔉⇢

The relation n2/v - n1/u = (n2 - n1)/R for image formation by a single spherical refracting surface, the building block of the lens maker's formula.

The Lens Maker's Formula 🔉⇢

The formula 1/f = (n21 - 1)(1/R1 - 1/R2) that gives the focal length of a thin lens from the refractive index and the two radii of curvature.

The Thin Lens Formula & Magnification 🔉⇢

The relation 1/v - 1/u = 1/f for a thin lens, with magnification m = v/u, valid for convex and concave lenses and for real and virtual images.

Power of a Lens & Combination of Lenses 🔉⇢

The power P = 1/f (in dioptres) measures how strongly a lens converges or diverges light; for thin lenses in contact the powers add, P = P1 + P2 + ...

Refraction Through a Prism & Minimum Deviation 🔉⇢

A prism deviates a ray by an angle that depends on the angle of incidence; at minimum deviation the ray passes symmetrically and n21 = sin[(A + Dm)/2] / sin(A/2).

The Simple Microscope (Magnifier) 🔉⇢

A single converging lens of short focal length used close to the eye to give an erect, magnified virtual image, with magnifying power m = 1 + D/f (image at the near point).

The Compound Microscope 🔉⇢

Two converging lenses in series — an objective forming a real magnified image and an eyepiece acting as a magnifier — giving m = (L/fo)(D/fe).

The Refracting & Reflecting Telescope 🔉⇢

An instrument for the angular magnification of distant objects, m = fo/fe, using a large-aperture objective (lens or mirror) and a short-focus eyepiece.

📖 Contents Full chapter — read straight through

The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.

Ray Optics and Optical Instruments
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What you are looking at

The chapter banner — the physics of this chapter in a single moving picture, with live symbols and values.

Every diagram below it is interactive: drag the controls and the numbers move with the drawing.

The Cartesian Sign Convention 🔉⇢

🎯 Slide the object around the pole and read its coordinates. ONE rule fixes every sign in this chapter: measured along the incident light (→) is +, against it −; up is +, down is −.
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Cartesian sign convention — origin at the pole; along the incident light (left→right) is +, against it is −; heights up are +, down are −.
this object sits at x = —, y = — — —
What you are looking at — the coordinate frame every mirror and lens problem secretly uses.
  • The pole P sits at the origin.
  • The teal beam is the incident light, always drawn travelling left→right. That direction defines +x.
  • The red arrow is your object; its foot is on the axis and its coordinates read out live.
What to do
  1. Drag the object to the left of the pole. x goes negative — which is exactly why a real object distance u is written as a negative number.
  2. Flip the height below the axis and watch y turn negative.
Why it matters — get this one convention right and the single mirror equation and single lens equation handle every case (real/virtual, erect/inverted) with no special rules. Get it wrong and every sign in the chapter is a coin toss.
Definition: A single set of rules for measuring distances and heights in reflection and refraction, so that one mirror formula and one lens formula cover every case. 🔉⇢

Before we can derive a single mirror formula or a single lens formula that survives every geometric case, we must first agree on how distances are measured along the principal axis. The Cartesian sign convention supplies exactly this bookkeeping discipline. According to it, all distances are measured from one reference point: the pole of a spherical mirror or the optical centre of a lens. Nothing is measured from the centre of curvature, from the focus, or from the object itself. Fixing a single origin is the first act of self-consistency, because it guarantees that the object distance, the image distance, and the focal length are all reported against the same ruler laid along the principal axis. 🔉⇢

The second ingredient is a chosen positive direction. We take the direction in which the incident light actually travels, conventionally left to right, as the positive $x$ direction. Any distance measured in the same direction as the incident light is therefore counted positive, while any distance measured against the direction of the incident light is counted negative. This is why, for a real object sitting in front of a mirror, the light must travel from the object to the reflecting surface, and the object distance comes out negative: to reach the object from the pole you walk opposite to the incident beam. The sign is not an arbitrary decoration; it encodes geometry. 🔉⇢

Heights are handled by a companion rule. Perpendicular to the principal axis, distances measured upward from the axis are taken as positive and distances measured downward are taken as negative. Thus an erect object of height $h$ is a positive number, and an inverted image of the same physical size carries a negative height $h'$. Because the magnification is defined as $m=h'/h$, this height convention is precisely what lets a single number report both the size ratio and the orientation of the image in one stroke, a point we will exploit when the mirror and lens magnifications are derived. 🔉⇢

It is worth stressing why one insists on a convention at all. Without it, every geometric configuration, an object beyond the centre of curvature, an object between the focus and the pole, a concave mirror, a convex mirror, would demand its own separately memorised formula with its own placement of plus and minus signs. That is a recipe for error under examination pressure. With one accepted convention, it turns out that a single formula for spherical mirrors and a single formula for spherical lenses can handle all the different cases. The convention buys generality: you learn one relation and let the signs do the case analysis for you. 🔉⇢

Consider how the convention assigns signs to the standard optical elements. For a concave mirror the focus and the centre of curvature lie in front of the mirror, on the same side as the incident light source, so both the focal length and the radius of curvature are negative. For a convex mirror the focus and the centre of curvature lie behind the reflecting surface, away from the incident light, so both the focal length and the radius of curvature are positive. These signs are consequences of the convention, not extra facts to remember, and the relation $f=R/2$ carries the sign automatically. 🔉⇢

The same logic disciplines the image distance. If the reflected rays actually converge and meet in front of the mirror, the image is real and lies on the incident-light side, so its distance is negative. If the reflected rays only appear to diverge from a point behind the mirror, the image is virtual and its distance is positive. Thus the mere sign of the computed image distance already tells you the nature of the image before you draw a single ray. A negative image distance for a mirror announces a real, and for a mirror also inverted, image; a positive one announces a virtual, erect image. 🔉⇢

A frequent source of confusion is the belief that the convention changes depending on whether the mirror is concave or convex, or whether the object is real or virtual. It does not. The convention is a fixed coordinate frame anchored at the pole with the incident light defining the positive axis. What changes from case to case are the numerical values and their resulting signs, not the rules that assign them. This invariance is exactly what makes the convention powerful: you set up the same axis every time, substitute the signed quantities, and read off a signed answer whose sign is itself physically meaningful. 🔉⇢

The convention also clarifies the treatment of a virtual object, a subtlety that appears in multi-element systems such as two lenses in contact. When converging rays are intercepted by a second surface before they meet, the point toward which they were heading acts as an object lying on the far side, measured in the direction of the incident light, and therefore carries a positive object distance. Beginners who apply the convention mechanically, measuring every distance from the same pole with the same positive direction, handle this case correctly, whereas those who memorise sign patterns for real objects alone stumble. 🔉⇢

It helps to rehearse the bookkeeping on a concrete concave mirror. Suppose an object stands $20\ \text{cm}$ in front of a concave mirror whose radius of curvature is $30\ \text{cm}$. By the convention the object distance is $u=-20\ \text{cm}$, the radius is $R=-30\ \text{cm}$, and hence the focal length is $f=R/2=-15\ \text{cm}$. Every quantity that lies on the incident-light side of the pole has picked up a minus sign purely from the geometry. When these signed numbers are fed into the mirror equation, the computed image distance emerges with a sign that we simply interpret, rather than argue about after the fact. 🔉⇢

Finally, note the deep economy the convention delivers across the whole chapter. The very same sign rules that govern reflection by spherical mirrors also govern refraction at a spherical surface and refraction by thin lenses; only the physical formula changes, never the coordinate discipline. Because distances measured with the incident light are positive and those against it are negative, and because heights above the axis are positive, one consistent frame threads reflection and refraction together. Mastering this convention is therefore not a preliminary chore but the single most reusable skill in ray optics, since $u$, $v$, $f$, and $R$ never again need case-by-case sign guessing. 🔉⇢

In practice a reliable examination habit is to draw the principal axis, mark the pole as the origin, draw an arrow showing the incident light as the positive direction, and only then translate each physical distance into a signed number. If a computed value such as an image distance satisfies $v\lt 0$ for a mirror, you immediately know the image is real and on the object side; if $v\gt 0$, the image is virtual and behind the mirror. Trusting the signs, rather than overriding them with intuition, is what separates a clean solution from a sign-error-riddled one under time pressure. 🔉⇢

Derivation 🔉⇢

  1. Step 1 — Fix the origin and axis. Place the origin of coordinates at the pole $P$ of the mirror (or optical centre of the lens) and let the principal axis be the $x$-axis. Orient the axis so that the incident light travels in the positive $x$ direction. Every distance in any optical formula is now a coordinate of a point measured from $P$ along this axis, so the object, image, focus, and centre of curvature each acquire a definite signed value rather than a bare magnitude.
  2. Step 2 — Assign the axial sign rule. A point whose displacement from $P$ is along the incident light gets a positive coordinate; a point whose displacement is opposite to the incident light gets a negative coordinate. Hence a real object in front of a mirror gives $u\lt 0$, and a real image formed in front of the mirror gives $v\lt 0$, because both lie on the side from which light is incident, reached by travelling against the incident beam from the pole.
  3. Step 3 — Assign the transverse sign rule. Measure heights perpendicular to the principal axis: upward is positive, downward is negative. An erect object therefore has $h\gt 0$, while a real image that is inverted has $h'\lt 0$. This single rule lets the magnification $m=h'/h$ report orientation through its sign, with $m\lt 0$ meaning inverted and $m\gt 0$ meaning erect, without any auxiliary statement.
  4. Step 4 — Derive the focal and radius signs. Because the focus $F$ and centre of curvature $C$ of a concave mirror lie in front of it, against the incident light from the pole, both $f$ and $R$ are negative; for a convex mirror they lie behind it, so both are positive. The relation $f=R/2$ then holds with signs intact, so knowing $R$ fixes $f$ including its sign automatically.
  5. Step 5 — Verify universality. Substitute the signed $u$, $v$, $f$, and $R$ into the mirror equation or the lens relations. Because a single origin and a single positive direction were used throughout, the same equation reproduces real and virtual images, concave and convex elements, and even virtual objects. The convention thus proves that one formula per element covers every case, which is exactly the economy it was designed to guarantee, and the resulting sign of $v$ is read directly as the image nature.
⚠️ JEE trap: A very common trap is to treat the sign convention as flexible, flipping the positive direction to 'make the answer come out right' or dropping the minus signs on $u$, $f$, and $R$ and reinserting them by intuition afterwards. This defeats the whole purpose. The convention is a fixed frame anchored at the pole with the incident light defining $+x$; you substitute signed values once and trust the signed result. If you later force $v\gt 0$ because you 'expect a real image', you have overwritten the physics with a guess. 🔉⇢

Reflection by Spherical Mirrors & Focal Length 🔉⇢

🎯 Build the image of a concave mirror with the two standard rays: one parallel to the axis (reflects through F), one to the pole (reflects symmetrically). Where they cross is the image.
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concave mirror · f = R/2 — the parallel ray reflects through F, the pole ray reflects at an equal angle; the image is where they meet.
object — the focus → image is —
What you are looking at — image formation in a concave mirror, drawn the textbook way.
  • The curved black line is the mirror; F (brown) and C (blue) are its focus and centre of curvature, and F always sits at exactly R/2.
  • The red arrow is the object.
  • The two teal rays are the standard construction rays; the purple arrow where they meet is the image (dashed when the image is virtual).
What to do
  1. Start with the object beyond C, then slide it inward. Watch the real image grow, flip through infinity as the object crosses F...
  2. ...and become a virtual, erect, magnified image once the object is inside F — the shaving/make-up mirror regime.
Why it matters — every case (real vs virtual, magnified vs diminished) comes from the SAME two rays and the same f = R/2; nothing about the mirror changes, only where you put the object.
Definition: The laws of reflection applied to concave and convex mirrors, with the focal length shown to be half the radius of curvature (f = R/2) for paraxial rays. 🔉⇢

Reflection by spherical mirrors rests on the same two laws of reflection that govern any reflecting surface. The angle of reflection, measured between the reflected ray and the normal, equals the angle of incidence, measured between the incident ray and the same normal; and the incident ray, the reflected ray, and the normal at the point of incidence all lie in one plane. What makes the spherical case special is the identity of the normal. For a curved reflecting surface the normal is taken along the radius, that is, along the line joining the centre of curvature of the mirror to the point of incidence, because the normal must be perpendicular to the tangent to the surface at that point. 🔉⇢

A spherical mirror carries a small vocabulary that must be used precisely. The geometric centre of the mirror is called its pole, usually labelled $P$. The centre of the sphere of which the reflecting surface is a part is the centre of curvature $C$, and the distance $PC$ is the radius of curvature $R$. The straight line joining the pole and the centre of curvature is the principal axis, the axis of symmetry about which the whole optical behaviour is organised. Every distance we later measure, object distance, image distance, focal length, is measured from the pole along this principal axis, which is exactly what the sign convention demands. 🔉⇢

Now send a parallel beam of light, travelling close to and nearly along the principal axis, onto the mirror. For a concave mirror the reflected rays actually converge and cross at a single point $F$ on the principal axis. For a convex mirror the reflected rays diverge outward but, when produced backward, appear to come from a single point $F$ behind the mirror. In both cases this point is the principal focus of the mirror, and the distance from the pole $P$ to the focus $F$ is the focal length $f$. The concave mirror thus has a real focus in front of it, while the convex mirror has a virtual focus behind it. 🔉⇢

The crucial quantitative fact is that for such a beam the focal length is exactly half the radius of curvature, $f=R/2$. This is not an empirical accident; it follows from the laws of reflection applied to a ray parallel to the principal axis, together with the small-angle behaviour of the paraxial regime. Because a single number $R$ then fixes $f$, and because the sign convention makes both negative for a concave mirror and both positive for a convex mirror, the relation $f=R/2$ is one of the most heavily used shortcuts in the entire chapter and deserves to be understood by derivation rather than merely memorised. 🔉⇢

The word paraxial carries real weight and must not be skipped over. A paraxial ray is one incident at a point close to the pole and making a small angle with the principal axis. Only for such rays do the reflected rays from a parallel beam meet at a single sharp focus, because only then are the small-angle approximations, in which the tangent of an angle is replaced by the angle itself, accurate. The result $f=R/2$ is therefore a paraxial result. It is the clean idealisation on which the mirror equation is subsequently built, and its limitations are physical, not mathematical carelessness. 🔉⇢

The aperture of a mirror, its lateral size or width, is what decides whether the paraxial assumption is safe. If the aperture is small compared with the radius of curvature, every ray that strikes the mirror is nearly paraxial and the beam focuses tightly. As the aperture grows, rays striking the outer zones of the mirror are far from the pole and make appreciable angles with the axis, so the small-angle approximation breaks down for them. This is why textbook derivations quietly assume a small aperture: it is the condition under which a spherical mirror behaves like an ideal focusing element with a well-defined single focus. 🔉⇢

When the aperture is not small, the failure of the paraxial approximation shows up as spherical aberration. Rays reflected from the outer zones of a wide concave mirror cross the axis closer to the mirror than the paraxial rays do, so instead of a single sharp focus one obtains a blurred region and a bright envelope known as the caustic. Spherical aberration is a defect of the spherical shape itself, not of the material or polish, which is precisely why large telescope objectives favour a paraboloidal mirror: a parabola brings a genuinely parallel axial beam to one exact focus, with no aberration of this kind. 🔉⇢

The contrasting behaviour of concave and convex mirrors follows directly from where the focus sits. A concave mirror, curving inward toward the incoming light, is a converging mirror: parallel rays are brought together at a real focus, and depending on the object position it can form real or virtual images. A convex mirror, curving away from the incoming light, is a diverging mirror: parallel rays spread apart and only appear to come from a virtual focus behind the mirror. This is why a convex mirror always yields an erect, diminished, virtual image and is chosen for wide-view rear and side mirrors on vehicles. 🔉⇢

It is illuminating to note how many rays actually take part. An infinite number of rays leave every point of an object in all directions, and the laws of reflection hold at each and every point of the reflecting surface. A point is a genuine image point only if every ray leaving the corresponding object point passes through it after reflection. This is also why covering part of a mirror does not chop the image in half: the uncovered portion still receives rays from the whole object and forms a complete, though dimmer, image, since the intensity, not the extent, of the image depends on the reflecting area. 🔉⇢

Two convenient rays make ray tracing for spherical mirrors quick. A ray from an object point travelling parallel to the principal axis reflects so as to pass through the focus of a concave mirror, or to appear to come from the focus of a convex mirror. A ray directed through, or toward, the centre of curvature strikes the mirror along its own normal and simply retraces its path. The intersection of any two such reflected rays locates the image point. These constructions, combined with the laws of reflection and the relation $f=R/2$, are the geometric backbone from which the algebraic mirror equation is derived. 🔉⇢

Pulling the strands together, reflection by a spherical mirror is completely specified once you know the laws of reflection, the geometry of pole, centre of curvature, principal axis, and principal focus, and the paraxial relation $f=R/2$. The concave mirror converges light to a real focus while the convex mirror diverges it from a virtual focus, and the small-aperture paraxial condition is what keeps that focus sharp. Everything that follows, the mirror equation and the magnification, is a quantitative reformulation of this geometric picture under the Cartesian sign convention, so a firm grasp here pays off across every image-formation problem. 🔉⇢

Derivation 🔉⇢

  1. Step 1 — Set up the geometry. Let $C$ be the centre of curvature and $P$ the pole of a concave mirror, so that $PC=R$, the radius of curvature. Consider a single ray travelling parallel to the principal axis and striking the mirror at a point $M$. Because the normal to a spherical surface lies along the radius, the line $CM$ is normal to the mirror at $M$, and this normal makes an angle $\theta$ with the incident ray.
  2. Step 2 — Apply the law of reflection. The angle of incidence between the incident ray and the normal $CM$ equals $\theta$. Since the incident ray is parallel to the principal axis, alternate angles give $\angle MCP=\theta$. The reflected ray crosses the principal axis at the focus $F$, and because the angle of reflection also equals $\theta$, the exterior-angle relation for the triangle yields $\angle MFP=2\theta$. Thus the reflected ray turns through twice the angle that the radius makes with the axis.
  3. Step 3 — Drop a perpendicular. Let $MD$ be the perpendicular from $M$ onto the principal axis, meeting it at $D$. From the right triangles formed, $\tan\theta=\dfrac{MD}{CD}$ and $\tan 2\theta=\dfrac{MD}{FD}$. These two relations express the same transverse height $MD$ in terms of the axial distances $CD$ and $FD$ from the centre of curvature and the focus respectively.
  4. Step 4 — Impose the paraxial approximation. For paraxial rays $\theta$ is small, so $\tan\theta\approx\theta$ and $\tan 2\theta\approx 2\theta$. Dividing the two expressions gives $\dfrac{MD}{FD}=2\,\dfrac{MD}{CD}$, and cancelling the common height $MD$ leaves $\dfrac{1}{FD}=\dfrac{2}{CD}$, that is $FD=\dfrac{CD}{2}$. The small-angle condition is precisely what allows this clean cancellation.
  5. Step 5 — Take the paraxial limit and conclude. Because the ray strikes close to the pole, the foot $D$ is very near $P$, so $FD\approx FP=f$ and $CD\approx CP=R$. Substituting gives $f=\dfrac{R}{2}$. This paraxial result, that the focal length is half the radius of curvature, holds with signs intact under the Cartesian convention: $f$ and $R$ are both negative for a concave mirror and both positive for a convex mirror.
  6. Step 6 — Note the breakdown. If the aperture is large, outer rays violate the small-angle assumption of Step 4, so they cross the axis at points where $FD$ differs from the paraxial value. The single focus smears into a caustic, which is spherical aberration; the derivation of $f=R/2$ is therefore valid strictly in the paraxial, small-aperture regime.
⚠️ JEE trap: Students often assume $f=R/2$ holds for every ray hitting the mirror, so that any concave mirror focuses a wide parallel beam to one perfect point. It does not: the relation is a paraxial result, valid only for rays close to the pole making small angles with the principal axis. Wide-aperture rays reflect to cross the axis nearer the mirror, producing spherical aberration and a caustic rather than a sharp focus. Treating $f=R/2$ as exact for all apertures, and expecting a flawless image, is the trap. 🔉⇢

The Mirror Equation & Magnification 🔉⇢

🎯 Drag the object distance u and watch the image distance v obey 1/v + 1/u = 1/f on the graph, while the magnification m = −v/u ticks over. Note the blow-up as u approaches f.
🔉⇢
mirror equation: 1/v + 1/u = 1/f  ·  (1/—) + (1/—) = 1/—
magnification m = −v/u = h′/h = —
What you are looking at — the mirror equation as a graph, not a formula.
  • Left: the object (red) and the image (purple) sliding on the axis.
  • Right: the curve of image distance v against object distance u for the focal length you set. The red dot is where you are now.
  • The dashed vertical line at u = f is where v blows up to infinity.
What to do
  1. Move the object slowly toward F. Predict first: what happens to the image as u → f?
  2. Cross inside F and watch v flip sign — the image goes virtual (positive v) and m becomes positive (erect).
Why it matters — m = −v/u ties the two distances to the size and orientation of the image in one stroke. A negative m means inverted and real; a positive m means erect and virtual. The magnitude tells you enlarged or shrunk.
Definition: The relation 1/v + 1/u = 1/f between object distance, image distance and focal length, together with the linear magnification m = -v/u. 🔉⇢

The mirror equation is the algebraic statement of image formation by a spherical mirror. It relates three signed quantities, the object distance $u$, the image distance $v$, and the focal length $f$, all measured from the pole along the principal axis, through the compact relation $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$. What makes this single equation so valuable is that, once the Cartesian sign convention is respected, it applies to concave and convex mirrors alike and to real and virtual images alike. There is no separate formula to memorise for each configuration; the signs of the numbers carry all the case-by-case information, which is exactly the economy the convention was designed to deliver. 🔉⇢

The derivation begins from a ray diagram for a concave mirror forming a real, inverted image, and it rests entirely on similar triangles. When an object $AB$ stands on the principal axis and its image $A'B'$ is formed after reflection, two pairs of triangles turn out to be similar: one pair built from the ray through the focus and one pair built from the ray reflecting at the pole. Ratios of corresponding sides of these similar triangles give relations among the magnitudes of the distances, and the sign convention is then imposed at the very end to convert those magnitude relations into the signed mirror equation. 🔉⇢

Linear magnification measures how large the image is relative to the object, and it is defined as the ratio of the height of the image to the height of the object, $m=\frac{h'}{h}$. Using the triangle formed by the ray incident at the pole, where the incident and reflected rays make equal angles with the principal axis, one finds that this height ratio equals $-\frac{v}{u}$. Hence the working formula $m=-\frac{v}{u}=\frac{h'}{h}$. Because heights above the axis are positive and those below are negative under the convention, the single number $m$ reports both how much the image is enlarged or diminished and whether it is erect or inverted. 🔉⇢

The sign of $m$ is therefore loaded with physical meaning. A negative magnification means the image height $h'$ has the opposite sign to the object height $h$, so the image is inverted; for a mirror this always accompanies a real image, whose reflected rays actually converge in front of the mirror. A positive magnification means the image is erect, which for a mirror accompanies a virtual image formed behind the mirror by rays that only appear to diverge. Thus the two numbers you compute, the signed image distance $v$ and the signed magnification $m$, together announce the position, the nature, and the orientation of the image without any further ray drawing. 🔉⇢

The magnitude of $m$ separates enlargement from diminution. If the magnitude of $m$ exceeds one the image is magnified, if it is less than one the image is diminished, and if it equals one the image is the same size as the object. Combining magnitude and sign lets a single computed value summarise the whole outcome: for instance $m=-3$ describes an image three times as tall as the object and inverted, hence real, while $m=+\tfrac{1}{2}$ describes an upright image half the object's height, hence virtual. Learning to read $m$ fluently is one of the quickest routes to correct answers under time pressure. 🔉⇢

Consider how the image of a concave mirror evolves as the object moves inward from far away. With the object beyond the centre of curvature, the image is real, inverted, and diminished, forming between the focus and the centre of curvature. As the object reaches the centre of curvature, the image is real, inverted, and the same size, formed at the centre of curvature itself. Bringing the object between the centre of curvature and the focus produces a real, inverted, magnified image lying beyond the centre of curvature. In each stage the image distance $v$ computed from the mirror equation comes out negative, signalling a real image on the object side. 🔉⇢

The behaviour changes character once the object crosses the focus of the concave mirror. With the object placed between the focus and the pole, the reflected rays diverge and no longer meet in front of the mirror; they only appear to come from a point behind it. The mirror equation then returns a positive image distance $v$, marking a virtual image, and the magnification turns positive and greater than one, so the image is erect and magnified. This is the shaving-mirror or make-up-mirror regime, in which a concave mirror close to the face yields an enlarged upright view, a direct algebraic consequence of the object lying inside the focal length. 🔉⇢

A convex mirror behaves far more simply, and the mirror equation makes the simplicity transparent. Its focal length is positive, and for any real object with negative $u$ the equation $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ forces $v$ to be positive, so the image is always virtual and located behind the mirror. The magnification $m=-\frac{v}{u}$ then always comes out positive and less than one, so the image is invariably erect and diminished, regardless of where the object sits. This is precisely why convex mirrors are chosen as vehicle rear-view and side mirrors: they give an upright, reduced image over a wide field of view. 🔉⇢

It is worth appreciating that the very same equation, derived from a concave mirror forming a real image, silently handles all these outcomes because the algebra respects the signs. You never decide in advance whether an image will be real or virtual; you substitute the signed $u$ and $f$, solve for $v$, and then read the sign. A negative $v$ for a mirror is a real image on the incident-light side; a positive $v$ is a virtual image behind the mirror. The equation is, in this sense, a small case-analysis engine that you feed with signed inputs and query with signed outputs. 🔉⇢

There is also a subtle relationship between the mirror equation and the paraxial focal relation $f=R/2$. The mirror equation presumes a well-defined single focal length, which itself only exists in the paraxial, small-aperture regime where a parallel beam converges to one point. For wide apertures the focus smears into a caustic through spherical aberration, and the notion of a single $v$ satisfying the mirror equation becomes an approximation. In routine problem solving the paraxial assumption is taken for granted, but it is worth remembering that the clean predictive power of $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ is inherited from that same idealisation. 🔉⇢

To use these tools reliably, adopt a fixed routine. Write down $f$ with its correct sign, negative for a concave mirror and positive for a convex mirror; write $u$ as a negative number for a real object; solve $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ for $v$; then compute $m=-\frac{v}{u}$. Interpret the results by their signs: $v\lt 0$ means a real, inverted image and $v\gt 0$ means a virtual, erect one, while the magnitude of $m$ compared with one settles enlargement. Trusting this signed output, rather than overriding it with expectation, is what turns the mirror equation into a dependable instrument. 🔉⇢

Derivation 🔉⇢

  1. Step 1 — Identify similar triangles. In the ray diagram of a concave mirror forming a real, inverted image, take the ray from the object top that travels parallel to the axis and reflects through the focus $F$. This ray produces two right triangles sharing the focus: triangle $A'B'F$ built on the image and triangle $MPF$ built at the pole, where $MP$ equals the object height $AB$ for paraxial rays. Their similarity gives $\dfrac{B'A'}{BA}=\dfrac{B'F}{FP}$, relating image and object heights to focal distances.
  2. Step 2 — Use the pole ray. The ray from the object top incident at the pole $P$ reflects with the angle of reflection equal to the angle of incidence, so $\angle APB=\angle A'PB'$. Hence the right triangles $A'B'P$ and $ABP$ are similar, giving $\dfrac{B'A'}{BA}=\dfrac{B'P}{BP}$. This provides a second, independent expression for the same height ratio, now in terms of the image and object distances from the pole.
  3. Step 3 — Equate the two ratios. Since both equal $\dfrac{B'A'}{BA}$, we set $\dfrac{B'F}{FP}=\dfrac{B'P}{BP}$. Writing $B'F=B'P-FP$ turns this into $\dfrac{B'P-FP}{FP}=\dfrac{B'P}{BP}$, a relation purely among the magnitudes of the distances from the pole to the image, the focus, and the object, ready for the sign convention to be applied.
  4. Step 4 — Apply the Cartesian sign convention. Light travels from object to mirror, taken as positive; the object, image, and focus are all reached from the pole by moving against the incident light, so $B'P=-v$, $FP=-f$, and $BP=-u$. Substituting, $\dfrac{-v+f}{-f}=\dfrac{-v}{-u}$, which simplifies to $\dfrac{v-f}{f}=\dfrac{v}{u}$. Rearranging gives $\dfrac{v}{f}=1+\dfrac{v}{u}$.
  5. Step 5 — Obtain the mirror equation. Dividing both sides of $\dfrac{v}{f}=1+\dfrac{v}{u}$ by $v$ yields $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$, that is $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$. Although derived for a concave mirror forming a real image, the use of signed distances makes it valid for convex mirrors and for virtual images as well.
  6. Step 6 — Derive the magnification. From the similar triangles $A'B'P$ and $ABP$, the height ratio equals $\dfrac{B'P}{BP}$. Applying the convention with image height $h'$ (downward, negative for a real image) and object height $h$ (upward, positive), $\dfrac{-h'}{h}=\dfrac{-v}{-u}$, giving $m=\dfrac{h'}{h}=-\dfrac{v}{u}$. The sign of $m$ then encodes orientation: $m\lt 0$ inverted and real, $m\gt 0$ erect and virtual.
⚠️ JEE trap: A recurring error is to insert $u$, $v$, or $f$ as bare positive magnitudes and then attach signs to the answer by intuition, or to memorise separate formulas for concave and convex mirrors. The single relation $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ works only if signed values are substituted once and the resulting sign of $v$ is trusted. Another trap is reading $m\gt 0$ as 'magnified'; the sign of $m$ gives orientation, while enlargement is judged by whether the magnitude of $m$ is greater or less than one. 🔉⇢

Refraction & Snell's Law 🔉⇢

Definition: The bending of light at an interface between two media, governed by Snell's law sin i / sin r = n21, with the refracted ray bending toward or away from the normal depending on relative optical density. 🔉⇢

Refraction is the phenomenon that occurs when a beam of light encounters another transparent medium at an interface. At the boundary a part of the light gets reflected back into the first medium while the rest enters the other. A ray of light represents such a beam, and the direction of propagation of an obliquely incident ray, with an angle of incidence between $0^\circ$ and $90^\circ$, changes as it crosses the interface between the two media. This bending of the ray at the surface separating two transparent media is what we call refraction of light. It arises because light travels with different speeds in different media, and it is governed by two simple, experimentally established laws first quantified by Snell. Note that if the ray strikes the interface normally, along the normal itself, it passes straight through without any bending; only an obliquely incident ray is deviated, and it is this oblique refraction that produces the familiar sight of a straight object appearing broken at a water surface. Throughout this discussion we treat light in the ray picture, tracing the incident ray and the refracted ray on either side of the point of incidence. 🔉⇢

Full derivation, worked example and interactive 3D on the Refraction & Snell's Law tab →

Total Internal Reflection & Optical Fibres 🔉⇢

Definition: When light travelling in a denser medium meets a rarer medium beyond the critical angle, it is reflected entirely back — the principle behind optical fibres and totally reflecting prisms. 🔉⇢

When a ray of light travels from an optically denser medium to a rarer medium, at the interface it is partly reflected back into the same denser medium and partly refracted into the rarer medium. This partial reflection into the originating medium is called internal reflection. Because the second medium is rarer, the refracted ray bends away from the normal, so the angle of refraction $r$ is larger than the angle of incidence $i$. As long as $i$ is modest, both a reflected ray and a refracted ray coexist, and the refracted ray carries away most of the light energy while the internally reflected ray remains comparatively feeble. This everyday situation, light escaping upward from water into air, is the starting point for understanding total internal reflection. 🔉⇢

Full derivation, worked example and interactive 3D on the Total Internal Reflection & Optical Fibres tab →

Refraction at a Spherical Surface 🔉⇢

🎯 A single curved boundary between two media already forms an image. Slide the object or raise the second medium's index and watch the refracted rays converge to a new point.
🔉⇢
n₂/v − n₁/u = (n₂ − n₁)/R  (with n₁ = 1, air on the left)
solving: v = — — a — image in the denser medium
What you are looking at — refraction at ONE spherical surface, the building block a lens is made from.
  • The pale left region is air (n₁ = 1); the tinted right region is a denser medium whose index n₂ you control.
  • The curved line is the single refracting surface; C is its centre of curvature.
  • The teal rays refract at the surface and cross at the purple image point.
What to do
  1. Leave the defaults (u = −100 cm, R = +20 cm, n₂ = 1.5): the image lands at v = +100 cm, the exact NCERT worked example.
  2. Raise n₂ and the surface bends light harder, pulling the image closer.
Why it matters — n₂/v − n₁/u = (n₂ − n₁)/R is the seed of the whole lens story: apply it at the two surfaces of a piece of glass in turn and out falls the lens maker's formula.
Definition: The relation n2/v - n1/u = (n2 - n1)/R for image formation by a single spherical refracting surface, the building block of the lens maker's formula. 🔉⇢

So far we have treated refraction only at a plane interface, where the two transparent media meet along a flat boundary. We now consider refraction at a single spherical surface separating two media of refractive index $n_1$ and $n_2$. An infinitesimal patch of a spherical surface can be regarded as planar, so the same laws of refraction apply at every point on the surface. Just as for a spherical mirror, the normal at the point of incidence is perpendicular to the tangent plane at that point and therefore passes through the centre of curvature $C$. This single refracting surface is the fundamental building block from which the lens maker's formula is later assembled. 🔉⇢

Consider a point object $O$ on the principal axis, lying in the medium of refractive index $n_1$. Light from $O$ strikes a spherical surface of radius of curvature $R$ whose centre of curvature is $C$, and refracts into the second medium of refractive index $n_2$, forming the image $I$ on the axis. A ray from $O$ meets the surface at a point $N$ close to the pole $M$, bends according to Snell's law, and proceeds toward $I$. We take the aperture, or the lateral size of the surface, to be small compared with the object and image distances, so that every ray remains paraxial and makes only a small angle with the principal axis. 🔉⇢

The paraxial approximation is the engine of the whole derivation. Because each ray stays near the axis, the point $N$ lies close to the pole $M$, and the perpendicular distance $MN$ is nearly equal to the arc height. For any small angle $\theta$ we may replace $\tan\theta$ and $\sin\theta$ by $\theta$ itself, since the correction terms are of higher order and negligible. This linearisation converts the awkward trigonometry of a curved refracting surface into simple ratios of lengths, exactly as it did for the focal length of a spherical mirror. Without this restriction the image would suffer aberration and no single sharp image point would exist. 🔉⇢

We now write the three relevant small angles as ratios. For the ray from the object, $\tan(\angle NOM)=\dfrac{MN}{OM}$; for the line to the centre of curvature, $\tan(\angle NCM)=\dfrac{MN}{MC}$; and for the ray reaching the image, $\tan(\angle NIM)=\dfrac{MN}{MI}$. The key geometric insight is the exterior-angle theorem: in triangle $NOC$ the angle of incidence $i$ is the exterior angle at $N$, so it equals the sum of the two remote interior angles, $i=\angle NOM+\angle NCM$. This clean relationship is what makes the spherical-surface derivation so much simpler than a brute-force ray trace. 🔉⇢

Applying the exterior-angle result and the small-angle ratios, the angle of incidence becomes $i=\dfrac{MN}{OM}+\dfrac{MN}{MC}$. For the refracted ray a similar triangle gives the angle of refraction as a difference of angles, $r=\angle NCM-\angle NIM=\dfrac{MN}{MC}-\dfrac{MN}{MI}$. Both $i$ and $r$ are now expressed purely through the perpendicular height $MN$ and the three axial distances $OM$, $MC$ and $MI$. Notice that the common factor $MN$ will cancel presently, which is why the final relation between object distance and image distance is independent of exactly where on the small aperture the ray happened to strike. 🔉⇢

The two media are linked by Snell's law, $n_1\sin i=n_2\sin r$. In the paraxial regime the sines reduce to the angles themselves, so this becomes simply $n_1 i=n_2 r$. Substituting the expressions for $i$ and $r$ and cancelling the common height $MN$ yields a relation among the magnitudes $OM$, $MI$ and $MC$: $\dfrac{n_1}{OM}+\dfrac{n_2}{MI}=\dfrac{n_2-n_1}{MC}$. At this stage every length is still a positive magnitude, because we have not yet decided on directions. The physics of refraction is complete; what remains is only the bookkeeping of signs so that one formula can serve every geometric case. 🔉⇢

We now impose the Cartesian sign convention consistently. All distances are measured from the pole of the surface. Distances measured in the same direction as the incident light are taken as positive, and those measured against it are negative. Here light travels from the object toward the surface, so to reach the object we travel opposite to the incident light, giving $OM=-u$. The image lies in the direction of the incident light, so $MI=+v$, and for the convex-toward-the-object surface shown the centre of curvature also lies that way, so $MC=+R$. Heights above the principal axis are positive and those below are negative, matching the mirror convention. 🔉⇢

Substituting $OM=-u$, $MI=+v$ and $MC=+R$ into the magnitude relation gives the compact and famous result $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$. This equation relates the object distance and the image distance to the refractive index of each medium and to the radius of curvature of the single spherical surface. It is strikingly general: nowhere did we assume the surface bulged one particular way. The same formula therefore governs a surface that is convex toward the incoming light and one that is concave toward it; only the algebraic sign of $R$ changes, and the mathematics automatically produces the correct real or virtual image. 🔉⇢

The sign of $R$ is fixed by the location of the centre of curvature relative to the pole. If $C$ lies on the outgoing side, in the direction of the incident light, then $R\gt 0$; if $C$ lies on the incoming side, then $R\lt 0$. Likewise a positive $v$ denotes a real image formed on the far side, while a negative $v$ denotes a virtual image on the same side as the object. Because the derivation nowhere assumed a real image, you may verify that the identical relation holds when the refracted rays only appear to diverge, provided the sign convention is applied faithfully throughout. 🔉⇢

This one relation for a single spherical surface is the seed of nearly all lens theory. A thin lens is a transparent medium bounded by two surfaces, at least one of which is spherical. By applying $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ at the first surface, treating its image as the object for the second, and then adding the two results in the thin-lens limit, one obtains the lens maker's formula and thereafter the thin lens formula. Mastering the paraxial approximation, the exterior-angle geometry and the Cartesian sign convention here therefore pays dividends across every optical instrument built from lenses. 🔉⇢

A useful sanity check is the limiting behaviour of the formula. If the two media have equal refractive index, $n_2=n_1$, the right-hand side vanishes and the surface bends nothing; the object and image distances satisfy $n/v=n/u$, i.e. no net refraction, exactly as a submerged lens of matching index would predict. Similarly, letting $R\to\infty$ turns the spherical surface into a plane interface, recovering the apparent-depth result for near-normal viewing. These consistency checks, together with the units, are quick tools in an examination to confirm that signs and reciprocals of the refractive index have been entered correctly before any arithmetic is attempted. 🔉⇢

Derivation 🔉⇢

  1. Step 1 (Geometry and small angles): Place the point object $O$ on the axis in medium $n_1$; a paraxial ray meets the spherical surface at $N$ near the pole $M$ and refracts into medium $n_2$ toward the image $I$, with $C$ the centre of curvature and radius $R$. Using the perpendicular height $MN$, write $\tan(\angle NOM)=\dfrac{MN}{OM}$, $\tan(\angle NCM)=\dfrac{MN}{MC}$ and $\tan(\angle NIM)=\dfrac{MN}{MI}$. Because every ray is paraxial, each angle is small and $\tan\theta\approx\theta$, so the three tangents may be replaced by the angles themselves in all that follows.
  2. Step 2 (Exterior-angle relations): In triangle $NOC$, the angle of incidence $i$ is the exterior angle at $N$, so $i=\angle NOM+\angle NCM=\dfrac{MN}{OM}+\dfrac{MN}{MC}$. For the refracted ray, treating triangle $NCI$, the angle of refraction is $r=\angle NCM-\angle NIM=\dfrac{MN}{MC}-\dfrac{MN}{MI}$. Both the incidence and refraction angles are now expressed entirely through the common perpendicular height $MN$ and the three axial distances, which is precisely what allows the height to cancel in the next step.
  3. Step 3 (Apply Snell's law): The two media obey $n_1\sin i=n_2\sin r$. In the paraxial limit the sines collapse to the angles, giving $n_1 i=n_2 r$. Substitute the expressions from Step 2: $n_1\left(\dfrac{MN}{OM}+\dfrac{MN}{MC}\right)=n_2\left(\dfrac{MN}{MC}-\dfrac{MN}{MI}\right)$. Every term now carries the same factor $MN$, confirming that the result will not depend on where on the small aperture the ray struck.
  4. Step 4 (Cancel and regroup): Divide throughout by $MN$ and collect the reciprocal distances. This gives $\dfrac{n_1}{OM}+\dfrac{n_2}{MI}=\dfrac{n_2}{MC}-\dfrac{n_1}{MC}=\dfrac{n_2-n_1}{MC}$. This is a clean relation among magnitudes only; no sign convention has yet been applied, so $OM$, $MI$ and $MC$ here are all positive lengths measured from the pole.
  5. Step 5 (Impose Cartesian signs): Light travels from object to surface, so measuring to the object runs opposite to the incident light: $OM=-u$. The image lies along the incident-light direction, $MI=+v$, and for the surface shown the centre of curvature lies that way too, $MC=+R$. Substituting, $\dfrac{n_1}{-u}+\dfrac{n_2}{v}=\dfrac{n_2-n_1}{R}$.
  6. Step 6 (Final relation and generality): Rearranging yields $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$. It holds for any curved spherical surface: a centre of curvature on the outgoing side gives $R\gt 0$, one on the incoming side gives $R\lt 0$, and a virtual image simply returns $v\lt 0$. Applied twice, once at each face of a thin lens, this equation directly generates the lens maker's formula.
⚠️ JEE trap: Students often force the mirror-style $\frac{1}{v}+\frac{1}{u}$ form onto a refracting surface, or use $R/2$ as a focal length. A single spherical surface has no simple $R/2$ focus; it obeys $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$, and the refractive index of each medium must sit on the matching term. Equally common is dropping the sign of $R$: whether $R\gt 0$ or $R\lt 0$ is decided only by where the centre of curvature lies relative to the pole, never by the shape alone. 🔉⇢

The Lens Maker's Formula 🔉⇢

🎯 A lens's focal length is designed, not given. Change the glass index n or the two surface radii and watch the focus F slide in and out as 1/f = (n−1)(1/R₁ − 1/R₂).
🔉⇢
lens maker: 1/f = (n − 1)(1/R₁ − 1/R₂)  (thin lens, in air)
= (— − 1)(1/— − 1/—) → f = —
What you are looking at — where a lens's focal length actually comes from.
  • The teal lens body redraws its curvature live as you change the two radii — tighter curves make a fatter, stronger lens.
  • The three parallel rays are a distant on-axis beam; they cross at the red focus F.
What to do
  1. Keep the defaults (n = 1.5, R₁ = +10 cm, R₂ = −15 cm): f comes out to +12 cm, the NCERT worked value.
  2. Raise the index n and F pulls in; flatten the surfaces (larger |R|) and F pushes out.
Why it matters — 1/f = (n − 1)(1/R₁ − 1/R₂) means a lens maker sets the focal length purely by choosing the glass and grinding the curvatures. The same formula, with (n − 1) → 0, shows why a lens submerged in a liquid of its own index simply disappears.
Definition: The formula 1/f = (n21 - 1)(1/R1 - 1/R2) that gives the focal length of a thin lens from the refractive index and the two radii of curvature. 🔉⇢

A thin lens is a transparent optical medium bounded by two surfaces, at least one of which is spherical. Having derived the relation for refraction at a single spherical surface, we can now build the lens maker's formula simply by applying that relation twice, once at each face, and adding the results in the thin-lens limit. The formula we shall obtain, $\dfrac{1}{f}=(n_{21}-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, connects the focal length of a lens to the two radii of curvature and to the refractive index of the lens material relative to its surroundings. It is the practical recipe an optician or manufacturer uses to grind lenses of any desired focal length. 🔉⇢

Consider a double convex lens of material refractive index $n_2$ immersed in a medium of refractive index $n_1$. Light from an object $O$ first refracts at the front surface of radius $R_1$, which would by itself form an intermediate image $I_1$. This image then serves as the object for the second surface of radius $R_2$, which refracts the light once more to produce the final image $I$. The whole action of the lens is thus the composition of two single-surface refractions. Because the lens is thin, the two surfaces are treated as essentially coincident at the optical centre, and the small thickness between them is neglected throughout. 🔉⇢

Apply the single-surface relation to the first interface, where light passes from medium $n_1$ into the glass $n_2$. Writing the object distance as $OB$ and the intermediate image distance as $BI_1$, we obtain $\dfrac{n_2}{BI_1}-\dfrac{n_1}{OB}=\dfrac{n_2-n_1}{R_1}$. This is nothing more than the spherical-surface formula with the appropriate media on each side. The intermediate image $I_1$ may be real or virtual, but its position is dictated entirely by this first refraction; it is only a stepping stone to the final image and never needs to be physically formed on a screen. 🔉⇢

Now treat the second interface, where light emerges from the glass $n_2$ back into the surrounding medium $n_1$. Here the roles of the two refractive index values are exchanged, and the intermediate image $I_1$ acts as the object. Applying the single-surface relation again gives $\dfrac{n_1}{DI}-\dfrac{n_2}{DI_1}=\dfrac{n_1-n_2}{R_2}$. In the thin-lens approximation the points $B$ and $D$ both collapse onto the optical centre, so the intermediate image distances satisfy $BI_1=DI_1$. This coincidence is exactly what allows the two equations to be combined cleanly, with the intermediate term cancelling in the sum. 🔉⇢

Adding the two surface equations, the intermediate image distance drops out entirely, leaving $\dfrac{n_1}{DI}-\dfrac{n_1}{OB}=(n_2-n_1)\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)$ once the algebra of signs on $R_2$ is respected. Dividing through by $n_1$ and using the relative refractive index $n_{21}=n_2/n_1$ tidies the expression. The remarkable feature is that the intermediate image $I_1$, which was central to the physical picture, has vanished from the final result; only the object distance, the final image distance, the two radii of curvature and the relative index survive. 🔉⇢

To extract the focal length we send the object to infinity, so that $OB\to\infty$ and the final image forms at the focus, $DI=f$. The reciprocal object term then vanishes, and we are left with $\dfrac{1}{f}=(n_{21}-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ after the sign convention converts the surface radii into their signed forms $R_1$ and $R_2$. This is the lens maker's formula. Combined with the accompanying result $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$, it lets us predict image formation for any thin lens once its geometry and material are known. 🔉⇢

The sign convention on the two radii is what distinguishes a converging lens from a diverging one. For the double convex lens the centre of curvature of the first surface lies on the outgoing side, giving $R_1\gt 0$, while the centre of curvature of the second surface lies on the incoming side, giving $R_2\lt 0$. With $n_{21}\gt 1$ the bracket $\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ is positive, so $f\gt 0$ and the lens converges light. A double concave lens reverses both radii, making $f\lt 0$, so it diverges the beam. The single formula thus handles convex and concave lenses alike. 🔉⇢

It is instructive to see how the shape maps onto the signs. A convex surface met by the incident light contributes a positive radius, a concave surface a negative one; the second face contributes with the opposite geometric sense because its centre of curvature sits on the other side. Feeding these signed radii into the bracket automatically yields a positive focal length for any net-converging shape and a negative focal length for any net-diverging shape. There is no need to memorise separate formulae for biconvex, plano-convex, or meniscus lenses; the Cartesian convention encodes every case in one compact expression. 🔉⇢

The lens maker's formula also reveals a strong dependence on the surrounding medium, through the factor $n_{21}-1$, where $n_{21}$ is the refractive index of the lens material relative to its surroundings. A glass lens that converges strongly in air converges far more weakly in water, because $n_{21}$ falls closer to unity and the bracket is multiplied by a smaller number, lengthening the focal length. In the striking limiting case where the lens material and the surrounding liquid share the same refractive index, $n_{21}=1$, the factor vanishes, $1/f=0$, and the lens becomes optically invisible, behaving like a flat sheet with no converging or diverging power at all. 🔉⇢

This medium dependence is not a mere curiosity; it underlies the classic demonstration of a glass lens vanishing when lowered into a liquid of matching index. It also explains why the focal length quoted for a lens is always tied to a stated surrounding medium, usually air. When solving problems you must recompute $n_{21}$ for the actual surroundings before using the formula: a lens of focal length $20\text{ cm}$ in air can acquire a focal length of nearly $78\text{ cm}$ in water. Overlooking this shift in focal length is one of the most common sources of error in lens combination and optical instrument problems, and examiners deliberately test it by immersing an otherwise familiar glass lens in a liquid of stated refractive index. 🔉⇢

Finally, note the assumptions baked into the derivation. The lens is thin, so its thickness is negligible and the two surfaces are coincident; the rays are paraxial, so aberrations are ignored; and the same relative refractive index describes both refractions. Real thick lenses give slightly coloured images through dispersion, since $n_{21}$ varies with wavelength, and multi-component designs are used in cameras, microscopes and telescopes to correct such defects. The thin lens maker's formula is nonetheless the indispensable first approximation, and every more sophisticated lens-design calculation is built upon this same spherical-surface foundation. 🔉⇢

Derivation 🔉⇢

  1. Step 1 (Refraction at the first surface): A thin lens of material index $n_2$ sits in a medium of index $n_1$. Light from object $O$ meets the first spherical surface of radius $R_1$ and would form an intermediate image $I_1$. The single-surface relation gives $\dfrac{n_2}{BI_1}-\dfrac{n_1}{OB}=\dfrac{n_2-n_1}{R_1}$, where $B$ is the pole of the first surface. This is the exact spherical-surface formula, with medium $n_1$ on the incidence side and glass $n_2$ on the transmission side.
  2. Step 2 (Refraction at the second surface): Light now leaves the glass $n_2$ into the surrounding medium $n_1$, and the intermediate image $I_1$ acts as the object for the second surface of radius $R_2$. Applying the single-surface relation with the media reversed gives $\dfrac{n_1}{DI}-\dfrac{n_2}{DI_1}=\dfrac{n_1-n_2}{R_2}$, where $D$ is the pole of the second surface and $DI$ locates the final image.
  3. Step 3 (Thin-lens limit and addition): Because the lens is thin, $B$ and $D$ coincide at the optical centre, so $BI_1=DI_1$. Adding the two equations makes the intermediate term cancel: $\dfrac{n_1}{DI}-\dfrac{n_1}{OB}=(n_2-n_1)\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)$. Dividing by $n_1$ and writing $n_{21}=n_2/n_1$ gives $\dfrac{1}{DI}-\dfrac{1}{OB}=(n_{21}-1)\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)$ in magnitude form.
  4. Step 4 (Object at infinity defines the focus): Set $OB\to\infty$ so the reciprocal object term vanishes and the image forms at the focus, $DI=f$. Then $\dfrac{1}{f}=(n_{21}-1)\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)$ in magnitudes. The point where an object at infinity images is, by definition, the focus, so $f$ here is precisely the focal length of the lens.
  5. Step 5 (Apply Cartesian signs to the radii): For the biconvex lens shown, the centre of curvature of the first surface lies on the outgoing side ($R_1\gt 0$) and that of the second on the incoming side ($R_2\lt 0$). Substituting the signed radii turns the sum into a difference, yielding the lens maker's formula $\dfrac{1}{f}=(n_{21}-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$.
  6. Step 6 (Converging, diverging, and medium dependence): With $n_{21}\gt 1$ a biconvex shape gives $f\gt 0$ (converging); reversing both radii gives $f\lt 0$ (diverging). The prefactor $n_{21}-1$ shows the power depends on the surrounding medium: as $n_{21}\to 1$ the focal length diverges and $1/f\to 0$, so a lens immersed in a liquid of equal index has no optical power at all.
⚠️ JEE trap: A frequent trap is treating $\frac{1}{R_1}-\frac{1}{R_2}$ as if the radii were always both positive, or plugging in an air-based focal length when the lens is actually in water. The radii carry Cartesian signs, and for a biconvex lens $R_1\gt 0$ while $R_2\lt 0$. Just as damaging is forgetting that $n_{21}$ is the lens index relative to its surroundings: in water the factor $n_{21}-1$ shrinks, so the same lens has a much longer focal length, not the air value. 🔉⇢

The Thin Lens Formula & Magnification 🔉⇢

Definition: The relation 1/v - 1/u = 1/f for a thin lens, with magnification m = v/u, valid for convex and concave lenses and for real and virtual images. 🔉⇢

A thin lens is a transparent optical medium bounded by two refracting surfaces, at least one of which is spherical, whose thickness is negligible compared with the object distance, the image distance and the radii of curvature involved. Because the thickness can be ignored, both refracting surfaces are treated as passing through a single point on the principal axis called the optical centre. The line through the optical centre and the two centres of curvature is the principal axis. When paraxial rays (rays close to the axis making small angles with it) from a point object are refracted by such a lens, they reconverge to, or appear to diverge from, a single image point. The single relation that ties the object distance $u$, the image distance $v$ and the focal length $f$ together is the thin lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, and it holds for every situation once the sign convention is respected. 🔉⇢

Full derivation, worked example and interactive 3D on the The Thin Lens Formula & Magnification tab →

Power of a Lens & Combination of Lenses 🔉⇢

🎯 Powers simply add. Set two thin lenses in contact, choose each one's power in dioptres, and watch the combined focus move as P = P₁ + P₂ — a plus and a minus lens can cancel to nothing.
🔉⇢
lenses in contact: P = P₁ + P₂ = — + — = —
and f = 1/P → f = —
What you are looking at — two thin lenses pushed into contact and treated as one.
  • The teal and purple glyphs are the two lenses; each fattens (convex) or pinches (concave) with the power you dial in.
  • The parallel beam converges to the red combined focus F.
What to do
  1. Add two converging lenses: the powers add and F pulls in (a stronger lens).
  2. Make one power negative and equal to the other: P = 0, the beam passes straight through — the pair behaves like a flat window.
Why it matters — power P = 1/f (in dioptres, f in metres) is defined precisely so that lenses in contact just ADD: P = P₁ + P₂ + … This is how opticians build a prescription to any strength from a small stock of standard lenses.
Definition: The power P = 1/f (in dioptres) measures how strongly a lens converges or diverges light; for thin lenses in contact the powers add, P = P1 + P2 + ... 🔉⇢

The power of a lens is a measure of the convergence or divergence that the lens introduces into a beam of light falling on it. A lens of shorter focal length bends the incident rays more sharply, converging them in the case of a convex lens and diverging them in the case of a concave lens. Quantitatively the power $P$ is defined as the tangent of the angle by which the lens deflects a ray that arrives parallel to the principal axis at unit distance from the optical centre. For small deflections this reduces to the beautifully simple relation $P=\dfrac{1}{f}$, tying the power directly to the reciprocal of the focal length. 🔉⇢

The SI unit of power is the dioptre, written $\text{D}$, defined by $1\,\text{D}=1\,\text{m}^{-1}$. A lens whose focal length is one metre therefore has a power of one dioptre. Because power is the reciprocal of focal length, a short-focal-length lens is a high-power lens. The sign of the power follows the sign of the focal length under the Cartesian convention: a converging convex lens has $f\gt 0$ and hence $P\gt 0$, while a diverging concave lens has $f\lt 0$ and hence $P\lt 0$. This signed quantity is exactly what an optician manipulates when prescribing corrective lenses. 🔉⇢

A concrete reading of the dioptre makes the idea tangible. When an optician prescribes a corrective lens of power $+2.5\,\text{D}$, the required lens is a convex lens of focal length $f=\dfrac{1}{2.5}\,\text{m}=+40\,\text{cm}$. Conversely a prescription of $-4.0\,\text{D}$ denotes a concave lens of focal length $-25\,\text{cm}$. The larger the magnitude of the power, the more strongly the lens bends light and the shorter its focal length. Expressing lens strength in dioptres is convenient precisely because, as we shall see, the powers of lenses placed in contact simply add as signed numbers. 🔉⇢

Consider now two thin lenses $A$ and $B$ of focal lengths $f_1$ and $f_2$ placed in contact with each other, their optical centres taken to be coincident at a common point $P$ since the lenses are thin. Let an object be placed beyond the focus of the first lens. Lens $A$ alone would form an image $I_1$, which then acts as the object for lens $B$, producing the final image $I$. The intermediate image $I_1$ need not be physically realised; it is only a device for locating the final image. This two-step picture is the same compositional idea used to derive the lens maker's formula. 🔉⇢

For the first lens the thin lens formula gives $\dfrac{1}{v_1}-\dfrac{1}{u}=\dfrac{1}{f_1}$, and for the second lens, taking $I_1$ as its object, $\dfrac{1}{v}-\dfrac{1}{v_1}=\dfrac{1}{f_2}$. Adding these two equations, the intermediate image distance $v_1$ cancels neatly, leaving $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$. If the combination is regarded as a single equivalent lens of focal length $f$, then $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$, and comparing the two expressions yields the central result $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$ for two thin lenses in contact. 🔉⇢

The derivation extends immediately to any number of thin lenses in contact. For lenses of focal length $f_1,f_2,f_3,\dots$ stacked together, the effective focal length of the combination is given by $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}+\dfrac{1}{f_3}+\cdots$. Because power is the reciprocal of focal length, this reciprocal sum becomes an ordinary sum of powers: $P=P_1+P_2+P_3+\cdots$. The powers add as a simple algebraic sum, in which some terms may be positive for convex lenses and others negative for concave lenses. This additivity is the single most useful property of the dioptre in practical optics. 🔉⇢

The additive rule makes lens design intuitive. Placing a convex lens of power $+5\,\text{D}$ in contact with a concave lens of power $-3\,\text{D}$ gives a net power of $+2\,\text{D}$, a weaker converging combination, without any lengthy focal-length arithmetic. Because the sum is algebraic, a strong convex lens can be trimmed by a weaker concave lens, or a diverging system built from an excess of concave power. Designers exploit this to reach a target power or focal length that no single available lens provides, simply by combining stock lenses of known dioptric strength in contact. 🔉⇢

Magnification behaves multiplicatively across a combination, in contrast to the additive powers. Since the image formed by the first lens becomes the object for the second, the linear magnification of the whole system is the product of the individual magnifications, $m=m_1\,m_2\,m_3\cdots$. Each factor may be positive for an erect image or negative for an inverted one, so the sign of the net magnification records whether the final image is erect or inverted, while its magnitude records the overall size ratio. This product rule, together with the additive power rule, completely characterises a stack of thin lenses in contact. 🔉⇢

The importance of lens combinations for optical instruments can hardly be overstated. A compound microscope uses an objective and an eyepiece in series, and its overall magnification is the product of the two individual magnifications, exactly as the product rule prescribes. Telescopes similarly combine an objective of long focal length with an eyepiece of short focal length. In every such instrument the freedom to add powers and multiply magnifications lets the designer reach magnifications far beyond what any single lens of realistic focal length could deliver, which is precisely why multi-lens systems dominate cameras, microscopes and telescopes. 🔉⇢

Beyond raw magnification, lens combinations are the primary tool for correcting aberrations. A single thick lens produces coloured images because its refractive index, and hence its power, varies with wavelength, a defect known as chromatic aberration. By cementing together a convex crown-glass lens and a concave flint-glass lens whose dispersions partly cancel, designers build an achromatic combination whose net power is correct while the colour spread is largely removed. Modern microscope objectives and eyepieces are multi-component assemblies for exactly this reason: combining lenses lets one keep the desired total power while minimising the various optical aberrations that degrade image quality. 🔉⇢

In problem solving the two rules should be applied with care over signs. When adding powers, convex contributions enter as positive dioptres and concave contributions as negative, so the net power may come out either sign, determining whether the combination converges or diverges. When forming the net magnification, multiply the signed magnifications rather than their magnitudes, so that the erect-or-inverted character of the final image emerges automatically. Finally, remember that the additive power rule assumes the lenses are thin and in contact; when lenses are separated by a finite distance the effective focal length picks up an extra term and the simple sum no longer holds exactly. 🔉⇢

Derivation 🔉⇢

  1. Step 1 (Set up the two-lens system): Place thin lenses $A$ and $B$, of focal lengths $f_1$ and $f_2$, in contact so their optical centres coincide at $P$. An object at distance $u$ from $P$ is imaged by lens $A$ alone at distance $v_1$; this image $I_1$ then acts as the object for lens $B$, which forms the final image at distance $v$. Because the lenses are thin and in contact, the same optical centre serves both, so all distances are measured from $P$.
  2. Step 2 (Thin lens formula for each lens): For lens $A$, $\dfrac{1}{v_1}-\dfrac{1}{u}=\dfrac{1}{f_1}$. For lens $B$, the intermediate image is the object, so $\dfrac{1}{v}-\dfrac{1}{v_1}=\dfrac{1}{f_2}$. Each equation is just the standard thin lens relation applied to one refraction stage, with the sign convention already built in through the signed distances.
  3. Step 3 (Add to eliminate the intermediate image): Adding the two equations, the intermediate distance cancels: $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$. Modelling the pair as one equivalent lens of focal length $f$ requires $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$, so comparison gives $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$.
  4. Step 4 (Generalise to many lenses): The same cancellation repeats for a third, fourth, and further lens in contact, giving $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}+\dfrac{1}{f_3}+\cdots$. This holds for any number of thin lenses in contact, each contributing its own reciprocal focal length to the sum.
  5. Step 5 (Recast in terms of power): Since $P=\dfrac{1}{f}$ with $f$ in metres and $P$ in dioptres, the reciprocal-focal-length sum becomes an additive power law $P=P_1+P_2+P_3+\cdots$. The sum is algebraic: convex lenses enter with $P\gt 0$ and concave lenses with $P\lt 0$, so the net power can be of either sign.
  6. Step 6 (Net magnification): Because the image of each lens is the object of the next, the linear magnifications multiply: $m=m_1\,m_2\,m_3\cdots$. Signed factors track erect ($m\gt 0$) versus inverted ($m\lt 0$) images, so the sign of the product reports the orientation of the final image while its magnitude gives the overall size ratio. This product rule underpins the magnification of compound microscopes and telescopes.
⚠️ JEE trap: A widespread error is to add focal lengths directly, writing $f=f_1+f_2$ for lenses in contact; it is the powers, or equivalently the reciprocals of focal length, that add. Another trap is adding magnifications instead of multiplying them, or ignoring the signs so that an inverted image is reported as erect. Also remember the additive power rule $P=P_1+P_2+\cdots$ holds only for thin lenses in contact; once the lenses are separated by a finite gap, an extra separation term appears and the simple sum fails. 🔉⇢

Refraction Through a Prism & Minimum Deviation 🔉⇢

Definition: A prism deviates a ray by an angle that depends on the angle of incidence; at minimum deviation the ray passes symmetrically and n21 = sin[(A + Dm)/2] / sin(A/2). 🔉⇢

When a narrow beam of light passes through a triangular prism it does not merely bend once — it is refracted twice and emerges travelling in a decidedly different direction from the one along which it entered. Consider a prism $ABC$ whose two polished refracting faces $AB$ and $AC$ meet along the refracting edge at $A$; the angle $A$ between these two faces is called the refracting angle, or simply the angle of the prism. A ray $PQ$ strikes the first face $AB$ at the point $Q$, making an angle of incidence $i$ with the normal there. Because glass is optically denser than air, the ray bends toward the normal as it enters, travelling inside the prism at an angle of refraction $r_1$. 🔉⇢

Full derivation, worked example and interactive 3D on the Refraction Through a Prism & Minimum Deviation tab →

The Simple Microscope (Magnifier) 🔉⇢

🎯 A magnifying glass is a single convex lens with the object just inside its focus. The eye sees a big, erect, virtual image; make the lens stronger (smaller f) and the magnification M = 1 + D/f climbs.
🔉⇢
simple microscope (image at the near point): M = 1 + D/f
= 1 + —/— = —
What you are looking at — a magnifying glass, the simplest microscope.
  • The teal lens is a single convex lens; the red object sits just inside its focus.
  • The dashed purple arrow on the left is the enlarged, erect, virtual image the eye actually sees at the near point.
What to do
  1. Keep f = 5 cm, D = 25 cm: M = 6×, the standard textbook value.
  2. Shorten f and the magnification rises — but only up to the point where the tiny lens is impractical, which is exactly why we move to a compound microscope.
Why it matters — M = 1 + D/f (image at the near point) or D/f (image at infinity). The gain comes from letting your eye sit far closer than 25 cm and still focus, so the object subtends a much larger angle.
Definition: A single converging lens of short focal length used close to the eye to give an erect, magnified virtual image, with magnifying power m = 1 + D/f (image at the near point). 🔉⇢

A simple microscope, also called a magnifier or magnifying glass, is nothing more than a single converging lens of small focal length held close to the eye. Its whole purpose is to let us examine small objects that the unaided eye cannot resolve. A normal relaxed eye can focus an object sharply only when the object lies at or beyond the near point, a distance $D\approx 25$ cm called the least distance of distinct vision. Bringing the object physically closer than $D$ makes it subtend a larger angle, but the eye can no longer converge the strongly diverging rays onto the retina, so the perceived image becomes blurred rather than clearer. 🔉⇢

The magnifier removes this limitation. When a converging lens is placed so that the object sits at a distance equal to or slightly less than its focal length $f$, the refracted rays emerge either parallel or only weakly diverging. The eye then interprets them as coming from a large, erect, virtual image located far beyond the object. Because this virtual image lies at or past the near point, it can be viewed comfortably, yet the object itself is held much nearer to the eye than $D$. This is the essential trick: the lens lets the tiny object subtend a much bigger angle at the eye than it ever could unaided. 🔉⇢

Two standard viewing configurations are used. In the first, the object is placed just inside the focal point so that the virtual image forms exactly at the near point, $v=-D$. This gives the greatest magnification but forces the eye muscles to accommodate, causing some strain during prolonged observation. In the second, the object is placed precisely at the focus, $u=-f$, so the image recedes to infinity and the eye views it in a fully relaxed state. The relaxed configuration sacrifices a little magnification for far greater viewing comfort, which is why most optical instruments are designed for image formation at infinity. 🔉⇢

It is important to distinguish linear magnification from angular magnification. Linear magnification $m=h'/h$ compares the size of the image with the size of the object. Angular magnification, or magnifying power, compares the angle subtended at the eye by the image seen through the lens with the angle the object would subtend if placed unaided at the near point. For a magnifier held close to the eye these two ratios turn out to be numerically equal, but conceptually the angular definition is the honest one, because what governs the apparent size of anything is the angle it subtends on the retina, not its absolute height. 🔉⇢

For the near-point setting the working relation is $m=1+\dfrac{D}{f}$. This expression shows that a shorter focal length yields a larger magnification, and that the magnifier always produces $m\gt 1$ since $D/f$ is positive for a converging lens. As a concrete figure, with $D=25$ cm a convex lens of focal length $f=5$ cm delivers $m=1+25/5=6$. To push the magnification higher one must shrink $f$ still further, which is exactly why magnifiers of high power are physically small, thick, strongly curved lenses that must be held very near both the object and the eye. 🔉⇢

For the relaxed-eye setting, with the image at infinity, the corresponding result is $m=\dfrac{D}{f}$. This is exactly one unit smaller than the near-point value, a difference that is usually negligible for high powers because $D/f$ greatly exceeds unity. The relaxed formula is the one carried forward when the magnifier serves as the eyepiece of a compound microscope or a telescope, since in those instruments the eyepiece is normally adjusted to throw the final image to infinity for the most comfortable viewing by the observer. 🔉⇢

Why does a short focal length translate into high magnifying power? A lens of small $f$ bends the incident light strongly, so it can accept rays from an object held extremely close and still render them parallel or gently diverging. The closer the object can be brought, the larger the angle $h/u$ it subtends, and that angle is what the eye ultimately magnifies. Since $D/f$ scales inversely with $f$, halving the focal length nearly doubles the magnification. This inverse dependence is the single most useful design fact about the simple microscope and about every magnifier built from one converging element. 🔉⇢

There are, however, firm practical limits. As $f$ is reduced, the lens surfaces must be ground with ever greater curvature and smaller aperture, and optical defects such as spherical and chromatic aberration grow rapidly, degrading the sharpness of the image. In practice a single-lens magnifier is limited to a magnification of roughly nine or ten before the image quality becomes unacceptable. Beyond that, one abandons the single element and compounds the effect of two lenses, an objective and an eyepiece, which is precisely the reasoning that leads to the compound microscope covered in the next topic. 🔉⇢

A subtle but frequently examined point is what actually changes when we look through the magnifier. The angular size of the object and the angular size of its virtual image are, strictly, equal. The instrument still helps because without it the smallest usable object distance is the near point $D$, whereas with it the object may be held at a distance of only $f$, far closer than $D$, so it subtends a much larger angle. The magnifier therefore does not enlarge the object in any absolute sense; it simply lets the eye exploit a viewing distance shorter than the near point while keeping the final image comfortably visible. 🔉⇢

The choice between the two magnification formulas is guided entirely by comfort and purpose. When the last ounce of magnification matters, the object is nudged just inside the focus so the virtual image lands at the near point and the eye accommodates to yield $1+D/f$. When the observation is prolonged, the object is set precisely at the focus so the image relaxes to infinity and the eye rests, yielding the slightly smaller $D/f$. The near point $D$ is itself a personal quantity, longer for a far-sighted observer and shorter for a near-sighted one, so the same converging lens delivers a somewhat different magnifying power to different eyes. 🔉⇢

In summary, the simple microscope is a converging lens of small focal length that forms an erect, magnified, virtual image of a nearby object. Its magnifying power is $1+D/f$ when the image is set at the near point and $D/f$ when the image is relaxed at infinity, with $D=25$ cm for a normal eye. The magnification rises as the focal length shrinks, but aberrations cap the single-lens instrument near tenfold. These same ideas of angular magnification, near point, focal length, and relaxed versus near-point viewing recur throughout the study of every refracting optical instrument. 🔉⇢

Derivation 🔉⇢

  1. Begin from the thin-lens formula written with the Cartesian sign convention, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$, where $u$ is the object distance, $v$ the image distance, and $f$ the focal length of the converging lens. The linear magnification produced by the lens is $m=\dfrac{v}{u}$. Our goal is to express this magnification purely in terms of the image position and the focal length, because the object is placed inside the focus and the image is what the eye actually views. Rearranging the lens formula gives $\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}$, which we substitute directly into the magnification.
  2. Substituting, $m=\dfrac{v}{u}=v\left(\dfrac{1}{v}-\dfrac{1}{f}\right)=1-\dfrac{v}{f}$. For the near-point setting the erect virtual image is formed on the same side as the object, so by the sign convention $v$ is negative and equal in magnitude to the least distance of distinct vision, that is $v=-D$. Inserting $v=-D$ yields $m=1-\dfrac{(-D)}{f}=1+\dfrac{D}{f}$. This is the magnifying power of the simple microscope when the image is placed at the near point $D\approx 25$ cm, and it confirms that shorter focal length gives larger magnification.
  3. Now derive the relaxed-eye result using angular magnification, which is the ratio of the angle subtended at the eye by the image to the angle subtended by the object placed unaided at the near point. Without the lens the largest angle a clearly visible object of height $h$ can subtend is at the near point, giving $\tan\theta_o=\dfrac{h}{D}$, and for small angles $\theta_o\approx\dfrac{h}{D}$. This is the reference angle against which the magnifier's performance must fairly be judged, since the near point is the closest the unaided eye can focus.
  4. With the object placed at the focus, $u=-f$, the refracted rays emerge parallel and the virtual image recedes to infinity. The angle now subtended at the eye equals the angle the object subtends at the lens, $\tan\theta_i=\dfrac{h}{f}$, so $\theta_i\approx\dfrac{h}{f}$ for paraxial rays. The angular magnification for the relaxed eye is therefore $m=\dfrac{\theta_i}{\theta_o}=\dfrac{h/f}{h/D}=\dfrac{D}{f}$.
  5. Comparing the two results, the near-point magnification $1+\dfrac{D}{f}$ exceeds the relaxed-eye magnification $\dfrac{D}{f}$ by exactly one. The near-point value is larger and useful when maximum magnification is wanted, while the infinity setting is preferred for comfort because the eye need not accommodate. Both expressions depend inversely on $f$, so a lens of small focal length is essential for high magnifying power; whenever $D/f\gg 1$ the two formulas give practically the same value, which is why the relaxed setting is adopted as the default in compound instruments.
⚠️ JEE trap: Students often claim a magnifier enlarges the object because its virtual image is bigger, and that the angle subtended by the image exceeds that subtended by the object. In fact those two angles are equal. The gain in magnifying power comes entirely from being able to hold the object at $f$, much nearer than the near point $D$, so it subtends a larger angle than it ever could unaided. Also note $m=1+D/f$ applies only for the image at $D$; the relaxed-eye value is $D/f$, one unit smaller, never larger. 🔉⇢

The Compound Microscope 🔉⇢

🎯 Two lenses, two stages: the objective makes a real magnified image inside the tube, then the eyepiece magnifies that. Shorten either focal length and the total M = (L/fₒ)(D/fₑ) jumps.
🔉⇢
compound microscope: M = mₒ × mₑ = (L/fₒ)(D/fₑ)  (D = 25 cm)
= (—/—)(25/—) = —
What you are looking at — the two-lens compound microscope.
  • The teal objective (short fₒ) forms a real, inverted, magnified first image (purple, pointing down) inside the tube.
  • The purple eyepiece then works on that first image like a simple magnifier for the eye.
What to do
  1. With fₒ = 2 cm, fₑ = 6.25 cm, L = 15 cm you get M ≈ 30×.
  2. Shorten fₒ or fₑ and watch M rise sharply — both appear in the denominator.
Why it matters — the total magnification is the PRODUCT of the two stages, M = mₒ × mₑ ≈ (L/fₒ)(D/fₑ). That product is why a compound microscope reaches hundreds of times, far beyond any single magnifying glass.
Definition: Two converging lenses in series — an objective forming a real magnified image and an eyepiece acting as a magnifier — giving m = (L/fo)(D/fe). 🔉⇢

A simple microscope built from one converging lens has a limited maximum magnification, typically no more than about nine or ten for realistic focal lengths, because shrinking the focal length further introduces severe aberrations. To achieve much larger magnification one uses two lenses, one compounding the effect of the other. This arrangement is the compound microscope. The lens nearest the object is called the objective and has a very short focal length $f_o$; the lens near the eye is the eyepiece with focal length $f_e$. The objective forms a real, inverted, magnified image, and the eyepiece then acts on that image exactly like a simple magnifier. 🔉⇢

The working sequence has two clear stages. Light from a small object placed just beyond the focus of the objective is refracted to form a real, inverted, magnified intermediate image inside the tube. This first image lands at or just within the focal plane of the eyepiece. The eyepiece, functioning as a magnifier, then produces the final image, which is enlarged and virtual, viewed by the eye. Because the objective inverts and the eyepiece preserves orientation, the final image is inverted with respect to the original object, which is acceptable for laboratory specimens that have no preferred up or down. 🔉⇢

The distance between the second focal point of the objective and the first focal point of the eyepiece is called the tube length $L$. It is this separation, rather than the object distance, that largely governs how much the objective magnifies. Because the object sits only just beyond $f_o$, the real intermediate image is thrown almost the full tube length away, so its linear magnification is close to $L/f_o$. A large tube length and a very short objective focal length therefore both increase the contribution of the objective to the total magnifying power of the instrument. 🔉⇢

The total magnification is the product of the magnification of the objective and the angular magnification of the eyepiece, $m=m_o\,m_e$. This multiplicative rule follows from the general principle that when the image formed by the first lens becomes the object for the second, the overall magnification is the product of the individual magnifications. The objective contributes $m_o=\dfrac{L}{f_o}$, a linear magnification, while the eyepiece contributes an angular magnification $m_e$ whose value depends on whether the final image is set at the near point or at infinity, exactly as for the simple magnifier. 🔉⇢

When the final image is formed at the near point of distinct vision, the eyepiece behaves as a near-point magnifier and contributes $m_e=1+\dfrac{D}{f_e}$, with $D\approx 25$ cm. The total magnifying power is then $m=\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$. This near-point setting gives the greatest possible magnification for a given pair of lenses, at the cost of some accommodation strain on the eye, because the eye must focus continually on an image held only $25$ cm away throughout the observation. 🔉⇢

When the final image is instead formed at infinity, the eyepiece is relaxed and contributes only $m_e=\dfrac{D}{f_e}$, so the total magnification reduces to the compact approximate form $m\approx\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$. This relaxed expression is the one usually quoted for the compound microscope, because comfortable viewing over long periods demands the image at infinity. It also cleanly separates the two design levers: a short objective focal length and a large tube length maximise $L/f_o$, while a short eyepiece focal length maximises $D/f_e$. 🔉⇢

As a worked figure, consider an objective with $f_o=1.0$ cm, an eyepiece with $f_e=2.0$ cm, and a tube length $L=20$ cm, with the final image relaxed at infinity. The magnification is $m=\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}=\dfrac{20}{1.0}\times\dfrac{25}{2.0}=250$. A magnification of two hundred fifty from lenses of centimetre focal length illustrates vividly why the compound design so decisively outperforms any single-lens magnifier, whose ceiling lies near tenfold. Had the same instrument been read at the near point instead, the eyepiece factor would rise from $D/f_e$ to $1+D/f_e$, lifting the total magnification modestly above the relaxed value, which confirms that the near-point setting always yields the larger magnifying power for a given objective and eyepiece. 🔉⇢

The expression makes plain why both focal lengths must be small. Since the objective magnification is $L/f_o$ and the eyepiece magnification is $D/f_e$, the total magnification varies inversely with the product $f_o f_e$. Reducing either focal length raises the magnifying power, so a good microscope demands both a short-focus objective and a short-focus eyepiece. In practice it is difficult to grind a lens with focal length much below one centimetre, and making the tube length very large requires correspondingly large lenses, so real instruments balance these competing constraints. 🔉⇢

Image quality depends on more than raw magnification. Illumination of the object, the numerical aperture of the objective, and the control of optical aberrations all shape the sharpness and visibility of the final image. For this reason modern microscopes replace each single lens with a multi-component objective and a multi-component eyepiece, groups of elements designed together to minimise spherical and chromatic aberration. A high magnification is worthless if the accompanying defects blur the very detail one is trying to resolve, so professional objectives are corrected assemblies rather than single thin lenses. The useful magnification is also bounded by the resolving power set by the objective aperture; magnifying beyond that limit produces a bigger but no sharper image, a condition known as empty magnification, so the objective is engineered for both a short focal length and a large numerical aperture together. 🔉⇢

A practical detail concerns where the object and the eye are placed. The object is positioned only just beyond the focal point of the objective, so that the real intermediate image is thrown far down the tube and magnified strongly by the factor $L/f_o$. At the other end, the eye is not pressed directly against the eyepiece but held a short distance behind it, at the position where all the emerging rays cross, called the eye ring. Placing the eye at this ring lets it collect the maximum light from the whole field of view, giving the brightest and widest final image the instrument can supply. 🔉⇢

In summary, the compound microscope cascades a short-focus objective that forms a real, inverted, magnified intermediate image with a short-focus eyepiece that magnifies it further as a virtual image. The total magnifying power is $m=m_o m_e=\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$ for the image at the near point and $m\approx\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$ for the relaxed eye. Both focal lengths must be small and the tube length large to secure high magnification, and multi-element lenses are used to keep aberrations in check while the objective and eyepiece do their compounding work. 🔉⇢

Derivation 🔉⇢

  1. Treat the compound microscope as two lenses whose magnifications multiply, since the real intermediate image formed by the objective serves as the object for the eyepiece. The total magnification is therefore $m=m_o\,m_e$, where $m_o$ is the linear magnification of the objective and $m_e$ is the angular magnification of the eyepiece acting as a simple magnifier. We derive each factor in turn and then combine them, taking $D\approx 25$ cm as the least distance of distinct vision for a normal eye.
  2. For the objective, let the object of height $h$ produce a first image of height $h'$. From the geometry of the ray diagram the angle the object subtends at the objective satisfies $\tan\beta=\dfrac{h}{f_o}$, while the same construction at the intermediate image gives $\tan\beta=\dfrac{h'}{L}$, where $L$ is the tube length measured between the second focal point of the objective and the first focal point of the eyepiece. Equating the two expressions, $\dfrac{h}{f_o}=\dfrac{h'}{L}$, so the objective's linear magnification is $m_o=\dfrac{h'}{h}=\dfrac{L}{f_o}$.
  3. For the eyepiece, the first inverted image lies at or within its focal plane, so it functions exactly as a simple microscope magnifying that image. Using the simple-magnifier result, when the final image is formed at the near point the eyepiece contributes an angular magnification $m_e=1+\dfrac{D}{f_e}$. This is the same near-point relation derived for a single converging lens, now applied to the intermediate image rather than to the original object.
  4. Combining the two factors for the near-point setting gives the exact total magnifying power $m=m_o\,m_e=\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$. This is the maximum magnification the instrument can deliver for a given tube length and pair of focal lengths, obtained when the eye accommodates to view the final virtual image at a distance $D$.
  5. For the relaxed eye the final image is thrown to infinity, so the eyepiece contributes only $m_e=\dfrac{D}{f_e}$ instead of $1+\dfrac{D}{f_e}$. The total magnification then reduces to the approximate form $m\approx\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$. Because this varies inversely with the product $f_o f_e$, both the objective and the eyepiece must have small focal lengths, and the tube length $L$ must be large, to achieve the high magnification for which the microscope is named.
⚠️ JEE trap: A common error is to add the two magnifications, writing $m=m_o+m_e$, instead of multiplying them. The correct rule is $m=m_o m_e$ because the intermediate image is the object for the eyepiece, so magnifications compound. A second trap is mixing the two eyepiece forms: use $1+D/f_e$ only for the final image at the near point and $D/f_e$ only for the relaxed eye at infinity. Finally, $L$ is the tube length between the objective and eyepiece focal points, not the separation between the lenses themselves. 🔉⇢

The Refracting & Reflecting Telescope 🔉⇢

🎯 A telescope is the microscope's opposite: for a distant object you want a LONG objective and a short eyepiece. Angular magnification is M = fₒ/fₑ and the tube is L = fₒ + fₑ long.
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telescope (normal adjustment): M = fₒ/fₑ  ·  tube length L = fₒ + fₑ
M = —/— = —  ·  L = —
What you are looking at — a refracting astronomical telescope in normal adjustment.
  • The teal parallel rays come from a distant object, arriving at a small angle.
  • The big teal objective focuses them to a small real image at its focal plane; the small purple eyepiece views that image at infinity for a relaxed eye.
What to do
  1. With fₒ = 144 cm and fₑ = 6 cm you get M = 24× and a tube 150 cm long.
  2. Shorten the eyepiece for more power — but the tube length is set almost entirely by the long objective.
Why it matters — opposite to the microscope: here you WANT a long objective focal length, because angular magnification is M = fₒ/fₑ. That is why big telescopes are physically long (L = fₒ + fₑ), and why observatories eventually switched to folded mirror designs.
Definition: An instrument for the angular magnification of distant objects, m = fo/fe, using a large-aperture objective (lens or mirror) and a short-focus eyepiece. 🔉⇢

A telescope is designed to provide angular magnification of distant objects rather than of small nearby ones. Like the compound microscope it has an objective and an eyepiece, but the roles are reversed in scale: here the objective has a large focal length and a much larger aperture than the eyepiece. Light from a distant object, arriving as an almost parallel beam, enters the objective and is refracted to form a real, inverted image at the second focal point of the objective, inside the tube. The eyepiece then magnifies this intermediate image, producing the final inverted image that the eye views. 🔉⇢

The quantity of interest for a telescope is the magnifying power, defined as the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ that the distant object itself subtends at the objective or the unaided eye. Because a distant object such as a star or a planet cannot be brought closer, its true angular size $\alpha$ is fixed; the telescope's task is to present the eye with a much larger apparent angle $\beta$. This is fundamentally different from the microscope, where the object's distance is ours to choose. 🔉⇢

In normal adjustment the telescope is arranged so that the final image is formed at infinity, which is the most comfortable setting for a relaxed eye observing the sky. The real image formed by the objective then falls exactly at the common focal point shared by objective and eyepiece, so the intermediate image sits at the first focal point of the eyepiece. Under this condition the magnifying power takes the clean form $m=\dfrac{f_o}{f_e}$, the ratio of the focal length of the objective to that of the eyepiece. A long-focus objective and a short-focus eyepiece together give high magnification. 🔉⇢

The overall length of the telescope tube in normal adjustment is simply $f_o+f_e$, the sum of the two focal lengths, because the intermediate image lies one objective focal length behind the objective and one eyepiece focal length in front of the eyepiece. This is the opposite arrangement to a microscope, whose short focal lengths are separated by a comparatively long tube length $L$. For example, an objective of focal length $100$ cm with an eyepiece of focal length $1$ cm gives a magnifying power of $100$ and a tube roughly $101$ cm long. A pair of stars whose true angular separation is one minute of arc would then appear separated by one hundred minutes, close to one and two-thirds of a degree, which is why even a modest refractor can split double stars that the unaided eye sees as a single point. 🔉⇢

Two other considerations dominate the design of an astronomical telescope: its light-gathering power and its resolving power. The light-gathering power depends on the area of the objective, so a larger diameter collects more light and allows fainter objects to be observed. The resolving power, the ability to distinguish two objects lying in very nearly the same direction, also improves with the diameter of the objective. For both reasons the desirable aim is an objective of large aperture. The largest lens objective ever used, at the Yerkes Observatory, has a diameter of about $1.02$ m. 🔉⇢

Large refracting objectives are, however, deeply problematic. A big lens is heavy and can be supported only around its rim, so it tends to sag under its own weight and distort the image. It is also difficult and expensive to grind a large lens whose two surfaces are figured accurately enough to be free from chromatic aberration, the coloured fringing caused because a lens brings different wavelengths to focus at slightly different points. These mechanical and optical limits set a practical ceiling on how large a refracting telescope can usefully be built. 🔉⇢

For these reasons modern large telescopes use a concave mirror instead of a lens as the objective; such instruments are called reflecting telescopes. A mirror has no chromatic aberration at all, because reflection does not depend on wavelength, so it forms images free of the colour defects that plague large lenses. A mirror also weighs far less than a lens of equivalent optical quality and, crucially, can be supported over its entire back surface rather than only at its rim, so mechanical sagging is far less of a problem and much larger apertures become feasible. 🔉⇢

The obvious difficulty with a reflecting telescope is that the concave objective mirror focuses the light back inside the tube, where an eyepiece and observer would obstruct the incoming beam. The Newtonian design solves this by placing a small plane mirror at forty-five degrees to divert the converging light sideways to an eyepiece mounted on the side of the tube. The Cassegrain design instead uses a convex secondary mirror to reflect the light back through a small hole bored in the centre of the objective primary mirror, giving a large effective focal length within a short, compact tube. 🔉⇢

These reflecting designs have made the world's giant telescopes possible. The Cassegrain arrangement in particular delivers a long focal length in a physically short instrument, which eases both mounting and housing. The largest telescope in India is a $2.34$ m diameter reflecting Cassegrain telescope at Kavalur, while the largest in the world are the pair of Keck telescopes in Hawaii, each with a reflector ten metres in diameter. No refracting design could reach such apertures, which is why every research-grade astronomical telescope today is a reflector. 🔉⇢

Two refinements are worth noting. If the eyepiece is drawn in slightly so the final image forms at the near point instead of infinity, the magnifying power rises to $m=\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)$, a modest increase bought at the cost of eye strain, so normal adjustment at infinity remains the preferred setting for astronomy. Secondly, the astronomical telescope gives an inverted final image, which is harmless for stars but unacceptable for viewing objects on Earth. Terrestrial telescopes therefore insert an extra pair of erecting lenses between objective and eyepiece to turn the final image upright, at the price of a longer tube. 🔉⇢

In summary, the astronomical refracting telescope forms a real image of a distant object with a large-aperture, long-focus objective and magnifies it with a short-focus eyepiece, giving magnifying power $m=f_o/f_e$ and tube length $f_o+f_e$ in normal adjustment. Large aperture is sought for light-gathering and resolving power, but the weight and chromatic aberration of big lenses force large instruments to use concave mirrors instead. Reflecting telescopes of the Newtonian and Cassegrain types avoid chromatic aberration, permit huge mirrors supported over their whole back, and so dominate modern astronomy. 🔉⇢

Derivation 🔉⇢

  1. Consider an astronomical refracting telescope in normal adjustment, meaning the final image is formed at infinity so that a relaxed eye can view it. Light from a distant object arrives as a nearly parallel beam making a small angle $\alpha$ with the principal axis; this $\alpha$ is the angle the object subtends at the objective and, effectively, at the unaided eye. The objective, of focal length $f_o$, refracts this beam to form a real, inverted intermediate image of height $h$ at its second focal point, which in normal adjustment coincides with the first focal point of the eyepiece.
  2. The intermediate image height $h$ can be related to the object angle. Since the incoming rays make angle $\alpha$ with the axis and are brought to focus one focal length behind the objective, the image subtends at the objective an angle equal to $\alpha$, giving $\tan\alpha=\dfrac{h}{f_o}$. For the small angles involved in viewing distant objects this becomes $\alpha\approx\dfrac{h}{f_o}$. This fixes the size of the real image formed inside the tube in terms of the objective focal length and the angular size of the distant object.
  3. The eyepiece, of focal length $f_e$, now views this intermediate image. Because in normal adjustment the image sits exactly at the first focal point of the eyepiece, the eyepiece renders the rays parallel and the final image is at infinity. The angle $\beta$ that this final image subtends at the eye is the angle the intermediate image subtends at the eyepiece, so $\tan\beta=\dfrac{h}{f_e}$, and for small angles $\beta\approx\dfrac{h}{f_e}$.
  4. The magnifying power is defined as the ratio of the angle subtended at the eye by the final image to that subtended by the object, $m=\dfrac{\beta}{\alpha}$. Substituting the two small-angle results, $m=\dfrac{\beta}{\alpha}=\dfrac{h/f_e}{h/f_o}=\dfrac{f_o}{f_e}$. The common image height $h$ cancels, leaving the magnifying power as the simple ratio of the objective focal length to the eyepiece focal length.
  5. Thus $m=\dfrac{f_o}{f_e}$ in normal adjustment, showing that a long-focus objective and a short-focus eyepiece are both needed for high magnification, which is the reverse of the microscope where both focal lengths are made small. The physical length of the telescope tube is the distance between the two lenses, namely $f_o+f_e$, since the intermediate image lies $f_o$ behind the objective and $f_e$ in front of the eyepiece. A large-diameter objective is additionally desired to raise the light-gathering power and the resolving power, though these do not enter the magnification formula itself.
⚠️ JEE trap: Many students confuse the microscope and telescope formulas, or think a bigger objective gives higher magnification. The telescope magnifying power is $m=f_o/f_e$, so it needs a long-focus objective and short-focus eyepiece; the objective diameter governs light-gathering and resolving power, not magnification. Another trap is writing the tube length as $f_o-f_e$ or $L$; in normal adjustment it is $f_o+f_e$. Also, the microscope uses two short focal lengths, whereas the telescope deliberately uses a long objective focal length, so their design logic is opposite. 🔉⇢

Refraction & Snell's Law 🔉⇢deep concept

Definition: The bending of light at an interface between two media, governed by Snell's law sin i / sin r = n21, with the refracted ray bending toward or away from the normal depending on relative optical density. 🔉⇢

🔬 Interactive 3D · A ray crosses from air into glass — drag the incidence angle and the refractive index and watch Snell's law bend the ray. angle of incidence i, refractive index n

Refraction is the phenomenon that occurs when a beam of light encounters another transparent medium at an interface. At the boundary a part of the light gets reflected back into the first medium while the rest enters the other. A ray of light represents such a beam, and the direction of propagation of an obliquely incident ray, with an angle of incidence between $0^\circ$ and $90^\circ$, changes as it crosses the interface between the two media. This bending of the ray at the surface separating two transparent media is what we call refraction of light. It arises because light travels with different speeds in different media, and it is governed by two simple, experimentally established laws first quantified by Snell. Note that if the ray strikes the interface normally, along the normal itself, it passes straight through without any bending; only an obliquely incident ray is deviated, and it is this oblique refraction that produces the familiar sight of a straight object appearing broken at a water surface. Throughout this discussion we treat light in the ray picture, tracing the incident ray and the refracted ray on either side of the point of incidence. 🔉⇢

The two laws of refraction, obtained experimentally, are stated as follows. First, the incident ray, the refracted ray and the normal to the interface at the point of incidence all lie in the same plane. Second, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media. Remembering that the angles of incidence $i$ and refraction $r$ are the angles that the incident and refracted rays make with the normal, we write $\dfrac{\sin i}{\sin r}=n_{21}$, the well-known Snell's law. Here $n_{21}$ is the refractive index of the second medium with respect to the first, a characteristic of the pair of media. 🔉⇢

It is important to appreciate what the refractive index $n_{21}$ represents. It is a characteristic of the pair of media and also depends on the wavelength of light, but it is independent of the angle of incidence. Thus, however obliquely or however nearly normally the ray strikes the interface, the ratio $\sin i / \sin r$ retains the same value for that pair of transparent media. When the first medium is vacuum, or to a good approximation air, the constant is called the absolute refractive index of the second medium, usually written simply as $n$. The refractive index of one medium with respect to another is then a relative refractive index built from these absolute values. 🔉⇢

Snell's law immediately tells us which way the refracted ray bends. From $\sin i/\sin r = n_{21}$, if $n_{21}\gt 1$ then $r\lt i$, so the refracted ray bends towards the normal. In such a case the second medium is said to be optically denser than the first. On the other hand, if $n_{21}\lt 1$ then $r\gt i$, and the refracted ray bends away from the normal; this is the situation when an incident ray in a denser medium refracts into a rarer medium. A useful rule follows from the vocabulary of the text: on entering an optically denser medium light bends toward the normal, while on entering a rarer medium it bends away from the normal. The angle of refraction is thus always smaller than the angle of incidence when the ray enters a denser medium, and always larger when it enters a rarer medium. 🔉⇢

The refractive index is intimately connected to the speed and the wavelength of light in the two media. The relative refractive index can be written as $n_{21}=\dfrac{v_1}{v_2}=\dfrac{\lambda_1}{\lambda_2}$, where $v_1,v_2$ are the speeds and $\lambda_1,\lambda_2$ the wavelengths of light in the first and second media respectively. When light passes into an optically denser medium its speed decreases and its wavelength shortens in the same proportion, so that a denser medium has a larger refractive index. A crucial point for problem solving is that the frequency of the light does not change on refraction; only the speed and the wavelength change, staying consistent with $v=\nu\lambda$. The absolute refractive index of a medium is just its value measured relative to vacuum, $n=c/v$, where $c$ is the highest speed attainable in nature, the speed of light in vacuum. Because the frequency is set by the source and is carried unchanged across the interface, the shortening of the wavelength inside a denser medium is precisely what accompanies the reduction in speed, and it is this dependence on wavelength that makes the refractive index slightly different for different colours of light. 🔉⇢

The text carefully warns that optical density must not be confused with mass density, which is mass per unit volume. Optical density is essentially the ratio of the speed of light in two media, and it is entirely possible that the mass density of an optically denser medium is actually less than that of an optically rarer medium. The standard example is turpentine and water: the mass density of turpentine is less than that of water, yet its optical density, and hence its refractive index, is higher. So when we call a medium denser in optics we mean that it slows light more strongly and bends the refracted ray towards the normal, irrespective of how heavy the substance happens to be. 🔉⇢

Several elementary but powerful relations follow at once from the laws of refraction. If $n_{21}$ is the refractive index of medium 2 with respect to medium 1, and $n_{12}$ that of medium 1 with respect to medium 2, then $n_{12}=\dfrac{1}{n_{21}}$. Furthermore, for three media the refractive indices chain together: if $n_{32}$ is the index of medium 3 with respect to medium 2, then $n_{32}=n_{31}\times n_{12}$, where $n_{31}$ is the index of medium 3 with respect to medium 1. These multiplicative relations let us combine the effect of several transparent media, and they are especially convenient when a ray passes successively through more than two media. 🔉⇢

Closely tied to Snell's law is the principle of reversibility of light, which states that the path of a ray of light is reversible: if the ray is made to retrace its route, it travels back along exactly the same path. If a ray travelling from medium 1 into medium 2 makes an angle of incidence $i$ and an angle of refraction $r$, then a ray sent backwards from medium 2 into medium 1 along the refracted direction emerges at the angle $i$ once more. This is fully consistent with $n_{12}=1/n_{21}$, since reversing the roles of the two media inverts the ratio of the sines. The same principle underlies the symmetry of the angles of incidence and emergence when a ray is traced back through a prism. 🔉⇢

A particularly instructive case is refraction through a rectangular, parallel-sided glass slab. Here refraction takes place at two interfaces, first air-to-glass and then glass-to-air. Applying Snell's law at both surfaces, and using the fact that the two faces are parallel so that the two normals are parallel, one finds that the angle of refraction at the second face equals the angle of incidence at the first, that is $r_2=i_1$. Consequently the emergent ray is parallel to the incident ray: there is no net deviation of the ray on passing through the slab. However, the emergent ray is displaced sideways with respect to the original incident ray, an effect called lateral displacement or lateral shift. 🔉⇢

The magnitude of this lateral shift depends on the thickness of the slab, on the angle of incidence and on the refractive index of the slab material. For a slab of thickness $t$, with angle of incidence $i$ and angle of refraction $r$ inside the glass, the perpendicular distance between the incident and emergent rays works out to $d=\dfrac{t\,\sin(i-r)}{\cos r}$. The shift grows with the thickness of the slab and with the obliquity of incidence, and it vanishes for normal incidence, where $i=r=0$. Because the ray suffers no angular deviation but only this sideways shift, an object viewed through a thick parallel slab appears displaced but not rotated, a fact that must be handled carefully in ray-tracing problems. 🔉⇢

Another familiar observation, noted directly in the text, is that the bottom of a tank filled with water appears to be raised. When we look at a coin or a needle lying at the bottom, the rays coming from it bend away from the normal as they pass from the denser water into the rarer air, and to our eye they seem to come from a point higher up. For viewing near the normal direction it can be shown that the apparent depth $h_1$ equals the real depth $h_2$ divided by the refractive index of the medium, that is $n=\dfrac{\text{real depth}}{\text{apparent depth}}$. Thus the denser the liquid, the more the bottom appears to be raised. 🔉⇢

Closely related to apparent depth is the normal shift produced by a transparent slab of thickness $t$ and refractive index $n$ placed in the line of sight. Since the apparent depth of an object seen through the slab is reduced by the factor $1/n$, the object appears to be shifted towards the observer by an amount $\text{shift}=t\left(1-\dfrac{1}{n}\right)$. For a glass slab of refractive index $1.5$ this normal shift is one-third of the thickness of the slab. A remarkable feature, which follows from the near-normal, paraxial treatment, is that this normal shift is independent of the position of the slab between the object and the eye; only the thickness and the refractive index of the slab matter. 🔉⇢

These ideas explain many everyday observations grounded in refraction. An object placed in water, such as a fish or a pebble, appears to be at a position different from the one it truly occupies, because the rays leaving the denser water bend away from the normal on entering the rarer air. A straight stick partly dipped in water looks bent at the surface for the same reason. Likewise, the apparent position of an object at the bottom of a swimming pool is higher than its real position, which is why the pool always looks shallower than it actually is. Every one of these effects is a direct manifestation of Snell's law at the water-air interface. Conversely, a person standing in water sees objects above the surface displaced as well, and a fish looking upward sees the entire outside world compressed into a bright cone, since rays arriving at every angle up to the horizon are refracted into the denser medium. In each case the amount by which the apparent position differs from the real one is governed by the refractive index of the water relative to air. 🔉⇢

On a much larger scale, the refraction of light by the earth's atmosphere causes the advance of sunrise and the delayed sunset. The density, and hence the refractive index, of air decreases with height, so a ray of light from the sun is refracted continuously and bends as it passes through progressively rarer layers of the atmosphere. As a result we can see the sun a little before it has actually risen above the horizon, and for a short while after it has actually set below it. The apparent shift in the sun's position near the horizon, produced by this atmospheric refraction, lengthens the effective daytime by a couple of minutes at both sunrise and sunset. 🔉⇢

The twinkling of stars is another beautiful consequence of atmospheric refraction. Starlight, on entering the earth's atmosphere, undergoes refraction continuously through layers of air whose refractive index keeps fluctuating because of physical changes such as varying temperature and density. Since the stars are so distant they behave as point sources, and the tiny, ever-changing refraction makes the apparent position and the apparent brightness of a star waver slightly, so that it appears to twinkle. Planets, being much closer, present an extended disc rather than a point, and the fluctuations from different parts of the disc average out, which is why planets generally do not twinkle. All these phenomena reinforce how pervasively Snell's law and refractive index shape what we see. 🔉⇢

To gather the core of this topic into one place: refraction is the change in direction of a ray as it crosses obliquely from one transparent medium into another, and Snell's law $\sin i/\sin r=n_{21}$ quantifies it exactly. The refracted ray bends towards the normal on entering a denser medium and away from the normal on entering a rarer medium; the refractive index equals the ratio of speeds $v_1/v_2$ and of wavelengths $\lambda_1/\lambda_2$ of light in the two media, while the frequency stays fixed. From these principles flow the reversibility of the ray, the zero net deviation but finite lateral shift through a parallel-sided slab, the apparent-depth and normal-shift formulae, and a host of natural phenomena from raised tank bottoms to twinkling stars. 🔉⇢

Derivation from first principles 🔉⇢

  1. Step 1 - Geometry of the parallel-sided slab. Consider a glass slab of thickness $t$ bounded by two parallel faces, with a ray incident on the first, air-to-glass, face at an angle of incidence $i$. Let the angle of refraction inside the glass be $r$. The refracted ray travels in a straight line across the slab and strikes the second, glass-to-air, face. Because the two faces are parallel, the normal at the second face is parallel to the normal at the first, so the angle of incidence at the second face is also equal to $r$. This single geometric fact is the key to the whole result.
  2. Step 2 - The emergent ray is parallel to the incident ray. Applying Snell's law at the first face gives $\sin i = n\sin r$, where $n$ is the refractive index of the glass with respect to air. At the second face the ray goes from the denser glass into the rarer air with angle of incidence $r$ and angle of emergence $e$, so $n\sin r=\sin e$. Comparing the two relations gives $\sin e=\sin i$, hence $e=i$. The emergent ray therefore makes the same angle with the normal as the incident ray and is parallel to it: the slab produces no net deviation, only a sideways shift.
  3. Step 3 - Identify the lateral shift. Let the ray strike the first face at the point $O$ and leave the second face at the point $B$. The lateral shift $d$ is the perpendicular distance between the emergent ray and the undeviated continuation of the incident ray. Dropping a perpendicular from $B$ onto the produced incident direction, meeting it at $M$, we have $d=OB\,\sin(i-r)$. This is because the actual path $OB$ inside the slab makes an angle $r$ with the normal, whereas the original incident direction makes an angle $i$ with the same normal, so the angle between the two directions is $(i-r)$.
  4. Step 4 - Express the path length inside the slab. The perpendicular thickness of the slab, measured along the normal, is $t$. If we drop the normal to the faces through the point $O$, then in the right triangle formed by the path $OB$ and this thickness the angle at $O$ is $r$, so that $\cos r = \dfrac{t}{OB}$. Rearranging gives $OB=\dfrac{t}{\cos r}$. Here $OB$ is the actual distance travelled by the refracted ray within the glass between the two parallel faces, and it is always longer than the perpendicular thickness $t$ for oblique incidence.
  5. Step 5 - Combine to obtain the lateral shift. Substituting $OB=t/\cos r$ into $d=OB\,\sin(i-r)$ yields $$d=\dfrac{t\,\sin(i-r)}{\cos r}.$$ This is the required lateral displacement of the ray. It confirms that $d=0$ when $i=r=0$, that is for normal incidence, and that the shift increases both with the thickness $t$ of the slab and with the obliquity of incidence, exactly as the qualitative discussion of the parallel-sided slab in the text predicts. The emergent ray is thus parallel to, but laterally displaced from, the incident ray.
  6. Step 6 - Geometry of apparent depth. Now consider a point object $O$ lying at a real depth $t$ below the plane surface of a denser medium of refractive index $n$, for example water, viewed from the rarer medium, air, above. Take a ray that leaves $O$ and meets the surface at a point $N$ close to the point $P$ situated directly above $O$, making a small angle of incidence $\theta_1$ with the normal drawn inside the denser medium. At the surface this ray refracts into the air and travels outward at an angle $\theta_2$ to the same normal, bending away from the normal since it enters a rarer medium.
  7. Step 7 - Apply Snell's law for near-normal viewing. Going from the denser medium into air, Snell's law gives $n\sin\theta_1=\sin\theta_2$. For viewing near the normal both angles are small, so we may use $\sin\theta_1\approx\tan\theta_1$ and $\sin\theta_2\approx\tan\theta_2$. The relation then simplifies to $n\tan\theta_1=\tan\theta_2$. This small-angle, paraxial step is exactly the condition, stated in the text, under which the simple apparent-depth result holds; for strongly oblique viewing the apparent depth would depend on the viewing angle.
  8. Step 8 - Introduce the image point. When produced backwards, the emergent ray appears to come from a point $I$ at a depth $t'$, the apparent depth, below the surface, lying on the vertical through $O$. Let $PN=x$ be the small horizontal distance of the emergence point $N$ from the foot of the vertical at $P$. Then from the right triangle with the real object we have $\tan\theta_1=\dfrac{x}{t}$, since $\theta_1$ is the angle subtended at $O$, and from the right triangle with the image we have $\tan\theta_2=\dfrac{x}{t'}$, since $\theta_2$ is the angle subtended at $I$.
  9. Step 9 - Solve for the apparent depth and the normal shift. Substituting these two results into $n\tan\theta_1=\tan\theta_2$ gives $n\cdot\dfrac{x}{t}=\dfrac{x}{t'}$, so that $t'=\dfrac{t}{n}$. Hence $n=\dfrac{\text{real depth}}{\text{apparent depth}}$: the apparent depth is the real depth divided by the refractive index of the medium. The object therefore appears raised towards the observer by the normal shift $$\text{shift}=t-t'=t\left(1-\dfrac{1}{n}\right).$$ Since the horizontal distance $x$ cancels out, the result is independent of the precise near-normal ray chosen, which is why the raised bottom of a tank is seen at a definite apparent depth.
⚠️ JEE trap: A common trap is to assume that the frequency of light changes on refraction: it does not. When light enters a denser medium its speed and its wavelength both decrease in the ratio $1/n$, but the frequency is fixed by the source, so $v=\nu\lambda$ stays consistent. A second frequent error is direction confusion, bending the ray away from the normal on entering a denser medium; remember that a denser medium means $r\lt i$, so the refracted ray bends towards the normal. In apparent-depth problems, dividing by the wrong quantity, using apparent over real depth, inverts the shift. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A small coin rests at the bottom of a tank. It is covered first by a layer of water of refractive index $n_1=4/3$ and thickness $8.0\ \text{cm}$, and above the water floats a layer of oil of refractive index $n_2=1.5$ and thickness $6.0\ \text{cm}$. An observer looks straight down from the air above, i.e. views the coin near the normal direction.
TARGET Find the apparent depth of the coin below the top oil surface, and the total normal shift by which the coin appears to be raised.
STRATEGY For near-normal viewing the apparent depth contributed by each transparent layer is its real thickness divided by its own refractive index, and the contributions of successive layers simply add. Compute $t_i/n_i$ for each layer, sum them to get the apparent depth measured from the top surface, then subtract this from the total real depth to obtain the normal shift.
EXECUTE The water layer contributes an apparent depth $\dfrac{t_1}{n_1}=\dfrac{8.0}{4/3}=8.0\times\dfrac{3}{4}=6.0\ \text{cm}$. The oil layer contributes $\dfrac{t_2}{n_2}=\dfrac{6.0}{1.5}=4.0\ \text{cm}$. Adding these, the apparent depth of the coin below the top oil surface is $6.0+4.0=10.0\ \text{cm}$. The true real depth is $t_1+t_2=8.0+6.0=14.0\ \text{cm}$. Hence the normal shift by which the coin appears raised is $\text{shift}=14.0-10.0=4.0\ \text{cm}$.
REFLECT The coin appears $4.0\ \text{cm}$ nearer the surface than it really is, purely because of refraction at the two interfaces where the ray bends away from the normal. Each denser layer raises the image through its own factor $1/n$, and the individual shifts add. The frequency of the light is unchanged throughout; only the speed and the wavelength differ from layer to layer. The result matches the near-normal apparent-depth relation $n=\text{real depth}/\text{apparent depth}$ applied layer by layer.

Source: JEE-pattern (NCERT Ch 9)

Total Internal Reflection & Optical Fibres 🔉⇢deep concept

Definition: When light travelling in a denser medium meets a rarer medium beyond the critical angle, it is reflected entirely back — the principle behind optical fibres and totally reflecting prisms. 🔉⇢

🔬 Interactive 3D · Light bouncing down an optical fibre by repeated total internal reflection — raise the angle past the critical angle to trap it. angle of incidence i, core refractive index n

When a ray of light travels from an optically denser medium to a rarer medium, at the interface it is partly reflected back into the same denser medium and partly refracted into the rarer medium. This partial reflection into the originating medium is called internal reflection. Because the second medium is rarer, the refracted ray bends away from the normal, so the angle of refraction $r$ is larger than the angle of incidence $i$. As long as $i$ is modest, both a reflected ray and a refracted ray coexist, and the refracted ray carries away most of the light energy while the internally reflected ray remains comparatively feeble. This everyday situation, light escaping upward from water into air, is the starting point for understanding total internal reflection. 🔉⇢

Now imagine slowly increasing the angle of incidence $i$ at this denser-to-rarer interface. Snell's law forces the angle of refraction $r$ to increase as well, and the refracted ray tilts progressively further from the normal, grazing ever closer to the interface. At the same time the internally reflected ray steadily grows brighter, stealing energy from the transmitted beam. There comes a special angle of incidence for which the refracted ray bends so much that it just grazes the surface, meaning the angle of refraction becomes exactly $90°$. This particular angle of incidence, corresponding to an angle of refraction of $90°$, is called the critical angle $i_c$ for the given pair of media. It marks the boundary between ordinary refraction and the dramatic phenomenon that follows. 🔉⇢

If the angle of incidence is increased still further, so that $i \gt i_c$, Snell's law of refraction can no longer be satisfied: there is simply no real angle of refraction that solves the equation, because the sine of the angle of refraction would have to exceed unity. Refraction into the rarer medium therefore becomes impossible, and the entire incident beam is thrown back into the denser medium, obeying the ordinary law of reflection. This complete return of light is called total internal reflection. The two conditions are indispensable and must both hold: light must be travelling from a denser to a rarer medium, and the angle of incidence must be greater than the critical angle. Remove either condition and total internal reflection cannot occur. 🔉⇢

It is worth dwelling on what makes total internal reflection so remarkable compared with ordinary reflection. Whenever light is reflected at a surface in the usual way, some fraction of it is always transmitted as well; the reflected ray is therefore always less intense than the incident ray, no matter how smooth and polished the reflecting surface may be. In total internal reflection, by contrast, no transmission of light takes place at all. There is no energy left in a refracted ray because there is no refracted ray. Essentially one hundred per cent of the incident light is returned into the denser medium. This lossless, mirror-like return, achieved without any metallic coating, is precisely why total internal reflection is so valuable in optical instruments and optical fibres. 🔉⇢

The critical angle is directly tied to the refractive index of the two media through Snell's law. Writing the law at the critical angle, where the angle of refraction is $90°$, leads to the compact result $\sin i_c = 1/n$, where $n$ is the refractive index of the denser medium with respect to the rarer medium. A larger refractive index of the denser medium makes $1/n$ smaller, and hence the critical angle $i_c$ smaller. In other words, the more optically dense a material is relative to its surroundings, the more easily it traps light by total internal reflection, since even modest angles of incidence already exceed its small critical angle. This single relation underlies every application discussed below, from the sparkle of diamond to signal transmission along an optical fibre. 🔉⇢

Some representative critical angles, measured with respect to air, make the trend concrete. Water, with refractive index about $1.33$, has a critical angle close to $48.75°$. Crown glass, of refractive index $1.52$, has a critical angle near $41.14°$, while dense flint glass at $1.62$ drops to about $37.31°$. Diamond, with an exceptionally high refractive index of $2.42$, has a critical angle of only about $24.41°$. Reading down this list, the critical angle falls steadily as the refractive index rises, exactly as $\sin i_c = 1/n$ predicts. The very small critical angle of diamond is the single most important fact behind its optical behaviour, because it means that light entering a well-cut diamond is very likely to strike its inner faces at an angle greater than the critical angle. 🔉⇢

Because the refractive index of a medium depends on the wavelength of light, the critical angle is not a single fixed number but varies slightly with colour. A transparent medium is typically more refracting for violet light than for red light, so violet, having the larger refractive index, has the smaller critical angle, while red light has a marginally larger one. Consequently, when white light approaches an interface near the critical angle, different colours can behave differently: some may still refract and escape while others are already totally internally reflected. This subtle dependence on wavelength contributes, together with dispersion in a prism, to the play of colour, or fire, seen in a cut diamond, where the low critical angle and strong dispersion act together to enrich its appearance. 🔉⇢

Total internal reflection also occurs in nature, most famously in the mirage seen on a hot day. On a sunny afternoon the air in contact with the ground becomes hot and therefore optically rarer, while the cooler air higher up is comparatively denser. Light from a distant object, or from the bright sky, travelling downward through these layers is refracted more and more away from the normal as it passes from denser to progressively rarer air. When it eventually meets a layer at an angle greater than the critical angle, it is totally internally reflected and curves back upward toward the observer's eye. The brain, assuming light travels in a straight line, perceives an inverted image below the object, giving the shimmering illusion of a pool of water on the ground. 🔉⇢

The familiar wet-road illusion is a direct example of this optical mirage. On a hot road the shimmering patch that looks like a puddle of water is in fact an image of the bright sky, brought to the eye by total internal reflection in the layer of hot, rarer air just above the surface. No water is present; the road never actually becomes wet. The apparent surface even seems to shift and quiver, because turbulent, unevenly heated air continually changes the local refractive index and hence the exact angle at which total internal reflection sets in. Recognising this shows that total internal reflection is not merely a laboratory curiosity produced with a glass beaker and a laser beam, but a phenomenon that shapes what we see outdoors. 🔉⇢

The brilliance and sparkle of a diamond is one of the most celebrated consequences of total internal reflection. Because diamond has a very high refractive index of $2.42$, its critical angle is only about $24.41°$, far smaller than for glass or water. A skilled gem cutter shapes and angles the many facets so that light entering through the top strikes the inner faces at angles greater than this small critical angle. The light therefore suffers repeated total internal reflections, bouncing about inside the stone before finally emerging from the upper facets toward the eye. Since each internal reflection is essentially lossless, very little light leaks out of the sides or bottom, and the diamond appears to blaze with returned light. Proper cutting, not mere polishing, is what unlocks this trapped brilliance. 🔉⇢

Totally reflecting prisms exploit the same principle in a controlled, engineered way. A prism of glass cut as a right-angled isosceles triangle, with angles of $45°$, $45°$ and $90°$, can turn a beam of light through $90°$ or through $180°$, or invert an image without changing its size. Light entering a short face normally strikes the hypotenuse at an angle of incidence of $45°$. For crown glass or dense flint glass the critical angle is smaller than $45°$, so this $45°$ incidence is greater than the critical angle and the light is totally internally reflected at the hypotenuse. Such prisms are preferred over ordinary silvered mirrors in periscopes, binoculars and other optical instruments, because total internal reflection returns essentially all the light without the tarnishing or dimming that afflicts a metallic reflecting coating. 🔉⇢

The optical fibre is perhaps the most far-reaching technological application of total internal reflection. Each fibre is fabricated from high-quality composite glass or quartz and consists of two parts: a central core surrounded by a coating called the cladding. The refractive index of the material of the core is deliberately made higher than that of the surrounding cladding, so that the core behaves as the denser medium and the cladding as the rarer one. When light travelling in the core strikes the core-cladding boundary at an angle greater than the critical angle for that pair, it is totally internally reflected and confined within the core. The cladding thus does not merely protect the core; it provides the rarer medium that makes repeated total internal reflection along the fibre possible. 🔉⇢

When a light signal is directed into one end of such a fibre at a suitable angle, it undergoes repeated total internal reflections along the entire length of the fibre and finally emerges at the far end. Because every reflection is essentially lossless, there is no appreciable loss in the intensity of the signal, and light can travel even around gentle bends; the fibre acts as an optical pipe, guiding light much as a pipe guides water. This makes optical fibres ideal for transmitting audio, video and other signals over long distances, and a bundle of fibres can also relay an image. In medicine, such a light pipe is used in endoscopy to carry light into, and images out of, internal organs like the esophagus, stomach and intestines for visual examination. 🔉⇢

Bringing these ideas together, total internal reflection is governed by two simple requirements, light passing from a denser to a rarer medium and an angle of incidence greater than the critical angle $i_c$ given by $\sin i_c = 1/n$, yet from these follow a rich variety of phenomena. The distinguishing feature throughout is that the reflection is total: unlike ordinary partial reflection, which always leaves some energy in a transmitted ray, total internal reflection returns essentially all the light with no refracted ray at all. This lossless behaviour, combined with the way the critical angle shrinks as the refractive index grows, explains the sparkle of diamond, the working of totally reflecting prisms in periscopes and binoculars, the shimmering mirage on a hot road, and the quiet efficiency of optical fibres. 🔉⇢

Total internal reflection can be demonstrated very simply. If clear water in a glass beaker is made slightly turbid with a few drops of milk and a laser beam is shone through it, the path of the beam inside the water becomes clearly visible. When the beam is directed at the upper water surface at a gentle angle, it undergoes partial reflection back into the water and partial refraction out into the air, seen as two separate spots. As the beam is made to strike the surface more and more obliquely, a point is reached where the refracted beam above the water vanishes entirely and the beam is thrown completely back into the water. That vanishing of the refracted ray, at incidence greater than the critical angle, is total internal reflection in its simplest form. If the same turbid water is poured into a long test tube and the laser beam is sent in from the top, the beam can be adjusted so that it is totally internally reflected every time it strikes the walls of the tube, travelling down the tube in a zig-zag path. This is exactly what happens inside an optical fibre, where the beam is guided by repeated total internal reflection along its whole length. 🔉⇢

For an optical fibre to actually guide light, the ray must strike the core-cladding wall at an angle greater than the critical angle, and this in turn restricts the angles at which light may enter the front face of the fibre. Only rays entering within a certain cone about the axis are refracted into the core steeply enough to then meet the wall beyond the critical angle; rays entering too obliquely strike the wall at less than the critical angle and leak into the cladding. The half-angle of this cone is called the acceptance angle of the fibre. Understanding it requires combining Snell's law at the entrance face with the critical-angle condition at the wall, which is exactly the kind of analysis carried out in the derivation that follows. 🔉⇢

Derivation from first principles 🔉⇢

  1. Begin with Snell's law of refraction at the boundary between a denser medium of refractive index $n$ (taken relative to the rarer medium) and the rarer medium. Treating the denser medium as medium 1 and the rarer as medium 2, and assigning the rarer medium a refractive index of unity for convenience, Snell's law reads $n \sin i = (1)\sin r$, where $i$ is the angle of incidence in the denser medium and $r$ is the angle of refraction in the rarer medium. Equivalently, $\sin i / \sin r = 1/n$, since here $n$ is the refractive index of the denser medium with respect to the rarer one.
  2. As the angle of incidence $i$ is increased, the angle of refraction $r$ must increase to keep Snell's law satisfied, and because the second medium is rarer the refracted ray bends away from the normal. The largest physically possible angle of refraction is $r = 90°$, at which the refracted ray grazes along the interface. The angle of incidence that produces this grazing refracted ray is, by definition, the critical angle $i_c$. We therefore impose the two conditions $r = 90°$ and $i = i_c$ together in Snell's law in order to locate this threshold angle.
  3. Substituting $i = i_c$ and $r = 90°$ into $n \sin i = \sin r$ gives $n \sin i_c = \sin 90° = 1$. Solving for the critical angle yields the central result $\sin i_c = 1/n$. Because $n$, the refractive index of the denser medium with respect to the rarer, is greater than one, the quantity $1/n$ is less than one, so a valid angle $i_c$ always exists. This relation also shows that the larger the refractive index $n$, the smaller $\sin i_c$ and hence the smaller the critical angle, in agreement with the tabulated values for water, crown glass, flint glass and diamond.
  4. Now consider an angle of incidence greater than the critical angle, $i \gt i_c$. Snell's law would then demand $\sin r = n \sin i \gt n \sin i_c = 1$. But the sine of a real angle can never be greater than unity, so there is no real angle of refraction that satisfies the equation. Refraction into the rarer medium is thus impossible, and the whole incident beam is reflected back into the denser medium according to the ordinary law of reflection. This is total internal reflection, and it confirms the two necessary conditions: a denser-to-rarer path, and an angle of incidence greater than the critical angle.
  5. Next analyse the totally reflecting right-angled prism, with angles $45°$, $45°$ and $90°$, made of glass in air. A ray entering normally through one of the short faces travels undeviated and meets the hypotenuse. The geometry of the isosceles right triangle makes the angle of incidence of this ray on the hypotenuse equal to $45°$, measured from the normal to that face. Whether the ray is totally internally reflected there depends entirely on how this $45°$ compares with the critical angle of the glass-air interface, so the next step is to compare $45°$ with $i_c$.
  6. For crown glass the refractive index is about $1.52$, giving $\sin i_c = 1/1.52$ and a critical angle of roughly $41.1°$; for dense flint glass the critical angle is even smaller, about $37.3°$. In both cases the critical angle is less than $45°$. Since the ray strikes the hypotenuse at $45°$, which is greater than the critical angle, the condition $i \gt i_c$ is satisfied and the ray is totally internally reflected. By choosing which faces the light enters and leaves, the same prism can deviate a beam by $90°$ or by $180°$, or invert an image, all with essentially no loss of light.
  7. Finally, derive the fibre acceptance condition qualitatively. Let the core have refractive index $n_1$ and the cladding $n_2$, with $n_1 \gt n_2$. A ray enters the flat end face of the core from air at an angle $\theta$ to the fibre axis and is refracted into the core. Applying Snell's law at the entrance face gives $\sin \theta = n_1 \sin \theta_1$, where $\theta_1$ is the refraction angle inside the core, again measured from the axis. This refracted ray then travels across the core to strike the core-cladding wall.
  8. At the cylindrical core-cladding wall the angle of incidence, measured from the normal to the wall, is $\phi = 90° - \theta_1$. For the ray to be guided along the fibre it must undergo total internal reflection there, which requires $\phi \gt i_c$, where the critical angle for the core-cladding pair satisfies $\sin i_c = n_2/n_1$. Thus a ray that enters too steeply, giving a large $\theta_1$, makes $\phi$ too small; it then falls below the critical angle and leaks into the cladding instead of being guided down the core.
  9. Combining the two conditions gives the acceptance requirement. The largest entry angle $\theta$ that is still guided corresponds to $\phi = i_c$, that is $\theta_1 = 90° - i_c$. Rays entering within the cone of half-angle equal to this maximum value of $\theta$, called the acceptance angle, strike the wall at an angle greater than the critical angle and are trapped by repeated total internal reflection along the fibre; rays entering outside the cone are lost into the cladding. Qualitatively, a larger difference between the core and cladding refractive indices lowers $i_c$, widens the acceptance cone, and lets the fibre gather light entering over a broader range of angles.
⚠️ JEE trap: A common trap is to assume total internal reflection can happen when light goes from a rarer to a denser medium; it cannot, because the ray must travel from the denser into the rarer medium, otherwise the refracted ray bends toward the normal and always escapes. Equally wrong is thinking it occurs at any large angle: it requires the angle of incidence to be strictly greater than the critical angle $i_c$. And do not imagine that 'total' still leaves a faint refracted ray; for $i \gt i_c$ there is no refracted ray at all and no energy is transmitted into the rarer medium. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A tiny point source of light lies at the bottom of a still pool of water of refractive index $n = 1.33$, at a depth $h = 2.0$ m below the flat water surface. Light spreads out from the source in all directions and strikes the water-air surface from below.
TARGET Find the radius of the circular patch of the surface through which light actually emerges into the air, and explain what happens to the surface outside this circle.
STRATEGY Light travels here from the denser water to the rarer air. A ray emerges only if its angle of incidence at the surface is less than the critical angle $i_c$; at exactly $i_c$ the refracted ray grazes the surface, and for incidence greater than $i_c$ the ray is totally internally reflected back into the water. The escaping rays therefore fill a cone of half-angle $i_c$ about the vertical, meeting the surface in a circle of radius $R = h\tan i_c$.
EXECUTE First, $\sin i_c = 1/n = 1/1.33 = 0.752$, so $i_c = 48.75°$. Then $\cos i_c = \sqrt{1 - 0.752^2} = \sqrt{1 - 0.565} = \sqrt{0.435} = 0.659$, giving $\tan i_c = 0.752/0.659 = 1.14$. The radius of the illuminated circle is $R = h\tan i_c = 2.0 \times 1.14 = 2.28$ m. The corresponding area is $\pi R^2 = 3.14 \times (2.28)^2 = 16.3$ m$^2$.
REFLECT Light escapes only through a circle of radius about $2.28$ m directly above the source; everywhere outside this circle the rays reach the surface at an angle greater than the critical angle and are totally internally reflected back down into the water, so the surface there acts like a perfect mirror. Note that the radius scales directly with depth, $R = h\tan i_c$, and depends only on the refractive index through $i_c$, not on the brightness of the source.

Source: JEE-pattern (NCERT Ch 9)

The Thin Lens Formula & Magnification 🔉⇢deep concept

Definition: The relation 1/v - 1/u = 1/f for a thin lens, with magnification m = v/u, valid for convex and concave lenses and for real and virtual images. 🔉⇢

🔬 Interactive 3D · Ray-tracing through a convex lens with three construction rays — drag the object across the focus and watch the image flip from virtual to real. object distance u, focal length f

A thin lens is a transparent optical medium bounded by two refracting surfaces, at least one of which is spherical, whose thickness is negligible compared with the object distance, the image distance and the radii of curvature involved. Because the thickness can be ignored, both refracting surfaces are treated as passing through a single point on the principal axis called the optical centre. The line through the optical centre and the two centres of curvature is the principal axis. When paraxial rays (rays close to the axis making small angles with it) from a point object are refracted by such a lens, they reconverge to, or appear to diverge from, a single image point. The single relation that ties the object distance $u$, the image distance $v$ and the focal length $f$ together is the thin lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, and it holds for every situation once the sign convention is respected. 🔉⇢

All distances in this formula are measured from the optical centre using the Cartesian sign convention. Distances measured in the same direction as the incident light are taken as positive, and those measured against the direction of incident light are negative. Since light is drawn as travelling left to right, a real object sitting to the left of the lens has a negative object distance $u\lt0$. Heights measured upward from the principal axis are positive and those measured downward are negative. This single convention is what allows one compact formula to describe a converging or a diverging lens forming a real or a virtual image; you never switch formulae, you only substitute signed numbers. Careless students memorise a version of the relation for one case and then contradict the convention when the geometry changes, which is the single largest source of error in lens problems. 🔉⇢

The sign of the focal length classifies the lens. A convex (double-convex or converging) lens is thicker at its centre than at its rim; it bends a parallel incident beam inward so that the refracted rays actually converge to a real principal focus on the far side, and by convention its focal length is positive, $f\gt0$. A concave (double-concave or diverging) lens is thinner at the centre; it spreads a parallel incident beam outward so the refracted rays appear to diverge from a virtual focus on the incoming side, and its focal length is negative, $f\lt0$. The magnitude of $f$ measures how strongly the lens bends light: a short focal length means strong convergence or divergence. The power $P=1/f$ (in dioptres, with $f$ in metres) is positive for a converging lens and negative for a diverging lens, and for thin lenses in contact the powers simply add. 🔉⇢

To locate an image graphically we trace two of three standard construction rays from a chosen off-axis object point. The first ray leaves the object parallel to the principal axis; after refraction it passes through the second principal focus $F'$ of a convex lens, or appears to diverge from the first focus $F$ of a concave lens. The second ray passes straight through the optical centre and emerges undeviated, because near the centre the two lens surfaces are effectively parallel like a thin slab, giving zero net deviation. The third ray travels through (or is directed toward) the first focus and emerges parallel to the principal axis. Any two of these rays intersect at the image point; the third is a useful check. Where the refracted rays physically cross, the image is real and can be caught on a screen; where only their backward extensions cross, the image is virtual. 🔉⇢

Consider a convex lens and follow the image as the object moves inward from far away. When the object is at infinity, the incident rays are parallel and the refracted rays converge exactly at the second focus, forming a real, inverted, highly diminished image in the focal plane. This is the situation used to define the focal length, since $u\to-\infty$ makes $\frac{1}{u}\to0$ and the formula gives $v=f$. Objects at large but finite distance therefore form small real inverted images just beyond the focus, which is why a camera or the objective of a telescope forms a tiny real picture of a distant scene near its focal plane. 🔉⇢

As the object is brought to a distance beyond twice the focal length (that is, beyond $2f$ on the object side), the real inverted image lies between $F'$ and $2F'$ on the far side and is diminished, with magnification magnitude less than one. When the object is placed exactly at $2f$, the image forms at $2f$ on the other side, is real and inverted, and is exactly the same size as the object, so the magnification is $-1$. This symmetric conjugate pair, object and image both at twice the focal length, is a convenient reference point that JEE problems frequently exploit when they ask for the separation between an object and its equal-sized real image. 🔉⇢

Moving the object into the region between $f$ and $2f$ pushes the real inverted image outward beyond $2F'$, and now the image is magnified, with magnification magnitude greater than one. This is the regime used by a slide or film projector, which is why the object (the slide) is placed just outside the focus so that a large real inverted image lands on a distant screen. When the object reaches the first focus itself, so that $u=-f$, the refracted rays emerge exactly parallel to one another; they never converge, and the image recedes to infinity. Right at the focus, therefore, no finite image exists, a boundary case that separates real-image formation from virtual-image formation. 🔉⇢

Finally, when the object is placed inside the focal length, between the optical centre and the first focus, the refracted rays diverge on emerging. They no longer meet in front of the lens, but their backward extensions meet on the same side as the object, producing a virtual, erect and magnified image. This is exactly how a convex lens works as a simple magnifier or magnifying glass: the object is held closer than one focal length so that the eye sees an enlarged upright virtual image. Thus a single convex lens spans the full range of behaviour, from a tiny real inverted image of a distant object to a large virtual erect image of a nearby one, purely through the position of the object relative to the focus. 🔉⇢

A concave (diverging) lens is far simpler, because it produces only one kind of image. For any real object, wherever it is placed, the diverging refraction bends the rays outward, and their backward extensions meet between the object and the lens on the incoming side. The image is therefore always virtual, always erect, and always diminished, and it always lies closer to the lens than the object. Algebraically, with $f\lt0$ and $u\lt0$, the thin lens formula forces $v$ to be negative and smaller in magnitude than $u$, so no substitution can ever yield a real image from a single concave lens acting on a real object. This is the principle behind spectacle lenses for short-sightedness, which shrink and bring the image within the eye's range. 🔉⇢

The linear (transverse) magnification produced by a thin lens is defined as the ratio of image height to object height, and it equals $m=\frac{v}{u}$. The sign of $m$ carries physical meaning that must never be dropped. A positive $m$ means the image is erect and, for a lens, virtual; a negative $m$ means the image is inverted and real. The magnitude of $m$ tells whether the image is enlarged ($|m|\gt1$) or diminished ($|m|\lt1$). For a convex lens forming a real image, $u$ and $v$ have opposite signs so $m$ is negative, consistent with the inverted real image; for the same lens acting as a magnifier, $u$ and $v$ share the same sign and $m$ is positive, consistent with the erect virtual image. For a concave lens $m$ is always positive and less than one. 🔉⇢

It is worth stressing how the sign structure differs from the mirror equation. For a spherical mirror the relation is $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ with a plus sign between the reciprocal distances, and the magnification is $m=-\frac{v}{u}$ with a minus sign. For a thin lens the relation is $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with a minus sign, and the magnification is $m=+\frac{v}{u}$ with a plus sign. The difference arises because in a mirror the reflected light returns to the same side as the incident light, whereas in a lens the refracted light continues forward to the opposite side, so a real image in a lens sits on the positive side while a real image in a mirror sits on the negative side. Confusing these two sign patterns is a classic trap and reliably produces wrong answers. 🔉⇢

Both the thin lens formula and the magnification relation are not independent postulates: they reduce from refraction at two spherical surfaces. When one applies the single-surface refraction relation $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$ successively at the two curved faces of the lens, treating the intermediate image formed by the first surface as the object for the second, and then invokes the thin-lens condition that both surfaces lie at the same optical centre, the two intermediate terms cancel. What remains is the lens maker's formula $\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, and substituting the object-at-infinity definition of $f$ collapses the whole two-step process into the single thin lens formula. This is why the same compact relation governs a lens regardless of the refractive index or the exact curvature of its faces, and why the derivation below matters. 🔉⇢

The idea of the focal plane sharpens the picture of image formation. A parallel incident beam that is inclined to the principal axis is brought by a convex lens not to the axial focus but to a point in the focal plane, the plane through the second focus perpendicular to the axis; the exact point is fixed by the undeviated ray through the optical centre, which for an inclined beam is the central ray of that beam. This is why a distant extended object, every point of which sends in a nearly parallel bundle from a slightly different direction, produces a small real inverted image spread across the focal plane rather than a single dot. The same reasoning applied to a diverging lens places the virtual foci of inclined beams in the focal plane on the incoming side, so that a concave lens forms a diminished erect virtual image of a distant scene near its own focal plane. 🔉⇢

The formulae extend cleanly to more than one lens, which is why they underpin every real instrument. For thin lenses placed in contact, the image formed by the first lens serves as the object for the second, and adding the individual thin lens relations shows that the reciprocals of the focal lengths add: $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\cdots$, equivalently the powers add, $P=P_1+P_2+\cdots$, as an algebraic sum in which converging contributions are positive and diverging contributions are negative. The total linear magnification of such a combination is the product $m=m_1 m_2 \cdots$ of the individual magnifications, since each stage magnifies the image handed to it by the previous stage. This is exactly the machinery a compound microscope or a telescope uses: an objective forms a real intermediate image which the eyepiece, acting as a simple magnifier, further enlarges, so the overall magnification is the product of the two. 🔉⇢

In practice, whenever you attack a numerical problem, adopt a disciplined routine. First fix the direction of incident light and mark distances against it with signs; second identify the focal length sign from whether the lens is converging or diverging; third substitute the signed values of $u$ and $f$ into $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ to obtain $v$ with its own sign; fourth compute $m=\frac{v}{u}$ and read off the nature of the image from the signs. A positive computed $v$ means a real image on the far side of a convex lens; a negative $v$ means a virtual image on the near side. This algebraic discipline reproduces every ray-diagram result automatically, and it extends without change to a system of thin lenses in contact where powers add and total magnification is the product of the individual magnifications. 🔉⇢

A final conceptual point ties the whole picture together. The focal length of a lens depends on both the shape of its surfaces (the radii of curvature) and on the relative refractive index of the lens material with respect to its surroundings. If a converging glass lens is immersed in a medium of higher refractive index, the term $(n_{21}-1)$ can change sign and the lens can actually behave as a diverging element; if the surrounding medium matches the lens material exactly, $n_{21}=1$ and the focal length becomes infinite, so the lens produces no convergence at all and behaves like a flat plate. This is a direct consequence of the same two-surface refraction that gives the thin lens formula, and it reminds us that whether a lens converges or diverges is not a fixed label but a relationship between the glass and the world around it. 🔉⇢

Derivation from first principles 🔉⇢

  1. Step 1 — The single-surface starting point. Refraction at one spherical surface separating a medium of refractive index $n_1$ (where the light starts) from a medium $n_2$ (which it enters), with the surface having radius of curvature $R$ and centre of curvature on the axis, obeys $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$ for paraxial rays. This relation was itself obtained by writing the exterior-angle relations for the small angles that the incident and refracted rays make with the axis and applying Snell's law in the small-angle form $n_1 i = n_2 r$. We take it as the established building block and apply it twice, once at each face of a double-convex lens of material index $n_2$ placed in a surrounding medium of index $n_1$.
  2. Step 2 — Refraction at the first surface. Let a point object lie on the axis at object distance $u$ from the (first) surface. The first surface, of radius $R_1$, refracts light from $n_1$ into $n_2$ and would by itself form an intermediate image at some distance $v_1$. Applying the building block to this surface gives $\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}$. This intermediate image $I_1$ is generally virtual for a strongly converging first surface, but its algebraic distance $v_1$ carries its own sign, so no special casing is needed; the equation is written once and used as is.
  3. Step 3 — Refraction at the second surface. The light now travelling inside the glass strikes the second surface, of radius $R_2$, and refracts from $n_2$ back into the surrounding medium $n_1$. The intermediate image $I_1$ acts as the object for this second refraction. Applying the same building block, but now with the roles of the indices reversed (light goes from $n_2$ to $n_1$), the second surface forms the final image at distance $v$: $\frac{n_1}{v}-\frac{n_2}{v_1}=\frac{n_1-n_2}{R_2}$. Note carefully the swapped order $n_1-n_2$ on the right, which is the direct consequence of the light now entering the medium of index $n_1$.
  4. Step 4 — The thin-lens condition and addition. For a thin lens the physical separation of the two surfaces is negligible, so the object distance for the second surface is the same intermediate image distance $v_1$ produced by the first; there is no gap to account for. We now add the two surface equations. The awkward intermediate terms $\frac{n_2}{v_1}$ and $-\frac{n_2}{v_1}$ are equal and opposite, so they cancel exactly. Adding gives $\frac{n_1}{v}-\frac{n_1}{u}=(n_2-n_1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, where the right-hand side has been regrouped by factoring out $(n_2-n_1)$ and noting that $\frac{n_1-n_2}{R_2}=-\frac{(n_2-n_1)}{R_2}$.
  5. Step 5 — Dividing out the surrounding index. Divide the whole equation by $n_1$. This yields $\frac{1}{v}-\frac{1}{u}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. The bracketed factor $\left(\frac{n_2}{n_1}-1\right)=(n_{21}-1)$ contains the relative refractive index of the lens material with respect to its surroundings, and the second bracket depends only on the geometry of the two spherical surfaces. Everything on the right is a constant fixed by the lens itself, independent of where the object is placed.
  6. Step 6 — Identifying the focal length. Send the object to infinity, so that the incident rays are parallel and $\frac{1}{u}\to0$. By definition the image then forms at the focus, $v=f$. Substituting, $\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, which is the lens maker's formula. Because the right-hand side of the general relation in Step 5 is exactly this constant, we may replace it, obtaining the thin lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. The derivation shows the formula is a compressed statement of two successive refractions, and that $f\gt0$ for a converging lens and $f\lt0$ for a diverging lens follows directly from the signs of the radii through the lens maker's expression.
  7. Step 7 — Setting up the magnification geometry. To find the transverse magnification, consider an extended object $AB$ of height $h$ standing perpendicular to the axis with its foot $B$ on the axis, and its image $A'B'$ of height $h'$. Use the undeviated construction ray: the ray from the object tip $A$ that passes straight through the optical centre continues in a straight line to the image tip $A'$. This ray, together with the principal axis, forms two triangles on either side of the optical centre.
  8. Step 8 — Similar triangles and $m=v/u$. The triangle formed by the object (height $h$, base equal to the object distance) and the triangle formed by the image (height $h'$, base equal to the image distance) share the vertical angle at the optical centre and each has a right angle at the axis, so they are similar. Equating ratios of corresponding sides gives $\frac{h'}{h}=\frac{\text{image distance}}{\text{object distance}}$. Introducing the Cartesian signs, with the object distance measured as $u$ and the image distance as $v$ from the optical centre, this becomes $m=\frac{h'}{h}=\frac{v}{u}$.
  9. Step 9 — Reading the sign of the magnification. Because $m=\frac{v}{u}$ carries the ratio of two signed quantities, its sign is automatic. For a convex lens forming a real image, $u\lt0$ while $v\gt0$, so $m\lt0$: the image is inverted, exactly as the ray diagram shows. For the same lens used as a magnifier with the object inside the focus, both $u$ and $v$ are negative, so $m\gt0$ and the image is erect. For a concave lens acting on a real object, $v$ and $u$ are both negative and $|v|\lt|u|$, giving $0\lt m\lt1$, an erect diminished virtual image. The lens magnification $m=+\frac{v}{u}$ differs deliberately from the mirror result $m=-\frac{v}{u}$, because a real image in a lens lies on the positive side whereas in a mirror it lies on the negative side.
⚠️ JEE trap: A very common JEE trap is to carry the mirror sign pattern into lens problems: writing $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $m=-\frac{v}{u}$ for a lens. The lens formula uses a minus sign, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, and its magnification uses a plus sign, $m=+\frac{v}{u}$. A second trap is assuming any inverted image must be real for both lens types; a single concave lens acting on a real object never gives a real or inverted image at all. Always assign signs first, then read the nature of the image from the sign of $v$ and $m$, never from intuition. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A thin convex lens has focal length $20\ \text{cm}$. An object $3.0\ \text{cm}$ tall is placed on the principal axis, first at $30\ \text{cm}$ from the lens and then moved to $15\ \text{cm}$ from the lens.
TARGET For each object position find the image distance $v$, the linear magnification $m$, the image height, and the nature of the image (real or virtual, erect or inverted, magnified or diminished).
STRATEGY Use the Cartesian sign convention: incident light travels toward the lens, so a real object gives $u\lt0$, and a converging lens has $f\gt0$. Apply $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ to solve for $v$, then compute $m=\frac{v}{u}$ and $h'=mh$. Read the nature of the image from the signs of $v$ and $m$.
EXECUTE Here $f=+20\ \text{cm}$ and $h=3.0\ \text{cm}$. Case 1: object beyond the focus, $u=-30\ \text{cm}$. $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}+\frac{1}{-30}=\frac{3-2}{60}=\frac{1}{60}$, so $v=+60\ \text{cm}$. $m=\frac{v}{u}=\frac{+60}{-30}=-2$, and $h'=mh=-2\times3.0=-6.0\ \text{cm}$. Since $v\gt0$ the image is real (far side); $m\lt0$ means it is inverted; $|m|=2\gt1$ means it is magnified. Result: real, inverted, $6.0\ \text{cm}$ tall. Case 2: object inside the focus, $u=-15\ \text{cm}$. $\frac{1}{v}=\frac{1}{20}+\frac{1}{-15}=\frac{3-4}{60}=-\frac{1}{60}$, so $v=-60\ \text{cm}$. $m=\frac{v}{u}=\frac{-60}{-15}=+4$, and $h'=+4\times3.0=+12\ \text{cm}$. Since $v\lt0$ the image is virtual (same side as object); $m\gt0$ means erect; $|m|=4\gt1$ means magnified. Result: virtual, erect, $12\ \text{cm}$ tall.
REFLECT The same convex lens flips behaviour as the object crosses the focus: beyond $f$ it forms a real inverted image ($m\lt0$), while inside $f$ it acts as a magnifier giving a virtual erect image ($m\gt0$). The sign of $v$ alone tells you real versus virtual, and the sign of $m$ tells you inverted versus erect, so no case-by-case memorisation is needed. Had we mistakenly used the mirror pattern $m=-\frac{v}{u}$, both natures would have come out wrong.

Source: JEE-pattern (NCERT Ch 9)

Refraction Through a Prism & Minimum Deviation 🔉⇢deep concept

Definition: A prism deviates a ray by an angle that depends on the angle of incidence; at minimum deviation the ray passes symmetrically and n21 = sin[(A + Dm)/2] / sin(A/2). 🔉⇢

🔬 Interactive 3D · A ray through a triangular prism — sweep the angle of incidence and watch the deviation fall to a minimum when the path is symmetric. angle of incidence i, prism angle A

When a narrow beam of light passes through a triangular prism it does not merely bend once — it is refracted twice and emerges travelling in a decidedly different direction from the one along which it entered. Consider a prism $ABC$ whose two polished refracting faces $AB$ and $AC$ meet along the refracting edge at $A$; the angle $A$ between these two faces is called the refracting angle, or simply the angle of the prism. A ray $PQ$ strikes the first face $AB$ at the point $Q$, making an angle of incidence $i$ with the normal there. Because glass is optically denser than air, the ray bends toward the normal as it enters, travelling inside the prism at an angle of refraction $r_1$. 🔉⇢

Inside the glass the ray travels in a straight line from $Q$ across to the second face $AC$, where it strikes at some point $R$. At this interface light is passing from the denser glass into the rarer air, so the situation is reversed: the angle of incidence inside the glass is $r_2$, and the ray now bends away from the normal as it escapes, leaving along $RS$ at the angle of emergence $e$. The net effect of the two refractions — one at entry, one at emergence — is that the emergent ray $RS$ is turned through a definite angle relative to the original direction of the incident ray $PQ$. This total turning is called the angle of deviation, written $\delta$, and it is the single most important quantity in prism optics. 🔉⇢

The geometry linking the two internal angles to the prism angle is remarkably clean. Look at the quadrilateral $AQNR$, where $N$ is the intersection of the two normals drawn at $Q$ and at $R$. Two of its angles — those at $Q$ and $R$ — are right angles, because each normal is perpendicular to its face. Since the four interior angles of any quadrilateral sum to $360^\circ$, the remaining pair must add to $180^\circ$, giving $A + \angle QNR = 180^\circ$. In the triangle $QNR$ the three angles obey $r_1 + r_2 + \angle QNR = 180^\circ$. Comparing these two statements immediately yields the prism relation $A = r_1 + r_2$: the refracting angle equals the sum of the two internal refraction angles. 🔉⇢

The deviation itself is built up from the two faces separately. At the first face the ray is turned by $(i - r_1)$, since it swings from the incident direction toward the normal. At the second face it is turned by a further $(e - r_2)$ as it swings away from the normal on emergence. The total deviation is the sum of these two contributions, $\delta = (i - r_1) + (e - r_2)$. Substituting the prism relation $r_1 + r_2 = A$ collapses this to the compact and famous result $\delta = i + e - A$. Notice how the geometry of the interior has been folded entirely into the single measurable constant $A$, leaving deviation expressed only through the externally observable incidence and emergence angles. 🔉⇢

The relation $\delta = i + e - A$ conceals an important behaviour: the angle of deviation is not fixed but varies with the angle of incidence. If you slowly rotate the prism, feeding the same face a steadily changing $i$, the emergent ray sweeps and $\delta$ changes with it. A plot of $\delta$ against $i$ is not a straight line but a curve that first falls, reaches a lowest point, and then rises again. This means that, in general, any chosen value of the deviation — except one special value — is produced by two distinct angles of incidence. For each such value there is a smaller $i$ paired with a larger $e$, and a larger $i$ paired with a smaller $e$, both giving the identical deviation. 🔉⇢

This two-to-one behaviour is not an accident; it is demanded by the symmetry of the formula. Because $\delta = i + e - A$ is symmetric under the interchange of $i$ and $e$, swapping the roles of the incidence and emergence angles leaves the deviation unchanged. Physically this is the principle of reversibility of light: the path drawn through the prism can be traced backwards, entering where it formerly emerged, and the same deviation results. The two branches of the $\delta$ versus $i$ curve are therefore mirror images of one another. Where the two branches meet — the single value of $\delta$ that corresponds to just one angle of incidence — the incidence and emergence angles must have become equal, and this is the turning point of the curve. 🔉⇢

That turning point is the angle of minimum deviation, denoted $D_m$. It is the smallest deviation the prism can impose on a ray for a given wavelength, and it occurs under a beautifully symmetric configuration. At minimum deviation the incidence and emergence angles are equal, $i = e$; by the symmetry this forces the two internal angles to be equal as well, $r_1 = r_2$. Combining $r_1 = r_2$ with the prism relation $A = r_1 + r_2$ gives $r_1 = r_2 = A/2$. In this symmetric passage the ray inside the prism runs parallel to the base $BC$ — a fact you can use in the laboratory to recognise the minimum-deviation setting simply by watching the internal ray straighten out relative to the base. 🔉⇢

The symmetric minimum-deviation geometry is precisely what makes the prism a precision instrument for measuring refractive index. Setting $i = e$ in $\delta = i + e - A$ with $\delta = D_m$ gives $D_m = 2i - A$, so the common angle of incidence is $i = (A + D_m)/2$. The internal angle is $r_1 = A/2$. Applying Snell's law at the first face, $n_{21} = \sin i / \sin r_1$, and substituting these two results gives the celebrated prism formula, $n_{21} = \dfrac{\sin\!\big(\frac{A + D_m}{2}\big)}{\sin(A/2)}$. Here $n_{21}$ is the refractive index of the prism material relative to the surrounding medium. Both $A$ and $D_m$ are angles that can be read off a spectrometer table with high accuracy, so the formula turns two protractor readings into a precise value of the refractive index. 🔉⇢

It is worth dwelling on why the formula is so useful. The refracting angle $A$ is a fixed property of the ground glass, measurable once and for all. The minimum deviation $D_m$ is found experimentally by rotating the prism and locating the position where the emergent ray retreats no further — the deviation reaches its least value and momentarily stops changing. Because near the minimum the deviation is stationary, small errors in setting the angle of incidence produce only second-order errors in $D_m$, which is exactly why the minimum-deviation method is so robust. Feeding the measured $A$ and $D_m$ into $n_{21} = \sin[(A + D_m)/2] / \sin(A/2)$ then delivers the refractive index of the material, and — since $n$ depends on wavelength — a different $D_m$ for each colour. 🔉⇢

A particularly clean limit emerges for a thin prism, one whose refracting angle $A$ is very small. When $A$ is small the minimum deviation $D_m$ is also small, and for small angles the sine of an angle is nearly equal to the angle itself. The prism formula then simplifies: $n_{21} \simeq \dfrac{(A + D_m)/2}{A/2} = \dfrac{A + D_m}{A}$. Rearranging gives the thin-prism result $D_m = (n_{21} - 1)A$, usually written simply as $\delta = (n - 1)A$. This tidy expression says a thin prism produces a deviation directly proportional to its refracting angle and to the excess of its refractive index over unity. It also makes plain that thin prisms deviate light only feebly, which is why they are the natural building block for analysing dispersion. 🔉⇢

Dispersion is the reason a prism does more than bend white light — it unfurls it into a band of colours. The refractive index $n$ of any transparent material is not a single number but depends on the wavelength of the light passing through it. For ordinary glass, $n$ is larger for shorter wavelengths and smaller for longer wavelengths, so violet light, with its short wavelength, is slowed and bent more strongly than red light, with its longer wavelength. Since the deviation produced by a prism grows with $n$, each wavelength is deviated by a slightly different amount. White light entering the prism as a single beam therefore leaves as a fan of overlapping coloured rays, each colour emerging along its own direction. The word dispersion names precisely this splitting of light into its constituent colours, and it is inseparable from refraction: without a wavelength-dependent refractive index there would be no separation at all, and white light would pass through deviated but still white. It is the tiny variation of $n$ across the visible band, from roughly $400\,\text{nm}$ to $750\,\text{nm}$, that the prism magnifies into a visible spread of colour. 🔉⇢

The spread of colours the prism produces is captured by the idea of angular dispersion — the difference in deviation between the two extreme colours of the beam. Using the thin-prism result $\delta = (n - 1)A$ for each colour, the deviation of violet light is $\delta_v = (n_v - 1)A$ and that of red light is $\delta_r = (n_r - 1)A$. The angular dispersion between them is the difference $\delta_v - \delta_r = (n_v - n_r)A$. Because $n_v$ is greater than $n_r$, this difference is positive: violet is deviated most and red least, with orange, yellow, green, blue and indigo arranged in order between them. This ordered band of colours cast on a screen is what we call the spectrum of white light. 🔉⇢

It is important to keep dispersion and mean deviation as separate ideas. The overall deviation of the beam is governed roughly by a middle colour such as yellow and grows with the refracting angle $A$; the dispersion — the fanning apart of the colours — is governed by the difference $n_v - n_r$, a property of the material sometimes described through its dispersive power. Two prisms can bend light through the same average angle yet spread the colours very differently if their glasses have different dispersive behaviour. This distinction is what allows optical designers to combine prisms of different glasses so as to cancel dispersion while retaining deviation, or vice versa — the principle behind achromatic components in quality instruments. In short, deviation answers "by how much is the beam bent?" while dispersion answers "by how much are the colours pulled apart?" — related through the refractive index, yet controlled by different features of it. 🔉⇢

The rainbow is nature's own demonstration of exactly this chain of refraction, dispersion and deviation, staged not in glass but in countless spherical raindrops. Sunlight entering a droplet is refracted on the way in, and because the refractive index of water depends on wavelength, the colours are separated just as they are in a prism; the light is then reflected once at the far inside surface and refracted again on the way out. Each wavelength emerges most intensely at its own characteristic angle, and the eye, gathering rays from many drops at slightly different heights, sees the colours ranged in a great coloured arc. The physics is identical to the prism's — dispersion turning a single incident direction into an ordered spread of colours. 🔉⇢

For problem solving it helps to hold the working relations together as one toolkit. The prism relation $A = r_1 + r_2$ ties the internal geometry to the fixed refracting angle; the deviation relation $\delta = i + e - A$ gives the turning of the ray from the two externally measured angles; the minimum-deviation conditions $i = e$ and $r_1 = r_2 = A/2$ pin down the symmetric configuration; the prism formula $n_{21} = \sin[(A + D_m)/2] / \sin(A/2)$ extracts refractive index; and the thin-prism limit $\delta = (n - 1)A$ handles small-angle and dispersion questions. Whenever a value of $\delta$ that is not the minimum is quoted, remember two angles of incidence can produce it, and Snell's law at each face closes the system. 🔉⇢

A final grounding remark ties the algebra to what the eye actually sees. Because the deviation curve is flat near its minimum, a prism placed in a spectrometer at the minimum-deviation setting gives the sharpest, most stable colour separation, which is why that setting is used both for measuring $n$ and for displaying a clean spectrum. The same wavelength dependence that produces this spectrum also explains why thick lenses fringe images with colour — an unwanted cousin of dispersion. Understanding the prism thus does double duty: it supplies a precise laboratory method for refractive index through $n_{21} = \sin[(A + D_m)/2]/\sin(A/2)$, and it explains the everyday splendour of spectra and rainbows through the single fact that refractive index varies with wavelength. 🔉⇢

Derivation from first principles 🔉⇢

  1. Step 1 — Set up the geometry. Draw the triangular prism $ABC$ with refracting angle $A$ at the apex. A ray meets the first face $AB$ at $Q$ with angle of incidence $i$ and refracts to angle $r_1$ inside the glass. It crosses to the second face $AC$, meeting it at $R$ with internal angle $r_2$, and emerges into air at the angle of emergence $e$. Let the normals at $Q$ and at $R$ meet at $N$. Our aim is to relate the four ray angles $i, r_1, r_2, e$ and the prism angle $A$ to the total angle of deviation $\delta$ between the incident and emergent rays.
  2. Step 2 — Establish the prism relation $A = r_1 + r_2$. Consider the quadrilateral $AQNR$. The normals are perpendicular to their respective faces, so the angles at $Q$ and $R$ are each $90^\circ$. The interior angles of any quadrilateral sum to $360^\circ$, hence the angle at $A$ plus the angle $\angle QNR$ must equal $180^\circ$: that is, $A + \angle QNR = 180^\circ$. This isolates the internal angle at $N$ in terms of the known refracting angle $A$, and sets up the comparison we make in the next step using the triangle $QNR$.
  3. Step 3 — Use the triangle $QNR$. Inside the prism the refracted ray $QR$ together with the two normals forms the triangle $QNR$. The angle this internal ray makes with the normal at $Q$ is $r_1$, and with the normal at $R$ is $r_2$. The three angles of the triangle sum to $180^\circ$, so $r_1 + r_2 + \angle QNR = 180^\circ$. Comparing this with $A + \angle QNR = 180^\circ$ from Step 2, the common term $\angle QNR$ cancels, leaving the prism relation $A = r_1 + r_2$. The refracting angle of the prism equals the sum of the two internal angles of refraction — a purely geometric result independent of the medium.
  4. Step 4 — Build the deviation face by face. At the first face the ray bends toward the normal, turning through $(i - r_1)$. At the second face the ray bends away from the normal on emergence, turning through a further $(e - r_2)$. Both turnings are in the same rotational sense, so the total angle of deviation is their sum: $\delta = (i - r_1) + (e - r_2)$. Regrouping the terms gives $\delta = (i + e) - (r_1 + r_2)$. This expresses the deviation entirely through the entry and exit angles at the two faces plus the internal angles that we have just related to $A$.
  5. Step 5 — Substitute the prism relation. Insert $r_1 + r_2 = A$ from Step 3 into the grouped expression $\delta = (i + e) - (r_1 + r_2)$. The internal angles disappear, and we obtain the deviation relation $\delta = i + e - A$. This is one of the central results: the deviation depends only on the incidence angle, the emergence angle, and the fixed refracting angle of the prism. Because $i$ and $e$ enter symmetrically, interchanging them leaves $\delta$ unchanged — the mathematical signature of the reversibility of the ray path through the prism, and the reason the deviation curve has two branches.
  6. Step 6 — Impose the minimum-deviation condition. As the angle of incidence is varied, $\delta$ passes through a single minimum value $D_m$. At this extremum the two symmetric branches of the deviation curve meet, which requires the incidence and emergence angles to be equal, $i = e$. Feeding $i = e$ into $\delta = i + e - A$ with $\delta = D_m$ gives $D_m = 2i - A$, so the common angle of incidence at minimum deviation is $i = (A + D_m)/2$. Furthermore, equality of $i$ and $e$ combined with the symmetry forces the internal angles to be equal too, $r_1 = r_2$.
  7. Step 7 — Derive the prism formula. With $r_1 = r_2$ and the prism relation $A = r_1 + r_2$, each internal angle equals half the prism angle: $r_1 = r_2 = A/2$. Snell's law at the first face reads $n_{21} = \dfrac{\sin i}{\sin r_1}$, where $n_{21}$ is the refractive index of the prism material relative to its surroundings. Substituting $i = (A + D_m)/2$ and $r_1 = A/2$ gives the prism formula $n_{21} = \dfrac{\sin\!\big(\frac{A + D_m}{2}\big)}{\sin(A/2)}$. Since both $A$ and $D_m$ are directly measurable, this equation provides a precise experimental route to the refractive index of the material of the prism.
  8. Step 8 — Take the thin-prism limit. For a prism with a very small refracting angle $A$, the minimum deviation $D_m$ is also small. When an angle is small, its sine is approximately equal to the angle itself in radians, so $\sin[(A + D_m)/2] \simeq (A + D_m)/2$ and $\sin(A/2) \simeq A/2$. The prism formula then reduces to $n_{21} \simeq \dfrac{(A + D_m)/2}{A/2} = \dfrac{A + D_m}{A}$. Solving for the deviation gives $D_m = (n_{21} - 1)A$, commonly written $\delta = (n - 1)A$. Thin prisms therefore deviate light only slightly, in direct proportion to the refracting angle.
  9. Step 9 — Extend to dispersion. Because the refractive index varies with wavelength, the thin-prism result applies colour by colour: $\delta_v = (n_v - 1)A$ for violet and $\delta_r = (n_r - 1)A$ for red. Subtracting, the angular dispersion between the extreme colours is $\delta_v - \delta_r = (n_v - n_r)A$. Since $n_v$ exceeds $n_r$ for ordinary glass, this difference is positive: violet is deviated more than red, and the intermediate colours fall in order between them. Thus the same geometry that gave $\delta = (n-1)A$ also predicts, without any new physics, the ordered spread of colours — the spectrum — that a prism casts from a beam of white light.
⚠️ JEE trap: A common trap is to assume that minimum deviation corresponds to normal incidence ($i = 0$); in fact it occurs when the passage is symmetric, $i = e$ with $r_1 = r_2 = A/2$, and the internal ray runs parallel to the base — not when the ray enters perpendicular to a face. Students also confuse the refracting angle $A$ of the prism with the angle of incidence $i$; they are independent quantities. Finally, do not believe $\delta$ always increases with $i$: the deviation first decreases to $D_m$, then increases, so a single $\delta$ (except $D_m$) is produced by two different incidence angles. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A prism with refracting angle $A = 60^\circ$ is made of crown glass. For yellow sodium light the angle of minimum deviation is measured to be $D_m = 39^\circ$. For a separate thin prism of the same glass with refracting angle $A' = 5^\circ$, the refractive indices are $n_v = 1.532$ for violet and $n_r = 1.514$ for red.
TARGET Find (a) the refractive index of the crown glass for yellow light, and (b) the angular dispersion produced by the thin $5^\circ$ prism between violet and red.
STRATEGY For part (a) use the prism formula $n_{21} = \sin[(A + D_m)/2]/\sin(A/2)$, valid at minimum deviation where $i = e$ and $r_1 = r_2 = A/2$. For part (b) use the thin-prism deviation $\delta = (n - 1)A'$ for each colour and take the difference to get the angular dispersion $(n_v - n_r)A'$.
EXECUTE Part (a): $\dfrac{A + D_m}{2} = \dfrac{60^\circ + 39^\circ}{2} = 49.5^\circ$ and $\dfrac{A}{2} = 30^\circ$. Then $n = \dfrac{\sin 49.5^\circ}{\sin 30^\circ} = \dfrac{0.7604}{0.5} = 1.52$. Part (b): the angular dispersion is $\delta_v - \delta_r = (n_v - n_r)A' = (1.532 - 1.514)\times 5^\circ = 0.018 \times 5^\circ = 0.09^\circ$. Equivalently, violet deviates by $\delta_v = (0.532)(5^\circ) = 2.66^\circ$ and red by $\delta_r = (0.514)(5^\circ) = 2.57^\circ$, whose difference is $0.09^\circ$.
REFLECT The yellow-light index $n \approx 1.52$ matches the tabulated value for crown glass, a good consistency check. The dispersion $0.09^\circ$ is far smaller than the mean deviation, confirming that thin prisms deviate strongly relative to how little they spread colour — the reason a single thin prism gives only a faint spectrum, and why designers combine glasses to control dispersion independently of deviation.

Source: JEE-pattern (NCERT Ch 9)

✍️ Worked Examples Polya 5-move · JEE tier

WE1 · The Cartesian Sign Convention · JEE Main 🔉⇢

SITUATION An object is placed $20\ \text{cm}$ in front of a concave mirror whose radius of curvature has magnitude $30\ \text{cm}$. Only the assignment of signs is required here, not the final image position.
TARGET State the correct signed values of the object distance $u$, the radius of curvature $R$, and the focal length $f$ under the Cartesian sign convention, and explain each sign.
STRATEGY Anchor the origin at the pole, take the incident light travelling toward the mirror as the positive direction, and translate each physical distance into a signed coordinate. Distances against the incident light are negative; the focus and centre of curvature of a concave mirror lie on that same negative side.
EXECUTE The object lies in front of the mirror, reached from the pole by moving against the incident light, so $u=-20\ \text{cm}$. The centre of curvature of a concave mirror also lies in front, giving $R=-30\ \text{cm}$. Using $f=R/2$ we get $f=\frac{-30}{2}\ \text{cm}=-15\ \text{cm}$. All three are negative because every one of these points sits on the incident-light side of the pole.
REFLECT The signs were forced entirely by geometry, not chosen. Since $f\lt 0$ confirms a concave mirror and $u\lt 0$ confirms a real object, any later solution using the mirror equation will yield an image distance whose own sign we simply interpret rather than impose.

Source: JEE-pattern (NCERT Ch 9)

WE2 · Reflection by Spherical Mirrors & Focal Length · JEE Main 🔉⇢

SITUATION A concave mirror is part of a sphere of radius $24\ \text{cm}$, and a narrow parallel beam of light travelling close to the principal axis is incident on it.
TARGET Find the signed focal length of the mirror and state where a parallel paraxial beam is brought to a focus relative to the pole.
STRATEGY Use the paraxial relation $f=R/2$ together with the Cartesian sign convention. For a concave mirror the centre of curvature and focus lie in front of the reflecting surface, on the incident-light side of the pole, so $R$ and hence $f$ are negative.
EXECUTE The centre of curvature of a concave mirror lies in front of it, giving $R=-24\ \text{cm}$. Then $f=\dfrac{R}{2}=\dfrac{-24}{2}\ \text{cm}=-12\ \text{cm}$. The negative sign places the focus $12\ \text{cm}$ in front of the mirror, on the same side as the incoming light, where the reflected paraxial rays actually converge to a real focus.
REFLECT The magnitude, $12\ \text{cm}$, is simply half of $24\ \text{cm}$, and the sign is fixed by geometry, not chosen. Since $f\lt 0$ confirms a converging concave mirror, a wide beam on the same mirror would instead show spherical aberration because those outer rays are no longer paraxial.

Source: JEE-pattern (NCERT Ch 9)

WE3 · The Mirror Equation & Magnification · JEE Main 🔉⇢

SITUATION An object is placed $30\ \text{cm}$ in front of a concave mirror of focal length $20\ \text{cm}$.
TARGET Find the image distance $v$ and the linear magnification $m$, and state the position, nature, and orientation of the image.
STRATEGY Assign signs with the Cartesian convention: for a concave mirror $f$ is negative and a real object gives $u$ negative. Substitute into $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, solve for $v$, then compute $m=-\frac{v}{u}$ and read the signs.
EXECUTE Here $u=-30\ \text{cm}$ and $f=-20\ \text{cm}$. From $\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{-20}-\dfrac{1}{-30}=-\dfrac{3}{60}+\dfrac{2}{60}=-\dfrac{1}{60}$, so $v=-60\ \text{cm}$. Then $m=-\dfrac{v}{u}=-\dfrac{-60}{-30}=-2$. The negative $v$ places the image $60\ \text{cm}$ in front of the mirror.
REFLECT Since $v\lt 0$ the image is real and lies on the object side, and since $m=-2$ it is inverted and twice the object's height, hence magnified. The object sits between the focus and the centre of curvature, so a real, inverted, magnified image beyond the centre of curvature is exactly what the geometry predicts.

Source: JEE-pattern (NCERT Ch 9)

WE4 · Refraction at a Spherical Surface · JEE Main 🔉⇢

SITUATION A point source in air ($n_1=1$) faces a convex spherical glass surface of refractive index $n_2=1.5$ and radius of curvature $20\text{ cm}$, its centre of curvature lying on the glass side. The source is $100\text{ cm}$ in front of the surface.
TARGET Find the position of the image formed inside the glass, and state whether it is real or virtual.
STRATEGY Use the single-surface relation $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$. Apply the Cartesian sign convention: the object is against the incident-light direction so $u\lt 0$, and the centre of curvature lies on the outgoing side so $R\gt 0$. Solve algebraically for $v$, then interpret its sign.
EXECUTE With $u=-100\text{ cm}$, $R=+20\text{ cm}$, $n_1=1$, $n_2=1.5$: $\frac{1.5}{v}-\frac{1}{-100}=\frac{1.5-1}{20}$, i.e. $\frac{1.5}{v}+\frac{1}{100}=\frac{0.5}{20}=0.025$. So $\frac{1.5}{v}=0.025-0.01=0.015\text{ cm}^{-1}$, giving $v=\frac{1.5}{0.015}=+100\text{ cm}$.
REFLECT Since $v=+100\text{ cm}\gt 0$, the image lies in the direction of the incident light, $100\text{ cm}$ inside the glass, and is real. The positive value confirms the refracted rays actually converge; had a sign slip produced $v\lt 0$ we would have wrongly reported a virtual image on the object side.

Source: JEE-pattern (NCERT Ch 9)

WE5 · The Lens Maker's Formula · JEE Main 🔉⇢

SITUATION A double convex lens is ground with faces of radii of curvature $10\text{ cm}$ and $15\text{ cm}$. When used in air its focal length is measured to be $12\text{ cm}$.
TARGET Determine the refractive index of the glass from which the lens is made.
STRATEGY Use the lens maker's formula $\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ with air as the surrounding medium, so $n_{21}=n$. Apply the Cartesian sign convention: for a biconvex lens the first radius is positive and the second negative, so $R_1=+10\text{ cm}$ and $R_2=-15\text{ cm}$. Solve algebraically for $n$.
EXECUTE Substitute $f=+12\text{ cm}$, $R_1=+10\text{ cm}$, $R_2=-15\text{ cm}$: $\frac{1}{12}=(n-1)\left(\frac{1}{10}-\frac{1}{-15}\right)=(n-1)\left(\frac{1}{10}+\frac{1}{15}\right)$. The bracket is $\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}$. Hence $\frac{1}{12}=(n-1)\cdot\frac{1}{6}$, so $n-1=\frac{6}{12}=0.5$, giving $n=1.5$.
REFLECT The result $n=1.5$ is the typical refractive index of crown glass, a reassuring check. Notice the sign of $R_2$ was decisive: had we wrongly taken $R_2=+15\text{ cm}$ the bracket would have been tiny and the deduced index absurdly large, a classic sign-convention error in lens maker problems.

Source: JEE-pattern (NCERT Ch 9)

WE6 · Power of a Lens & Combination of Lenses · JEE Main 🔉⇢

SITUATION A convex lens of focal length $30\text{ cm}$ is placed in contact with a concave lens of focal length $20\text{ cm}$, their principal axes coincident and their thickness negligible.
TARGET Find the effective focal length and power of the combination, and state whether the system is converging or diverging.
STRATEGY Apply the Cartesian sign convention: the convex lens has $f_1=+30\text{ cm}$ and the concave lens has $f_2=-20\text{ cm}$. Use the combination rule $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$, or equivalently add the powers $P=P_1+P_2$ in dioptres. Interpret the sign of the result to classify the system.
EXECUTE $\frac{1}{f}=\frac{1}{30}+\frac{1}{-20}=\frac{2}{60}-\frac{3}{60}=-\frac{1}{60}\text{ cm}^{-1}$, so $f=-60\text{ cm}$. In dioptres, $P_1=\frac{1}{0.30}=+3.33\,\text{D}$ and $P_2=\frac{1}{-0.20}=-5.0\,\text{D}$, giving $P=3.33-5.0=-1.67\,\text{D}$, consistent with $P=\frac{1}{-0.60\text{ m}}$.
REFLECT Both methods agree: $f=-60\text{ cm}$ and $P\approx-1.67\,\text{D}$. Because $f\lt 0$ and $P\lt 0$, the combination is a diverging system overall; the concave lens dominates. Note we added powers, not focal lengths, and preserved the negative sign of the concave lens throughout.

Source: JEE-pattern (NCERT Ch 9)

WE7 · The Simple Microscope (Magnifier) · JEE Main 🔉⇢

SITUATION A jeweller wants a simple microscope, a single converging lens held close to the eye, that gives an angular magnification of $5$ when the final virtual image is formed at the near point of a normal eye.
TARGET Find the required focal length $f$ of the converging lens, and state the magnifying power the same lens would give when used by a relaxed eye with the image at infinity.
STRATEGY Use the near-point magnifying power $m=1+\dfrac{D}{f}$ with $D=25$ cm to solve for $f$. Then substitute that focal length into the relaxed-eye relation $m'=\dfrac{D}{f}$ to obtain the second magnification.
EXECUTE Near point: $5=1+\dfrac{25}{f}\Rightarrow \dfrac{25}{f}=4\Rightarrow f=\dfrac{25}{4}=6.25$ cm. Relaxed eye with the same lens: $m'=\dfrac{D}{f}=\dfrac{25}{6.25}=4$. Thus $f=6.25$ cm and the relaxed-eye magnifying power is $4$, exactly one less than the near-point value.
REFLECT The relaxed-eye magnification $4$ is one unit below the near-point value $5$, matching $m_{near}-m_{infinity}=1$. Because $D/f\gt 1$ here, the difference is not negligible, but for a high-power lens where $D/f\gg 1$ the two settings would give almost the same magnification.

Source: JEE-pattern (NCERT Ch 9)

WE8 · The Compound Microscope · JEE Main 🔉⇢

SITUATION A compound microscope has an objective of focal length $f_o=1.25$ cm and an eyepiece of focal length $f_e=5.0$ cm. The tube length, the separation between the second focal point of the objective and the first focal point of the eyepiece, is $L=15$ cm, and the observer has a normal near point $D=25$ cm.
TARGET Find the total magnifying power when the final image is formed (a) at the near point and (b) at infinity for a relaxed eye.
STRATEGY Compute the objective magnification $m_o=\dfrac{L}{f_o}$. For the near point multiply by $m_e=1+\dfrac{D}{f_e}$; for infinity multiply by $m_e=\dfrac{D}{f_e}$.
EXECUTE Objective: $m_o=\dfrac{15}{1.25}=12$. Near point: $m_e=1+\dfrac{25}{5.0}=6$, so $m=12\times 6=72$. Infinity: $m_e=\dfrac{25}{5.0}=5$, so $m=12\times 5=60$.
REFLECT The near-point magnification $72$ exceeds the relaxed value $60$, as expected since $1+D/f_e\gt D/f_e$. The relaxed setting trades a modest loss of magnification for comfortable viewing, and both results confirm that the short objective focal length dominates the magnifying power.

Source: JEE-pattern (NCERT Ch 9)

WE9 · The Refracting & Reflecting Telescope · JEE Main 🔉⇢

SITUATION A small astronomical refracting telescope has an objective lens of focal length $f_o=144$ cm and an eyepiece of focal length $f_e=6.0$ cm, and it is used in normal adjustment to view distant stars with a relaxed eye.
TARGET Find the magnifying power of the telescope and the separation between the objective and the eyepiece.
STRATEGY In normal adjustment the final image is at infinity, so the magnifying power is $m=\dfrac{f_o}{f_e}$ and the tube length equals $f_o+f_e$. Substitute the given focal lengths into each relation.
EXECUTE Magnifying power: $m=\dfrac{f_o}{f_e}=\dfrac{144}{6.0}=24$. Tube length, the separation between the lenses: $f_o+f_e=144+6.0=150$ cm.
REFLECT A magnifying power of $24$ from a $150$ cm tube is typical of a small refractor. Note the objective focal length is far larger than the eyepiece focal length, giving $m\gt 1$; a larger objective diameter would improve light-gathering and resolving power but would leave this magnification unchanged.

Source: JEE-pattern (NCERT Ch 9)

WE10 · Concave mirror: real image of a candle · JEE Main 🔉⇢

SITUATION A candle $2.5$ cm tall is placed $27$ cm in front of a concave mirror of radius of curvature $36$ cm.
TARGET Find the image distance, its nature and its size, and state how a screen must move if the candle is brought closer.
STRATEGY Use $f = R/2$ with Cartesian signs (concave mirror $f$ negative), then the mirror equation $1/v + 1/u = 1/f$ and magnification $m = -v/u$.
EXECUTE $f = -36/2 = -18$ cm, $u = -27$ cm. From $1/v = 1/f - 1/u = -1/18 + 1/27 = (-3+2)/54 = -1/54$, so $v = -54$ cm. Magnification $m = -v/u = -(-54)/(-27) = -2$, image height $= -2 \times 2.5 = -5$ cm. Boxed: $v = -54$ cm, real, inverted, $5$ cm tall.
REFLECT The object lies beyond the centre of curvature ($|u| \gt R = 36$ would give a diminished image; here $27$ cm is between $f$ and $2f$... check: $2f = 36$, so $27$ is between $f=18$ and $2f=36$), giving a magnified real image beyond $2f$. Moving the candle toward $f$ pushes the image farther away, so the screen must move away from the mirror.

Source: JEE-pattern (NCERT Ch 9)

WE11 · Convex mirror: virtual diminished image · JEE Main 🔉⇢

SITUATION An object $3.0$ cm tall stands $20$ cm from a convex mirror of focal length $15$ cm.
TARGET Locate the image, and give its magnification, size and nature.
STRATEGY Convex mirror has $f$ positive. Apply $1/v + 1/u = 1/f$ and $m = -v/u$.
EXECUTE $f = +15$ cm, $u = -20$ cm. $1/v = 1/f - 1/u = 1/15 + 1/20 = (4+3)/60 = 7/60$, so $v = 60/7 \approx 8.57$ cm. $m = -v/u = -(8.57)/(-20) = +0.43$, image height $= 0.43 \times 3.0 \approx 1.3$ cm. Boxed: $v \approx +8.6$ cm behind the mirror, virtual, erect, $\approx 1.3$ cm.
REFLECT A convex mirror always gives a virtual, erect, diminished image located between pole and focus. As the object recedes, the image shrinks toward $F$; it never becomes real. Common error: forgetting $f \gt 0$ for convex mirrors.

Source: JEE-pattern (NCERT Ch 9)

WE12 · Concave mirror: object inside focus (magnifier) · JEE Main 🔉⇢

SITUATION An object is placed $10$ cm in front of a concave mirror of radius of curvature $30$ cm.
TARGET Find the image position, magnification and nature.
STRATEGY With $f = R/2 = 15$ cm (negative for concave), the object at $|u| \lt |f|$ should give a virtual erect magnified image. Use $1/v + 1/u = 1/f$.
EXECUTE $f = -15$ cm, $u = -10$ cm. $1/v = 1/f - 1/u = -1/15 + 1/10 = (-2+3)/30 = 1/30$, so $v = +30$ cm. $m = -v/u = -(30)/(-10) = +3$. Boxed: $v = +30$ cm behind the mirror, virtual, erect, magnified $3\times$.
REFLECT This is the shaving/makeup-mirror regime: object between pole and focus yields an enlarged upright virtual image. The positive $v$ correctly signals the image is behind the mirror.

Source: JEE-pattern (NCERT Ch 9)

WE13 · Speed of image in a car side-view mirror · JEE Advanced 🔉⇢

SITUATION A jogger runs at $5$ m/s directly toward a convex side-view mirror of radius $2$ m. At one instant the jogger is $9$ m from the mirror.
TARGET Find the instantaneous speed of the image at that instant.
STRATEGY Longitudinal image velocity magnitude $= m^2 \times$ object speed, where $m = -v/u$. First get $v$ from $v = fu/(u-f)$, then $m$, then multiply. This uses $\mathrm{d}v/\mathrm{d}u = (f/(u-f))^2$.
EXECUTE $f = R/2 = +1$ m, $u = -9$ m. $v = fu/(u-f) = (1)(-9)/(-9-1) = -9/-10 = 0.9$ m. $m = -v/u = -(0.9)/(-9) = 0.1$. Image speed $= m^2 \times 5 = (0.1)^2 \times 5 = 0.01 \times 5 = 0.05$ m/s. Boxed: $0.05$ m/s.
REFLECT Although the object moves at a steady $5$ m/s, the image crawls at only $0.05$ m/s and speeds up sharply as the jogger nears the mirror (since $m$ grows). This is why 'objects in mirror are closer than they appear'. A discrete $1$-s step overestimates it because $m$ changes over the interval.

Source: JEE-pattern (NCERT Ch 9)

WE14 · Focal length from R and image location rule · JEE Main 🔉⇢

SITUATION A concave mirror has radius of curvature $20$ cm; an object is placed $30$ cm from it.
TARGET Find the focal length and the image, and verify the object-beyond-$2f$ rule.
STRATEGY Use $f = R/2$ (paraxial result derived from $\tan\theta \approx \theta$), then the mirror equation.
EXECUTE $f = -20/2 = -10$ cm, $u = -30$ cm. $1/v = 1/f - 1/u = -1/10 + 1/30 = (-3+1)/30 = -2/30 = -1/15$, so $v = -15$ cm. $m = -v/u = -(-15)/(-30) = -0.5$. Boxed: $f = -10$ cm, $v = -15$ cm, real, inverted, diminished.
REFLECT The object at $30$ cm lies beyond $2f = 20$ cm, so the image forms between $f$ and $2f$ and is diminished, exactly as the ray-diagram rule predicts. The $f = R/2$ relation holds only for paraxial rays; wide-aperture mirrors show spherical aberration.

Source: JEE-pattern (NCERT Ch 9)

WE15 · Snell's law at a plane air-glass interface · JEE Main 🔉⇢

SITUATION Light in air strikes a flat glass surface (refractive index $1.5$) at an angle of incidence $60^\circ$.
TARGET Find the angle of refraction inside the glass.
STRATEGY Apply Snell's law $n_1 \sin i = n_2 \sin r$; since glass is optically denser ($n_{21} \gt 1$), the ray bends toward the normal so $r \lt i$.
EXECUTE $1 \times \sin 60^\circ = 1.5 \times \sin r$. $\sin r = 0.8660/1.5 = 0.5774$, so $r = \sin^{-1}(0.5774) = 35.3^\circ$. Boxed: $r \approx 35.3^\circ$.
REFLECT As expected for a rarer-to-denser transition, $r \lt i$ (bent toward the normal). A quick sanity check: $\sin r$ must stay at most $1$, which it does. Reversing the ray (glass to air at $35.3^\circ$) would return exactly $60^\circ$.

Source: JEE-pattern (NCERT Ch 9)

WE16 · Apparent depth of a needle in water · JEE Main 🔉⇢

SITUATION A tank holds water ($n = 1.33$) to a real depth of $12.5$ cm; a needle lies at the bottom, viewed from nearly overhead.
TARGET Find the apparent depth and how much the needle appears raised.
STRATEGY For near-normal viewing, apparent depth $=$ real depth $/ n$, a special case of refraction from denser to rarer medium.
EXECUTE Apparent depth $= 12.5/1.33 = 9.40$ cm. Apparent raise $= 12.5 - 9.40 = 3.1$ cm. Boxed: apparent depth $\approx 9.4$ cm, raised by $\approx 3.1$ cm.
REFLECT The bottom looks shallower than it is, a routine denser-to-rarer effect. If a liquid of higher $n$ replaced the water, the apparent depth would shrink further, so the microscope would need to be lowered to refocus.

Source: JEE-pattern (NCERT Ch 9)

WE17 · Lateral shift through a glass slab · JEE Advanced 🔉⇢

SITUATION A ray hits a parallel-sided glass slab (thickness $6.0$ cm, $n = 1.5$) at an angle of incidence $60^\circ$.
TARGET Find the perpendicular lateral displacement of the emergent ray.
STRATEGY The emergent ray is parallel to the incident ray (since $r_2 = i_1$ for a slab) but laterally shifted by $d = t\,\sin(i-r)/\cos r$. First get $r$ from Snell's law.
EXECUTE $\sin r = \sin 60^\circ/1.5 = 0.8660/1.5 = 0.5774$, so $r = 35.26^\circ$. Then $d = t\,\sin(i-r)/\cos r = 6.0 \times \sin(24.74^\circ)/\cos(35.26^\circ) = 6.0 \times 0.4186/0.8165 = 6.0 \times 0.5127 = 3.08$ cm. Boxed: $d \approx 3.1$ cm.
REFLECT No net deviation, only a sideways slide, confirming a slab cannot form a deviated beam. The shift grows with thickness and incidence angle and vanishes at normal incidence, a useful limiting check.

Source: JEE-pattern (NCERT Ch 9)

WE18 · Critical angle for a water-air interface · JEE Main 🔉⇢

SITUATION Light travels inside water ($n = 1.33$) and meets the water-air boundary.
TARGET Find the critical angle for total internal reflection.
STRATEGY Total internal reflection needs denser-to-rarer travel; at the critical angle the refraction angle is $90^\circ$, giving $\sin i_c = 1/n$.
EXECUTE $\sin i_c = 1/1.33 = 0.752$, so $i_c = \sin^{-1}(0.752) = 48.8^\circ$. Boxed: $i_c \approx 48.8^\circ$.
REFLECT This matches the tabulated $48.75^\circ$ for water. For any incidence exceeding $i_c$ no light escapes, which is why a diver sees the outside world compressed into a bright circular 'Snell's window'. Denser media (diamond, $n=2.42$) have far smaller $i_c$.

Source: JEE-pattern (NCERT Ch 9)

WE19 · Acceptance angle of an optical fibre · JEE Advanced 🔉⇢

SITUATION A glass-fibre light pipe has core index $1.68$ and cladding index $1.44$.
TARGET Find the maximum angle a ray at the end face may make with the fibre axis so it is totally internally reflected at the core-cladding wall.
STRATEGY TIR at the wall needs the wall-incidence angle to be at least $i_c$ where $\sin i_c = n_{clad}/n_{core}$. A ray entering at axis-angle $r$ inside hits the wall at $90^\circ - r$; the entry face refraction links the external angle $i$ to $r$ by $\sin i = n_{core}\sin r$.
EXECUTE $\sin i_c = 1.44/1.68 = 0.857$, so $i_c = 59.0^\circ$ and $\cos i_c = \sqrt{1 - 0.857^2} = \sqrt{0.266} = 0.516$. Wall TIR requires axis-angle inside $r \le 90^\circ - i_c = 31.0^\circ$. Max entry angle: $\sin i_{max} = n_{core}\sin r_{max} = 1.68 \times \sin 31.0^\circ = 1.68 \times 0.515 = 0.865$, so $i_{max} = 59.9^\circ$. Boxed: rays within about $60^\circ$ of the axis are guided.
REFLECT The wide $\pm 60^\circ$ acceptance cone is why fibres capture light easily even when bent. Removing the cladding ($n_{clad}=1$) would make $i_c$ smaller and the acceptance cone essentially the full hemisphere, but real fibres use cladding to keep TIR clean at the wall.

Source: JEE-pattern (NCERT Ch 9)

WE20 · Refraction at a single spherical surface · JEE Main 🔉⇢

SITUATION A point source in air is $100$ cm from a convex spherical glass surface of radius $20$ cm; glass has $n = 1.5$.
TARGET Find where the image forms inside the glass.
STRATEGY Use the single-surface refraction relation $n_2/v - n_1/u = (n_2 - n_1)/R$ with Cartesian signs ($u$ negative, $R$ positive for a surface convex toward the source).
EXECUTE $n_1 = 1$, $n_2 = 1.5$, $u = -100$ cm, $R = +20$ cm. $1.5/v - 1/(-100) = (1.5-1)/20 = 0.5/20 = 0.025$. So $1.5/v = 0.025 - 0.010 = 0.015$, giving $v = 1.5/0.015 = +100$ cm. Boxed: $v = +100$ cm (inside the glass).
REFLECT The positive $v$ places a real image $100$ cm into the glass, along the incident-light direction. Note this formula uses $n$ on each side directly, not $(n-1)$, a frequent slip when students confuse it with the lens-maker's formula.

Source: JEE-pattern (NCERT Ch 9)

WE21 · Lens maker's formula: radius for a symmetric lens · JEE Main 🔉⇢

SITUATION A double-convex lens with both faces of equal radius is ground from glass of index $1.55$ to give a focal length of $20$ cm.
TARGET Find the required radius of curvature.
STRATEGY Apply the lens maker's formula $1/f = (n-1)(1/R_1 - 1/R_2)$ with $R_1 = +R$, $R_2 = -R$ for a symmetric biconvex lens.
EXECUTE $1/f = (n-1)(1/R + 1/R) = (n-1)(2/R)$. So $1/20 = (0.55)(2/R) = 1.1/R$, giving $R = 1.1 \times 20 = 22$ cm. Boxed: $R = 22$ cm.
REFLECT Both surfaces bend light the same way, so their curvatures add, halving the radius needed compared with a plano-convex lens of the same $f$. Higher index $n$ would allow a flatter (larger $R$) lens for the same focal length.

Source: JEE-pattern (NCERT Ch 9)

WE22 · Lens maker's formula: finding the refractive index · JEE Main 🔉⇢

SITUATION A double-convex lens has radii $10$ cm and $15$ cm and a measured focal length of $12$ cm in air.
TARGET Find the refractive index of the glass.
STRATEGY Use $1/f = (n-1)(1/R_1 - 1/R_2)$ with $R_1 = +10$ cm, $R_2 = -15$ cm.
EXECUTE $1/12 = (n-1)(1/10 - 1/(-15)) = (n-1)(1/10 + 1/15) = (n-1)(3+2)/30 = (n-1)(5/30) = (n-1)/6$. So $n - 1 = 6/12 = 0.5$, giving $n = 1.5$. Boxed: $n = 1.5$.
REFLECT A typical crown-glass value. The sign discipline matters: writing $R_2 = +15$ would wrongly cancel the terms and give a nonsensical index. Both radii contribute constructively because the lens is convex on both sides.

Source: JEE-pattern (NCERT Ch 9)

WE23 · Focal length of a lens immersed in water · JEE Advanced 🔉⇢

SITUATION A glass lens ($n = 1.5$) has focal length $20$ cm in air. It is submerged in water ($n = 1.33$).
TARGET Find its new focal length in water.
STRATEGY The lens maker's formula scales with $(n_{lens}/n_{medium} - 1)$. Form the ratio of $1/f$ in the two media, cancelling the geometric factor $(1/R_1 - 1/R_2)$.
EXECUTE In air: $1/f_a = (1.5 - 1)\,S = 0.5\,S$, so $S = 1/(20 \times 0.5) = 0.1$ cm$^{-1}$. In water: $1/f_w = (1.5/1.33 - 1)\,S = (1.1278 - 1)(0.1) = 0.12782 \times 0.1 = 0.012782$. Thus $f_w = 1/0.012782 = 78.2$ cm. Boxed: $f_w \approx 78.2$ cm.
REFLECT The focal length nearly quadruples because the glass-to-water contrast is far weaker than glass-to-air, so the lens bends light much less. If the medium index equalled the lens index, $1/f$ would vanish and the lens would 'disappear' optically.

Source: JEE-pattern (NCERT Ch 9)

WE24 · Convex lens intercepting a converging beam · JEE Advanced 🔉⇢

SITUATION A beam already converging toward a point $P$ meets a convex lens of focal length $20$ cm placed $12$ cm before $P$; it also considers a concave lens of focal length $16$ cm in the same spot.
TARGET Find where the beam finally converges for each lens.
STRATEGY The point $P$ acts as a virtual object at $u = +12$ cm (measured along the light direction). Apply $1/v - 1/u = 1/f$ for each lens.
EXECUTE Convex, $f = +20$: $1/v = 1/f + 1/u = 1/20 + 1/12 = (3+5)/60 = 8/60 = 2/15$, so $v = 7.5$ cm beyond the lens. Concave, $f = -16$: $1/v = -1/16 + 1/12 = (-3+4)/48 = 1/48$, so $v = 48$ cm beyond the lens. Boxed: convex $\to 7.5$ cm; concave $\to 48$ cm.
REFLECT A virtual object (converging incoming rays) is the key twist: $u$ is positive here. The convex lens converges the beam sooner ($7.5$ cm), the concave lens delays convergence to $48$ cm. Treating $P$ as a real object would give wrong signs.

Source: JEE-pattern (NCERT Ch 9)

WE25 · Concave lens image · JEE Main 🔉⇢

SITUATION An object $3.0$ cm tall is placed $14$ cm in front of a concave lens of focal length $21$ cm.
TARGET Describe the image (position, size, nature).
STRATEGY Use the thin-lens formula $1/v - 1/u = 1/f$ with $f$ negative for a diverging lens, then $m = v/u$.
EXECUTE $f = -21$ cm, $u = -14$ cm. $1/v = 1/f + 1/u = -1/21 - 1/14$. With LCD $42$: $-2/42 - 3/42 = -5/42$, so $v = -8.4$ cm. $m = v/u = (-8.4)/(-14) = +0.6$, image height $= 0.6 \times 3.0 = 1.8$ cm. Boxed: $v = -8.4$ cm, virtual, erect, $1.8$ cm.
REFLECT A concave lens always yields a virtual, erect, diminished image on the same side as the object, for any object position. Moving the object farther away pushes the image toward the focus while it stays virtual and shrinks.

Source: JEE-pattern (NCERT Ch 9)

WE26 · Two thin lenses in contact · JEE Main 🔉⇢

SITUATION A convex lens of focal length $30$ cm is placed in contact with a concave lens of focal length $20$ cm; lens thickness is ignored.
TARGET Find the equivalent focal length and state whether the system converges or diverges.
STRATEGY For lenses in contact, powers add: $1/f = 1/f_1 + 1/f_2$.
EXECUTE $1/f = 1/30 + 1/(-20) = (2 - 3)/60 = -1/60$, so $f = -60$ cm. Boxed: $f = -60$ cm, a diverging system.
REFLECT Even though the convex lens is 'stronger' geometrically, the shorter-focal-length concave lens ($20$ cm) has the larger magnitude of power, so the net system diverges. Checking with powers: $P = +3.33\,\mathrm{D} - 5\,\mathrm{D} = -1.67\,\mathrm{D}$, consistent with $f=-60$ cm.

Source: JEE-pattern (NCERT Ch 9)

WE27 · Image through a chain of three lenses · JEE Advanced 🔉⇢

SITUATION Three thin lenses lie along one axis: convex $f_1 = +10$ cm, then concave $f_2 = -10$ cm at $5$ cm, then convex $f_3 = +30$ cm. An object is $30$ cm to the left of the first lens.
TARGET Find the position of the final image.
STRATEGY Apply $1/v - 1/u = 1/f$ lens by lens; each image becomes the (possibly virtual) object for the next, shifting the reference by the inter-lens distance.
EXECUTE Lens 1: $1/v_1 = 1/10 + 1/(-30) = (3-1)/30 = 1/15$, so $v_1 = +15$ cm (right of L1). L2 is $5$ cm right, so this image is $15 - 5 = 10$ cm right of L2, a virtual object $u_2 = +10$. Lens 2: $1/v_2 = 1/(-10) + 1/10 = 0$, so $v_2 = \infty$. Lens 3 then receives a parallel beam ($u_3 = \infty$): $1/v_3 = 1/30 + 0$, so $v_3 = +30$ cm. Boxed: final image $30$ cm to the right of the third lens.
REFLECT The middle lens exactly collimates the beam ($v_2 = \infty$), a neat intermediate that makes the last step trivial. The recurring pitfall is mis-signing the virtual object for lens 2; here it is positive because the rays were still converging.

Source: JEE-pattern (NCERT Ch 9)

WE28 · Power of a lens combination · JEE Main 🔉⇢

SITUATION A convex lens of focal length $40$ cm is combined in contact with a concave lens of focal length $25$ cm.
TARGET Find the net power and equivalent focal length.
STRATEGY Convert each focal length (in metres) to power $P = 1/f$, then add algebraically: $P = P_1 + P_2$.
EXECUTE $P_1 = 1/0.40 = +2.5$ D, $P_2 = 1/(-0.25) = -4.0$ D. $P = 2.5 - 4.0 = -1.5$ D. Then $f = 1/P = 1/(-1.5) = -0.667$ m $= -66.7$ cm. Boxed: $P = -1.5$ D, $f \approx -66.7$ cm (diverging).
REFLECT Powers add as signed quantities; the concave lens dominates so the pair is net diverging. Working in dioptres avoids the reciprocal-of-a-sum arithmetic that trips people up when adding focal lengths directly.

Source: JEE-pattern (NCERT Ch 9)

WE29 · Refractive index of a prism from minimum deviation · JEE Main 🔉⇢

SITUATION A prism of refracting angle $60^\circ$ gives a minimum deviation of $40^\circ$ for a parallel beam.
TARGET Find the refractive index of the prism material.
STRATEGY At minimum deviation the ray is symmetric ($r_1 = r_2 = A/2$), giving $n = \sin[(A + D_m)/2]/\sin(A/2)$.
EXECUTE $(A + D_m)/2 = (60 + 40)/2 = 50^\circ$ and $A/2 = 30^\circ$. $n = \sin 50^\circ/\sin 30^\circ = 0.766/0.5 = 1.532$. Boxed: $n \approx 1.53$.
REFLECT The value sits between crown and flint glass, as expected. The symmetry condition at $D_m$ is what makes the single-formula measurement possible; away from $D_m$ two incidence angles give the same deviation.

Source: JEE-pattern (NCERT Ch 9)

WE30 · Prism deviation when immersed in water · JEE Advanced 🔉⇢

SITUATION The $60^\circ$ prism of index $1.532$ from the previous problem is now placed in water ($n = 1.33$).
TARGET Find the new angle of minimum deviation.
STRATEGY Replace $n$ by the relative index $n_{rel} = n_{prism}/n_{water}$ in $n_{rel} = \sin[(A + D_m')/2]/\sin(A/2)$ and solve for $D_m'$.
EXECUTE $n_{rel} = 1.532/1.33 = 1.152$. Then $\sin[(A + D_m')/2] = n_{rel}\sin(A/2) = 1.152 \times \sin 30^\circ = 1.152 \times 0.5 = 0.576$. So $(A + D_m')/2 = \sin^{-1}(0.576) = 35.2^\circ$, giving $A + D_m' = 70.3^\circ$ and $D_m' = 70.3 - 60 = 10.3^\circ$. Boxed: $D_m' \approx 10^\circ$.
REFLECT Deviation collapses from $40^\circ$ to about $10^\circ$ because the glass-water contrast is far smaller than glass-air. This is the same reason a glass object is nearly invisible in a matched liquid, and it warns that 'the' refractive index of a prism is only defined relative to its surroundings.

Source: JEE-pattern (NCERT Ch 9)

WE31 · Incidence angle for grazing total internal reflection in a prism · JEE Advanced 🔉⇢

SITUATION A prism of refracting angle $60^\circ$ is made of glass with index $1.524$.
TARGET Find the angle of incidence on the first face for which the ray just suffers total internal reflection at the second face.
STRATEGY 'Just TIR' at the second face means $r_2 = i_c$ where $\sin i_c = 1/n$. Use $r_1 + r_2 = A$ to get $r_1$, then Snell's law at the first face $\sin i = n\sin r_1$.
EXECUTE $\sin i_c = 1/1.524 = 0.6562$, so $i_c = 41.0^\circ$. Then $r_1 = A - r_2 = 60 - 41.0 = 19.0^\circ$. At the first face $\sin i = n\sin r_1 = 1.524 \times \sin 19.0^\circ = 1.524 \times 0.3256 = 0.496$, so $i = \sin^{-1}(0.496) = 29.75^\circ$. Boxed: $i \approx 29.75^\circ$.
REFLECT For incidence smaller than this, $r_2$ exceeds $i_c$ and the ray is trapped (TIR); for larger incidence it emerges. This threshold underlies the $45^\circ$ totally reflecting prisms used in binoculars, where $i_c \lt 45^\circ$ guarantees reflection.

Source: JEE-pattern (NCERT Ch 9)

WE32 · Simple microscope (magnifying glass) power · JEE Main 🔉⇢

SITUATION A converging lens of focal length $5.0$ cm is used as a simple magnifier; the near point is $D = 25$ cm.
TARGET Find the magnifying power when the image is at the near point and when it is at infinity.
STRATEGY For a simple microscope, image at near point gives $m = 1 + D/f$; image at infinity (relaxed eye) gives $m = D/f$.
EXECUTE Near point: $m = 1 + D/f = 1 + 25/5 = 1 + 5 = 6$. Infinity: $m = D/f = 25/5 = 5$. Boxed: $m = 6$ (near point), $m = 5$ (infinity).
REFLECT The near-point setting gives one extra unit of magnification but strains the eye; the infinity setting is more comfortable for only a small loss. The gain comes entirely from letting the eye view the object closer than $25$ cm at a large angle.

Source: JEE-pattern (NCERT Ch 9)

WE33 · Compound microscope: object position and magnifying power · JEE Advanced 🔉⇢

SITUATION A compound microscope has objective $f_o = 2.0$ cm and eyepiece $f_e = 6.25$ cm separated by $15$ cm; the final image is at the near point $D = 25$ cm.
TARGET Find how far the object sits from the objective and the total magnifying power.
STRATEGY Work backward: place the final image at $-25$ cm to find the eyepiece object $u_e$; the objective image distance is (separation $- |u_e|$); then find the objective object distance and combine $m = m_o \times m_e$.
EXECUTE Eyepiece: $1/u_e = 1/v_e - 1/f_e = 1/(-25) - 1/6.25 = -0.04 - 0.16 = -0.20$, so $u_e = -5.0$ cm. Objective image distance $v_o = 15 - 5 = 10$ cm. Objective: $1/u_o = 1/v_o - 1/f_o = 1/10 - 1/2 = 0.1 - 0.5 = -0.4$, so $u_o = -2.5$ cm. Magnifications: $m_o = v_o/u_o = 10/(-2.5) = -4$; $m_e = 1 + D/f_e = 1 + 25/6.25 = 5$. Total $|m| = 4 \times 5 = 20$. Boxed: object $2.5$ cm from objective, $m \approx 20$.
REFLECT The magnitudes multiply to give strong magnification from two modest lenses. Both focal lengths must be short: a short $f_o$ makes $m_o$ large, and a short $f_e$ makes $m_e$ large. The overall image is inverted, consistent with $m_o$ being negative.

Source: JEE-pattern (NCERT Ch 9)

WE34 · Astronomical telescope: magnifying power and tube length · JEE Main 🔉⇢

SITUATION A telescope has an objective of focal length $144$ cm and an eyepiece of focal length $6.0$ cm, in normal adjustment (final image at infinity).
TARGET Find the magnifying power and the separation between the two lenses.
STRATEGY In normal adjustment $m = f_o/f_e$, and the tube length equals $f_o + f_e$ (the objective's focal point coincides with the eyepiece's focal point).
EXECUTE $m = f_o/f_e = 144/6.0 = 24$. Tube length $= f_o + f_e = 144 + 6.0 = 150$ cm. Boxed: $m = 24$, separation $= 150$ cm.
REFLECT A telescope wants a long-focus objective and short-focus eyepiece, the opposite balance to a microscope's tube-length product. Because the object is effectively at infinity, magnifying power is purely the ratio of focal lengths, independent of object distance.

Source: JEE-pattern (NCERT Ch 9)

On the concept tabs

These worked examples are taught in full alongside their interactive scene:

📐 Formula Sheet Printable · every formula cited

Sign Convention & Spherical Mirrors

QuantityFormulaWhat it means / when to useSource
Focal length and radius of curvature 🔉⇢$f=\dfrac{R}{2}$Here $f$ is the focal length and $R$ the radius of curvature of a spherical mirror. Valid for paraxial rays; the focus lies midway between pole and centre of curvature.NCERT Class XII Physics, Ch. 9
Mirror equation 🔉⇢$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$Relates object distance $u$, image distance $v$ and focal length $f$, all measured from the pole using the Cartesian sign convention. Use it for both concave and convex mirrors.NCERT Class XII Physics, Ch. 9
Linear magnification (mirror) 🔉⇢$m=\dfrac{h'}{h}=-\dfrac{v}{u}$Here $h'$ is image height, $h$ object height, $v$ image distance and $u$ object distance. A negative $m$ means a real inverted image; a positive $m$ means a virtual erect image.NCERT Class XII Physics, Ch. 9
Image distance solved from mirror equation 🔉⇢$v=\dfrac{fu}{u-f}$A rearrangement of the mirror equation giving image distance $v$ directly from object distance $u$ and focal length $f$; convenient when tracking how the image moves as the object moves.NCERT Class XII Physics, Ch. 9

Refraction & Snell's Law

QuantityFormulaWhat it means / when to useSource
Snell's law 🔉⇢$n_{21}=\dfrac{\sin i}{\sin r}$Here $i$ is the angle of incidence, $r$ the angle of refraction and $n_{21}$ the refractive index of medium 2 with respect to medium 1. It is constant for a given pair of media and independent of the angle of incidence.NCERT Class XII Physics, Ch. 9
Refractive index from speeds of light 🔉⇢$n_{21}=\dfrac{v_1}{v_2}$Here $v_1$ and $v_2$ are the speeds of light in medium 1 and medium 2. Optical density is the ratio of speeds, so a larger $n_{21}$ means light travels slower in the second medium.NCERT Class XII Physics, Ch. 9
Reciprocal relation of refractive indices 🔉⇢$n_{12}=\dfrac{1}{n_{21}}$Here $n_{12}$ is the refractive index of medium 1 with respect to medium 2, which is the reciprocal of $n_{21}$. Use it when reversing the direction of light between the same pair of media.NCERT Class XII Physics, Ch. 9
Chain rule for three media 🔉⇢$n_{32}=n_{31}\times n_{12}$Here $n_{32}$, $n_{31}$ and $n_{12}$ are pairwise refractive indices. This multiplicative relation lets you combine indices across three media 1, 2 and 3 when a common reference medium is used.NCERT Class XII Physics, Ch. 9
Apparent depth 🔉⇢$h_1=\dfrac{h_2}{n}$Here $h_1$ is the apparent depth, $h_2$ the real depth and $n$ the refractive index of the medium. Valid for near-normal viewing; it explains why a tank bottom appears raised.NCERT Class XII Physics, Ch. 9
Lateral shift through a parallel slab 🔉⇢$d=\dfrac{t\,\sin(i-r)}{\cos r}$Here $t$ is slab thickness, $i$ the angle of incidence and $r$ the angle of refraction. The emergent ray stays parallel to the incident ray but is displaced sideways by $d$ with no net deviation.NCERT Class XII Physics, Ch. 9

Total Internal Reflection

QuantityFormulaWhat it means / when to useSource
Critical angle condition 🔉⇢$\sin i_c=\dfrac{1}{n}$Here $i_c$ is the critical angle and $n$ the refractive index of the denser medium with respect to the rarer one. For incidence angles greater than $i_c$ from denser to rarer medium, total internal reflection occurs.NCERT Class XII Physics, Ch. 9
Refractive index from critical angle 🔉⇢$n=\dfrac{1}{\sin i_c}$Here $n$ is the refractive index of the denser medium with respect to the rarer medium and $i_c$ the critical angle. Use it to compute the index when the critical angle for a pair of media is known.NCERT Class XII Physics, Ch. 9

Refraction at a Spherical Surface & Lens Maker's Formula

QuantityFormulaWhat it means / when to useSource
Refraction at a single spherical surface 🔉⇢$\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$Here $n_1$ and $n_2$ are the refractive indices on the incidence and refraction sides, $u$ and $v$ the object and image distances, and $R$ the radius of curvature. It holds for any single curved refracting surface.NCERT Class XII Physics, Ch. 9
Lens maker's formula (general) 🔉⇢$\dfrac{1}{f}=\dfrac{(n_2-n_1)}{n_1}\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$Here $f$ is focal length, $n_1$ the surrounding medium index, $n_2$ the lens material index, and $R_1$, $R_2$ the radii of the two lens surfaces. Use it to design lenses of a desired focal length.NCERT Class XII Physics, Ch. 9
Lens maker's formula (lens in air) 🔉⇢$\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$Here $n$ is the refractive index of the lens material relative to air and $R_1$, $R_2$ the surface radii. This simplified form applies when the lens is placed in air, where the surrounding index is unity.NCERT Class XII Physics, Ch. 9

Thin Lens, Power & Combination of Lenses

QuantityFormulaWhat it means / when to useSource
Thin lens formula 🔉⇢$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$Here $u$ is object distance, $v$ image distance and $f$ focal length measured from the optical centre. Valid for both convex and concave lenses and for real as well as virtual images.NCERT Class XII Physics, Ch. 9
Linear magnification (lens) 🔉⇢$m=\dfrac{h'}{h}=\dfrac{v}{u}$Here $h'$ is image height, $h$ object height, $v$ image distance and $u$ object distance. A positive $m$ denotes an erect virtual image; a negative $m$ denotes an inverted real image.NCERT Class XII Physics, Ch. 9
Power of a lens 🔉⇢$P=\dfrac{1}{f}$Here $P$ is the power in dioptres and $f$ the focal length in metres, with $1\,\text{D}=1\,\text{m}^{-1}$. Power is positive for a converging lens and negative for a diverging lens.NCERT Class XII Physics, Ch. 9
Equivalent focal length of lenses in contact 🔉⇢$\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}+\dfrac{1}{f_3}+\dots$Here $f$ is the effective focal length of thin lenses of focal lengths $f_1$, $f_2$, $f_3$ placed in contact. The reciprocals add; it is derived by applying the lens formula successively.NCERT Class XII Physics, Ch. 9
Net power of lenses in contact 🔉⇢$P=P_1+P_2+P_3+\dots$Here $P$ is the net power of the combination and $P_1$, $P_2$, $P_3$ the individual powers. This is an algebraic sum, so convex powers are positive and concave powers are negative.NCERT Class XII Physics, Ch. 9
Total magnification of a lens combination 🔉⇢$m=m_1\,m_2\,m_3\dots$Here $m$ is the overall magnification and $m_1$, $m_2$, $m_3$ the magnifications of individual lenses. Because each image serves as the object for the next lens, the magnifications multiply.NCERT Class XII Physics, Ch. 9

Refraction Through a Prism

QuantityFormulaWhat it means / when to useSource
Angle of deviation 🔉⇢$\delta=i+e-A$Here $\delta$ is the deviation, $i$ the angle of incidence, $e$ the angle of emergence and $A$ the refracting angle of the prism. It shows deviation depends on incidence and is symmetric in $i$ and $e$.NCERT Class XII Physics, Ch. 9
Prism angle and refraction angles 🔉⇢$A=r_1+r_2$Here $A$ is the prism angle, $r_1$ the refraction angle at the first face and $r_2$ the incidence angle at the second face inside the prism. It follows from the geometry of the quadrilateral.NCERT Class XII Physics, Ch. 9
Refraction angle at minimum deviation 🔉⇢$r=\dfrac{A}{2}$Here $r$ is the common refraction angle at each face and $A$ the prism angle at minimum deviation, when $r_1=r_2$ and the ray inside the prism runs parallel to its base.NCERT Class XII Physics, Ch. 9
Incidence angle at minimum deviation 🔉⇢$i=\dfrac{A+D_m}{2}$Here $i$ is the angle of incidence, $A$ the prism angle and $D_m$ the angle of minimum deviation, for which $i=e$. It is obtained by combining the deviation and prism-angle relations.NCERT Class XII Physics, Ch. 9
Prism formula for refractive index 🔉⇢$n_{21}=\dfrac{\sin\!\left[\dfrac{A+D_m}{2}\right]}{\sin\!\left[\dfrac{A}{2}\right]}$Here $n_{21}$ is the refractive index of the prism relative to the surrounding medium, $A$ the prism angle and $D_m$ the minimum deviation. Measuring $A$ and $D_m$ gives the material's refractive index.NCERT Class XII Physics, Ch. 9
Thin (small-angle) prism deviation 🔉⇢$D_m=(n_{21}-1)A$Here $D_m$ is the minimum deviation, $n_{21}$ the refractive index and $A$ the small refracting angle. It shows thin prisms deviate light only slightly, in proportion to their angle.NCERT Class XII Physics, Ch. 9

Optical Instruments (Microscope & Telescope)

QuantityFormulaWhat it means / when to useSource
Simple microscope, image at near point 🔉⇢$m=1+\dfrac{D}{f}$Here $m$ is the magnifying power, $D\approx 25\,\text{cm}$ the least distance of distinct vision and $f$ the focal length. Use it when the virtual image is formed at the near point for maximum magnification.NCERT Class XII Physics, Ch. 9
Simple microscope, image at infinity 🔉⇢$m=\dfrac{D}{f}$Here $m$ is the angular magnification, $D$ the least distance of distinct vision and $f$ the focal length. This applies for relaxed-eye viewing with the image at infinity, giving one less than the near-point value.NCERT Class XII Physics, Ch. 9
Compound microscope, image at near point 🔉⇢$m=\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$Here $L$ is the tube length, $f_o$ the objective focal length, $f_e$ the eyepiece focal length and $D$ the near-point distance. It combines objective magnification with near-point eyepiece magnification.NCERT Class XII Physics, Ch. 9
Compound microscope, image at infinity 🔉⇢$m=\dfrac{L}{f_o}\times\dfrac{D}{f_e}$Here $L$ is the tube length, $f_o$ and $f_e$ the objective and eyepiece focal lengths and $D$ the near-point distance. Small $f_o$ and $f_e$ yield large magnification for relaxed-eye viewing at infinity.NCERT Class XII Physics, Ch. 9
Telescope magnifying power (normal adjustment) 🔉⇢$m=\dfrac{f_o}{f_e}$Here $m$ is the magnifying power, $f_o$ the objective focal length and $f_e$ the eyepiece focal length, with the final image at infinity. A large $f_o$ and small $f_e$ give high angular magnification.NCERT Class XII Physics, Ch. 9
Telescope tube length 🔉⇢$L=f_o+f_e$Here $L$ is the separation between objective and eyepiece in normal adjustment, $f_o$ the objective focal length and $f_e$ the eyepiece focal length. The objective's real image lies at the common focal point.NCERT Class XII Physics, Ch. 9
Telescope magnifying power (image at near point) 🔉⇢$m=\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)$Here $f_o$ and $f_e$ are the objective and eyepiece focal lengths and $D$ the least distance of distinct vision. Use it when the final image is formed at the near point rather than at infinity.NCERT Class XII Physics, Ch. 9

📜 Previous-Year Questions Authentic NTA · 82 questions

Every PYQ traces to an official NTA/JAB paper (year + shift + paper). Answer keys match the official key. OCR-extracted questions are flagged for SME verification.

IIT-JEE 2008 Paper 1 Q27 Answer: $30^\circ$ for both the colours

Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is $60^\circ$). In the position of minimum deviation, the angle of refraction will be

  • $30^\circ$ for both the colours
  • greater for the violet colour
  • greater for the red colour
  • equal but not $30^\circ$ for both the colours
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2008 Paper 1 Q27, source page 9). Answer per official NTA/JAB key: $30^\circ$ for both the colours.
IIT-JEE 2008 Paper 2 Q31 Answer: $\sin^{-1}\left(\dfrac{1}{8}\right)$

A light beam is traveling from Region I to Region IV through four parallel slabs placed one after another. The refractive index in Regions I, II, III and IV are $n_0$, $\dfrac{n_0}{2}$, $\dfrac{n_0}{6}$ and $\dfrac{n_0}{8}$, respectively. The angle of incidence $\theta$ (at the Region I – Region II boundary) for which the beam just misses entering Region IV is

  • $\sin^{-1}\left(\dfrac{3}{4}\right)$
  • $\sin^{-1}\left(\dfrac{1}{8}\right)$
  • $\sin^{-1}\left(\dfrac{1}{4}\right)$
  • $\sin^{-1}\left(\dfrac{1}{3}\right)$
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2008 Paper 2 Q31, source page 12). Answer per official NTA/JAB key: $\sin^{-1}\left(\dfrac{1}{8}\right)$.
IIT-JEE 2009 Paper 1 Q43 Answer: 16 m/s

A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as [Take $g = 10$ m/s$^2$.]

  • 9 m/s
  • 12 m/s
  • 16 m/s
  • 21.33 m/s
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2009 Paper 1 Q43, source page 12). Answer per official NTA/JAB key: 16 m/s.
IIT-JEE 2009 Paper 1 Q50 Answer: $(66, 33)$; $(78, 39)$

A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of $(u, v)$ values recorded by the student (in cm) are: $(42, 56)$, $(48, 48)$, $(60, 40)$, $(66, 33)$, $(78, 39)$. The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are)

  • $(42, 56)$
  • $(48, 48)$
  • $(66, 33)$
  • $(78, 39)$
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2009 Paper 1 Q50, source page 14). Answer per official NTA/JAB key: $(66, 33)$; $(78, 39)$.
IIT-JEE 2010 Paper 2 Q41 Answer: real and at a distance of 16 cm from the mirror

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

  • virtual and at a distance of 16 cm from the mirror
  • real and at a distance of 16 cm from the mirror
  • virtual and at a distance of 20 cm from the mirror
  • real and at a distance of 20 cm from the mirror
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2010 Paper 2 Q41, source page 15). Answer per official NTA/JAB key: real and at a distance of 16 cm from the mirror.
IIT-JEE 2010 Paper 2 Q47 Answer: 3

Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from $\dfrac{25}{3}$ m to $\dfrac{50}{7}$ m in 30 seconds. What is the speed of the object in km per hour?

Solution + reasoning
Official IIT-JEE question (IIT-JEE 2010 Paper 2 Q47, source page 17). Answer per official NTA/JAB key: 3.
IIT-JEE 2010 Paper 2 Q48 Answer: 6

A large glass slab ($\mu=5/3$) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius $R$ cm. What is the value of $R$?

Solution + reasoning
Official IIT-JEE question (IIT-JEE 2010 Paper 2 Q48, source page 18). Answer per official NTA/JAB key: 6.
IIT-JEE 2010 Paper 1 Q75 Answer: 6

The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\dfrac{m_{25}}{m_{50}}$ is

Solution + reasoning
Official IIT-JEE question (IIT-JEE 2010 Paper 1 Q75, source page 23). Answer per official NTA/JAB key: 6.
IIT-JEE 2011 Paper 2 Q36 Answer: 2

Water (with refractive index $=\dfrac{4}{3}$) in a tank is 18 cm deep. Oil of refractive index $\dfrac{7}{4}$ lies on water making a convex surface of radius of curvature $R=6$ cm. Consider oil to act as a thin lens. An object S is placed 24 cm above water surface. The location of its image is at $x$ cm above the bottom of the tank. Then $x$ is

Solution + reasoning
Official IIT-JEE question (IIT-JEE 2011 Paper 2 Q36, source page 16). Answer per official NTA/JAB key: 2.
IIT-JEE 2012 Paper 1 Q10 Answer: $40.0$ cm.

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surfaces are of the same radius of curvature $R=14$ cm. For this bi-convex lens, for an object distance of $40$ cm, the image distance will be

  • $-280.0$ cm.
  • $40.0$ cm.
  • $21.5$ cm.
  • $13.3$ cm.
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2012 Paper 1 Q10, source page 6). Answer per official NTA/JAB key: $40.0$ cm..
IIT-JEE 2012 Paper 2 Q10 Answer: The speed of light in the meta-material is $v=\frac{c}{|n|}$

Paragraph for Questions 9 and 10: Most materials have the refractive index, $n>1$. So, when a light ray from air enters a naturally occurring material, then by Snell's law, $\frac{\sin\theta_1}{\sin\theta_2}=\frac{n_2}{n_1}$, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, $n=\left(\frac{c}{v}\right)=\pm\sqrt{\varepsilon_r\mu_r}$, where $c$ is the speed of electromagnetic waves in vacuum, $v$ its speed in the medium, $\varepsilon_r$ and $\mu_r$ are the relative permittivity and permeability of the medium respectively. In normal materials, both $\varepsilon_r$ and $\mu_r$ are positive, implying positive $n$ for the medium. When both $\varepsilon_r$ and $\mu_r$ are negative, one must choose the negative root of $n$. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behavior, without violating any physical laws. Since $n$ is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. Choose the correct statement.

  • The speed of light in the meta-material is $v=c|n|$
  • The speed of light in the meta-material is $v=\frac{c}{|n|}$
  • The speed of light in the meta-material is $v=c$.
  • The wavelength of the light in the meta-material $(\lambda_m)$ is given by $\lambda_m=\lambda_{air}|n|$, where $\lambda_{air}$ is the wavelength of the light in air.
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2012 Paper 2 Q10, source page 6). Answer per official NTA/JAB key: The speed of light in the meta-material is $v=\frac{c}{|n|}$.
JEE Advanced 2013 Paper 1 Q2 Answer: $30^\circ$

A ray of light travelling in the direction $\frac{1}{2}\left(\hat{i}+\sqrt{3}\,\hat{j}\right)$ is incident on a plane mirror. After reflection, it travels along the direction $\frac{1}{2}\left(\hat{i}-\sqrt{3}\,\hat{j}\right)$. The angle of incidence is

  • $30^\circ$
  • $45^\circ$
  • $60^\circ$
  • $75^\circ$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2013 Paper 1 Q2, source page 1). Answer per official NTA/JAB key: $30^\circ$.
JEE Advanced 2013 Paper 1 Q10 Answer: $3$ m

The image of an object, formed by a plano-convex lens at a distance of $8$ m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $\frac{2}{3}$ times the wavelength in free space. The radius of the curved surface of the lens is

  • $1$ m
  • $2$ m
  • $3$ m
  • $6$ m
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2013 Paper 1 Q10, source page 6). Answer per official NTA/JAB key: $3$ m.
JEE Advanced 2014 Paper 2 Q5 Answer: $1.36$

A point source $S$ is placed at the bottom of a transparent block of height $10$ mm and refractive index $2.72$. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter $11.54$ mm on the top of the block. The refractive index of the liquid is

  • $1.21$
  • $1.30$
  • $1.36$
  • $1.42$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2014 Paper 2 Q5, source page 3). Answer per official NTA/JAB key: $1.36$.
JEE Advanced 2014 Paper 1 Q9 Answer: $|f_1| = 3R$; $|f_2| = 2R$

A transparent thin film of uniform thickness and refractive index $n_1 = 1.4$ is coated on the convex spherical surface of radius $R$ at one end of a long solid glass cylinder of refractive index $n_2 = 1.5$. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance $f_1$ from the film, while rays of light traversing from glass to air get focused at distance $f_2$ from the film. Then

  • $|f_1| = 3R$
  • $|f_1| = 2.8R$
  • $|f_2| = 2R$
  • $|f_2| = 1.4R$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2014 Paper 1 Q9, source page 5). Answer per official NTA/JAB key: $|f_1| = 3R$; $|f_2| = 2R$.
JEE Advanced 2015 Paper 2 Q6 Answer: 2

A monochromatic beam of light is incident at $60^\circ$ on one face of an equilateral prism of refractive index $n$ and emerges from the opposite face making an angle $\theta(n)$ with the normal (see the figure). For $n = \sqrt{3}$ the value of $\theta$ is $60^\circ$ and $\dfrac{d\theta}{dn} = m$. The value of $m$ is

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 2 Q6, source page 3). Answer per official NTA/JAB key: 2.
JEE Advanced 2015 Paper 1 Q8 Answer: 7

Consider a concave mirror and a convex lens (refractive index $= 1.5$) of focal length $10$ cm each, separated by a distance of $50$ cm in air (refractive index $= 1$) as shown in the figure. An object is placed at a distance of $15$ cm from the mirror. Its erect image formed by this combination has magnification $M_1$. When the set-up is kept in a medium of refractive index $7/6$, the magnification becomes $M_2$. The magnitude $\left|\dfrac{M_2}{M_1}\right|$ is [From the figure: the concave mirror and the convex lens face each other on a common principal axis, $50$ cm apart, and the object stands on the axis between them at $15$ cm from the mirror.]

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 1 Q8, source page 4). Answer per official NTA/JAB key: 7.
JEE Advanced 2015 Paper 1 Q14 Answer: $70$ cm

Two identical glass rods $S_1$ and $S_2$ (refractive index $= 1.5$) have one convex end of radius of curvature $10$ cm. They are placed with the curved surfaces at a distance $d$ as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light $P$ is placed inside rod $S_1$ on its axis at a distance of $50$ cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside $S_2$. The distance $d$ is [From the figure: the two rods lie end to end on a common horizontal axis with their convex faces facing each other across the air gap of width $d$; $P$ lies inside $S_1$, $50$ cm to the left of the convex face of $S_1$.]

  • $60$ cm
  • $70$ cm
  • $80$ cm
  • $90$ cm
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 1 Q14, source page 8). Answer per official NTA/JAB key: $70$ cm.
JEE Advanced 2015 Paper 2 Q19 Answer: NA of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\dfrac{16}{3\sqrt{15}}$; NA of $S_1$ placed in air is the same as that of $S_2$ immersed in liquid of refractive index $\dfrac{4}{\sqrt{15}}$

PARAGRAPH: Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $n_1$ surrounded by a medium of lower refractive index $n_2$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $n_1$ and $n_2$. All rays with the angle of incidence $i$ less than a particular value $i_m$ are confined in the medium of refractive index $n_1$. The numerical aperture (NA) of the structure is defined as $\sin i_m$. For two structures namely $S_1$ with $n_1 = \sqrt{45}/4$ and $n_2 = 3/2$, and $S_2$ with $n_1 = 8/5$ and $n_2 = 7/5$ and taking the refractive index of water to be $4/3$ and that of air to be $1$, the correct option(s) is(are)

  • NA of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\dfrac{16}{3\sqrt{15}}$
  • NA of $S_1$ immersed in liquid of refractive index $\dfrac{6}{\sqrt{15}}$ is the same as that of $S_2$ immersed in water
  • NA of $S_1$ placed in air is the same as that of $S_2$ immersed in liquid of refractive index $\dfrac{4}{\sqrt{15}}$
  • NA of $S_1$ placed in air is the same as that of $S_2$ placed in water
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 2 Q19, source page 13). Answer per official NTA/JAB key: NA of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\dfrac{16}{3\sqrt{15}}$; NA of $S_1$ placed in air is the same as that of $S_2$ immersed in liquid of refractive index $\dfrac{4}{\sqrt{15}}$.
JEE Advanced 2015 Paper 2 Q20 Answer: $NA_2$

PARAGRAPH: Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $n_1$ surrounded by a medium of lower refractive index $n_2$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $n_1$ and $n_2$. All rays with the angle of incidence $i$ less than a particular value $i_m$ are confined in the medium of refractive index $n_1$. The numerical aperture (NA) of the structure is defined as $\sin i_m$. If two structures of same cross-sectional area, but different numerical apertures $NA_1$ and $NA_2$ ($NA_2 < NA_1$) are joined longitudinally, the numerical aperture of the combined structure is

  • $\dfrac{NA_1\,NA_2}{NA_1 + NA_2}$
  • $NA_1 + NA_2$
  • $NA_1$
  • $NA_2$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 2 Q20, source page 13). Answer per official NTA/JAB key: $NA_2$.
JEE Advanced 2016 Paper 1 Q4 Answer: $15^\circ$

A parallel beam of light is incident from air at an angle $\alpha$ on the side $PQ$ of a right angled triangular prism of refractive index $n = \sqrt{2}$. Light undergoes total internal reflection in the prism at the face $PR$ when $\alpha$ has a minimum value of $45^\circ$. The angle $\theta$ of the prism is [From the figure: the prism is the right triangle $PQR$ with the right angle at $Q$; $PQ$ is the vertical face on which the light is incident, $QR$ is the horizontal base, and $\theta$ is the angle of the prism at the vertex $P$, between the faces $PQ$ and $PR$.]

  • $15^\circ$
  • $22.5^\circ$
  • $30^\circ$
  • $45^\circ$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2016 Paper 1 Q4, source page 3). Answer per official NTA/JAB key: $15^\circ$.
JEE Advanced 2016 Paper 2 Q5 Answer: $(25,\ 25\sqrt{3})$

A small object is placed $50$ cm to the left of a thin convex lens of focal length $30$ cm. A convex spherical mirror of radius of curvature $100$ cm is placed to the right of the lens at a distance of $50$ cm. The mirror is tilted such that the axis of the mirror is at an angle $\theta = 30^\circ$ to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point $(x, y)$ at which the image is formed are [From the figure: the lens is at the origin $(0,0)$ with its axis along the $x$-axis and the object at $(-50, 0)$; the pole of the convex mirror is on that axis at $(50, 0)$ and its centre of curvature is at $(50 + 50\sqrt{3},\ -50)$, so the mirror axis is tilted $30^\circ$ below the lens axis.]

  • $(0, 0)$
  • $(50 - 25\sqrt{3},\ 25)$
  • $(25,\ 25\sqrt{3})$
  • $(125/3,\ 25/\sqrt{3})$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2016 Paper 2 Q5, source page 3). Answer per official NTA/JAB key: $(25,\ 25\sqrt{3})$.
JEE Advanced 2016 Paper 1 Q9 Answer: The refractive index of the lens is $2.5$; The focal length of the lens is $20$ cm

A plano-convex lens is made of a material of refractive index $n$. When a small object is placed $30$ cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of $10$ cm away from the lens. Which of the following statement(s) is(are) true?

  • The refractive index of the lens is $2.5$
  • The radius of curvature of the convex surface is $45$ cm
  • The faint image is erect and real
  • The focal length of the lens is $20$ cm
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2016 Paper 1 Q9, source page 7). Answer per official NTA/JAB key: The refractive index of the lens is $2.5$; The focal length of the lens is $20$ cm.
JEE Advanced 2016 Paper 1 Q13 Answer: $n_1\sin\theta_i = n_2\sin\theta_f$; $l$ is independent of $n_2$; $l$ is dependent on $n(z)$

A transparent slab of thickness $d$ has a refractive index $n(z)$ that increases with $z$. Here $z$ is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices $n_1$ and $n_2$ ($> n_1$), as shown in the figure. A ray of light is incident with angle $\theta_i$ from medium 1 and emerges in medium 2 with refraction angle $\theta_f$ with a lateral displacement $l$. Which of the following statement(s) is(are) true?

  • $n_1\sin\theta_i = n_2\sin\theta_f$
  • $n_1\sin\theta_i = (n_2 - n_1)\sin\theta_f$
  • $l$ is independent of $n_2$
  • $l$ is dependent on $n(z)$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2016 Paper 1 Q13, source page 11). Answer per official NTA/JAB key: $n_1\sin\theta_i = n_2\sin\theta_f$; $l$ is independent of $n_2$; $l$ is dependent on $n(z)$.
JEE Advanced 2017 Paper 1 Q7 Answer: For the angle of incidence $i_1 = A$, the ray inside the prism is parallel to the base of the prism; At minimum deviation, the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1 = (i_1/2)$; For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1 = \sin^{-1}\left[\sin A\sqrt{4\cos^2\dfrac{A}{2} - 1} - \cos A\right]$

For an isosceles prism of angle $A$ and refractive index $\mu$, it is found that the angle of minimum deviation $\delta_m = A$. Which of the following options is/are correct?

  • For the angle of incidence $i_1 = A$, the ray inside the prism is parallel to the base of the prism
  • For this prism, the refractive index $\mu$ and the angle of prism $A$ are related as $A = \dfrac{1}{2}\cos^{-1}\left(\dfrac{\mu}{2}\right)$
  • At minimum deviation, the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1 = (i_1/2)$
  • For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1 = \sin^{-1}\left[\sin A\sqrt{4\cos^2\dfrac{A}{2} - 1} - \cos A\right]$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2017 Paper 1 Q7, source page 6). Answer per official NTA/JAB key: For the angle of incidence $i_1 = A$, the ray inside the prism is parallel to the base of the prism; At minimum deviation, the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1 = (i_1/2)$; For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1 = \sin^{-1}\left[\sin A\sqrt{4\cos^2\dfrac{A}{2} - 1} - \cos A\right]$.
JEE Advanced 2017 Paper 1 Q10 Answer: 8

A monochromatic light is travelling in a medium of refractive index $n = 1.6$. It enters a stack of glass layers from the bottom side at an angle $\theta = 30^\circ$. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $n_m = n - m\,\Delta n$, where $n_m$ is the refractive index of the $m^{\text{th}}$ slab and $\Delta n = 0.1$. The ray is refracted out parallel to the interface between the $(m-1)^{\text{th}}$ and $m^{\text{th}}$ slabs from the right side of the stack. What is the value of $m$?

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2017 Paper 1 Q10, source page 8). Answer per official NTA/JAB key: 8.
JEE Advanced 2018 Paper 1 Q13 Answer: 130.00

Sunlight of intensity $1.3\ \mathrm{kW\,m^{-2}}$ is incident normally on a thin convex lens of focal length $20\ \mathrm{cm}$. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in $\mathrm{kW\,m^{-2}}$, at a distance $22\ \mathrm{cm}$ from the lens on the other side is __________.

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2018 Paper 1 Q13, source page 7). Answer per official NTA/JAB key: 130.00.
JEE Advanced 2019 Paper 2 Q6 Answer: $H_2 > H_1$; $H_2 > H_3$

Three glass cylinders of equal height $H = 30$ cm and same refractive index $n = 1.5$ are placed on a horizontal surface as shown in figure. Cylinder I has a flat top, cylinder II has a convex top and cylinder III has a concave top. The radii of curvature of the two curved tops are same ($R = 3$ m). If $H_1$, $H_2$, and $H_3$ are the apparent depths of a point $X$ on the bottom of the three cylinders, respectively, the correct statement(s) is/are:

  • $H_2 > H_1$
  • $H_3 > H_1$
  • $H_2 > H_3$
  • $0.8\ \mathrm{cm} < (H_2 - H_1) < 0.9\ \mathrm{cm}$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2019 Paper 2 Q6, source page 5). Answer per official NTA/JAB key: $H_2 > H_1$; $H_2 > H_3$.
JEE Advanced 2019 Paper 1 Q10 Answer: For $n = 1.5$, $\Delta n = 10^{-3}$ and $f = 20$ cm, the value of $|\Delta f|$ will be 0.02 cm (round off to 2nd decimal place).; If $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$; The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.

A thin convex lens is made of two materials with refractive indices $n_1$ and $n_2$, as shown in figure. The radius of curvature of the left and right spherical surfaces are equal. $f$ is the focal length of the lens when $n_1 = n_2 = n$. The focal length is $f + \Delta f$ when $n_1 = n$ and $n_2 = n + \Delta n$. Assuming $\Delta n \ll (n-1)$ and $1 < n < 2$, the correct statement(s) is/are, [Figure: the biconvex lens is divided into two halves by the plane through its centre perpendicular to the optic axis; the half bounded by the left spherical surface has refractive index $n_1$ and the half bounded by the right spherical surface has refractive index $n_2$.]

  • $\left|\frac{\Delta f}{f}\right| < \left|\frac{\Delta n}{n}\right|$
  • For $n = 1.5$, $\Delta n = 10^{-3}$ and $f = 20$ cm, the value of $|\Delta f|$ will be 0.02 cm (round off to 2nd decimal place).
  • If $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$
  • The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2019 Paper 1 Q10, source page 8). Answer per official NTA/JAB key: For $n = 1.5$, $\Delta n = 10^{-3}$ and $f = 20$ cm, the value of $|\Delta f|$ will be 0.02 cm (round off to 2nd decimal place).; If $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$; The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature..
JEE Advanced 2019 Paper 2 Q11 Answer: 1.50

A monochromatic light is incident from air on a refracting surface of a prism of angle $75^\circ$ and refractive index $n_0 = \sqrt{3}$. The other refracting surface of the prism is coated by a thin film of material of refractive index $n$ as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of $\theta \le 60^\circ$. The value of $n^{2}$ is ____.

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2019 Paper 2 Q11, source page 8). Answer per official NTA/JAB key: 1.50.
JEE Advanced 2019 Paper 1 Q18 Answer: 50

A planar structure of length $L$ and width $W$ is made of two different optical media of refractive indices $n_1 = 1.5$ and $n_2 = 1.44$ as shown in figure. If $L \gg W$, a ray entering from end AB will emerge from end CD only if the total internal reflection condition is met inside the structure. For $L = 9.6$ m, if the incident angle $\theta$ is varied, the maximum time taken by a ray to exit the plane CD is $t \times 10^{-9}$ s, where $t$ is ____. [Speed of light $c = 3 \times 10^{8}$ m/s] [Figure: a slab of medium $n_1$ of width $W$ and length $L$ is sandwiched between layers of medium $n_2$ above and below it; AB is the left end face of the $n_1$ slab and CD is its right end face, and the ray is incident from air on face AB at angle $\theta$ to the normal of AB (the normal being along the length $L$ of the structure).]

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2019 Paper 1 Q18, source page 11). Answer per official NTA/JAB key: 50.
JEE Advanced 2020 Paper 2 Q7 Answer: $R = \dfrac{h^2 + r^2}{2h}$; Apparent depth of the bottom of the beaker is close to $\dfrac{3H}{4}\left(1 + \dfrac{\omega^2 H}{4g}\right)^{-1}$

A beaker of radius $r$ is filled with water (refractive index $\dfrac{4}{3}$) up to a height $H$ as shown in the figure on the left. The beaker is kept on a horizontal table rotating with angular speed $\omega$. This makes the water surface curved so that the difference in the height of water level at the center and at the circumference of the beaker is $h$ ($h \ll H$, $h \ll r$), as shown in the figure on the right. Take this surface to be approximately spherical with a radius of curvature $R$. Which of the following is/are correct? ($g$ is the acceleration due to gravity)

  • $R = \dfrac{h^2 + r^2}{2h}$
  • $R = \dfrac{3r^2}{2h}$
  • Apparent depth of the bottom of the beaker is close to $\dfrac{3H}{2}\left(1 + \dfrac{\omega^2 H}{2g}\right)^{-1}$
  • Apparent depth of the bottom of the beaker is close to $\dfrac{3H}{4}\left(1 + \dfrac{\omega^2 H}{4g}\right)^{-1}$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2020 Paper 2 Q7, source page 4). Answer per official NTA/JAB key: $R = \dfrac{h^2 + r^2}{2h}$; Apparent depth of the bottom of the beaker is close to $\dfrac{3H}{4}\left(1 + \dfrac{\omega^2 H}{4g}\right)^{-1}$.
JEE Advanced 2021 Paper 1 Q3 Answer: $0.8$

An extended object is placed at point O, 10 cm in front of a convex lens $L_1$ and a concave lens $L_2$ is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens system is

  • $0.4$
  • $0.8$
  • $1.3$
  • $1.6$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2021 Paper 1 Q3, source page 2). Answer per official NTA/JAB key: $0.8$.
JEE Advanced 2021 Paper 2 Q3 Answer: $\Delta e$ is proportional to $\Delta n$; $\Delta e$ lies between 2.0 and 3.0 milliradians, if $\Delta n = 2.8 \times 10^{-3}$

For a prism of prism angle $\theta = 60^\circ$, the refractive indices of the left half and the right half are, respectively, $n_1$ and $n_2$ ($n_2 \geq n_1$) as shown in the figure. The angle of incidence $i$ is chosen such that the incident light rays will have minimum deviation if $n_1 = n_2 = n = 1.5$. For the case of unequal refractive indices, $n_1 = n$ and $n_2 = n + \Delta n$ (where $\Delta n \ll n$), the angle of emergence $e = i + \Delta e$. Which of the following statement(s) is(are) correct?

  • The value of $\Delta e$ (in radians) is greater than that of $\Delta n$
  • $\Delta e$ is proportional to $\Delta n$
  • $\Delta e$ lies between 2.0 and 3.0 milliradians, if $\Delta n = 2.8 \times 10^{-3}$
  • $\Delta e$ lies between 1.0 and 1.6 milliradians, if $\Delta n = 2.8 \times 10^{-3}$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2021 Paper 2 Q3, source page 3). Answer per official NTA/JAB key: $\Delta e$ is proportional to $\Delta n$; $\Delta e$ lies between 2.0 and 3.0 milliradians, if $\Delta n = 2.8 \times 10^{-3}$.
JEE Advanced 2021 Paper 1 Q12 Answer: The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 < n_1$; The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 > n_1$; The light ray is reflected back into the medium of refractive index $n_1$ if $n_2 = 1$

A wide slab consisting of two media of refractive indices $n_1$ and $n_2$ is placed in air as shown in the figure. A ray of light is incident from medium $n_1$ to $n_2$ at an angle $\theta$, where $\sin\theta$ is slightly larger than $1/n_1$. Take refractive index of air as 1. Which of the following statement(s) is(are) correct?

  • The light ray enters air if $n_2 = n_1$
  • The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 < n_1$
  • The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 > n_1$
  • The light ray is reflected back into the medium of refractive index $n_1$ if $n_2 = 1$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2021 Paper 1 Q12, source page 6). Answer per official NTA/JAB key: The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 < n_1$; The light ray is finally reflected back into the medium of refractive index $n_1$ if $n_2 > n_1$; The light ray is reflected back into the medium of refractive index $n_1$ if $n_2 = 1$.
JEE Main 2021 (August 31 Shift 1) Paper 1 Q5 Answer: magnification 1/2; image erect and diminished⚑ verify

An object is placed at the focus of concave lens having focal length f. What is the magnification and distance of the image from the optical centre of the lens?

Solution + reasoning
JEE Main 2021 (August 31 Shift 1) Paper 1 Q5 (source page 2). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2021 (August 27 Shift 1) Paper 1 Q7 Answer: 1⚑ verify

An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is $d_{1}$ from C and the distance of the image formed is $d_{2}$ from C, the radius of curvature of this mirror is :

Solution + reasoning
JEE Main 2021 (August 27 Shift 1) Paper 1 Q7 (source page 2). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2021 (September 1 Shift 2) Paper 1 Q12 Answer: 10 cm⚑ verify

A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water ($\mu$ = 4/3) into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?

Solution + reasoning
JEE Main 2021 (September 1 Shift 2) Paper 1 Q12 (source page 4). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Advanced 2022 Paper 1 Q4 Answer: 6

A rod of length $2\ cm$ makes an angle $\dfrac{2\pi}{3}\ rad$ with the principal axis of a thin convex lens. The lens has a focal length of $10\ cm$ and is placed at a distance of $\dfrac{40}{3}\ cm$ from the object. The height of the image is $\dfrac{30\sqrt{3}}{13}\ cm$ and the angle made by it with respect to the principal axis is $\alpha\ rad$. The value of $\alpha$ is $\dfrac{\pi}{n}\ rad$, where $n$ is _____.

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2022 Paper 1 Q4, source page 10). Answer per official NTA/JAB key: 6.
JEE Advanced 2022 Paper 2 Q5 Answer: 4

Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h = \dfrac{1}{3}\ cm$, and the second and third layers are of equal thickness $d = \dfrac{\sqrt{3}-1}{2}\ cm$, and refractive indices $\mu_1 = \sqrt{\dfrac{3}{2}}$ and $\mu_2 = \sqrt{3}$, respectively. A light source $O$ is placed on the top of the first unit. A ray of light from $O$ is incident on the second layer of the first unit at an angle of $\theta = 60^{\circ}$ to the normal. For a specific value of $n$, the ray of light emerges from the bottom of the configuration at a horizontal distance $l = \dfrac{8}{\sqrt{3}}\ cm$. The value of $n$ is _____.

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2022 Paper 2 Q5, source page 10). Answer per official NTA/JAB key: 4.
JEE Advanced 2022 Paper 2 Q8 Answer: 3

An object and a concave mirror of focal length $f = 10\ cm$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_0 = 15\ cm\ s^{-1}$ towards the mirror with respect to a laboratory frame. The distance between the object and the mirror at a given moment is denoted by $u$. When $u = 30\ cm$, the speed of the mirror $V_m$ is such that the image is instantaneously at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of $V_m$ is _____ $cm\ s^{-1}$.

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2022 Paper 2 Q8, source page 11). Answer per official NTA/JAB key: 3.
JEE Advanced 2022 Paper 1 Q18 Answer: (I) $\to$ P; (II) $\to$ R; (III) $\to$ Q; (IV) $\to$ T

List I contains four combinations of two lenses (1 and 2) whose focal lengths (in $cm$) are indicated below. In all cases, the object is placed $20\ cm$ from the first lens on the left, and the distance between the two lenses is $5\ cm$. List II contains the positions of the final images. List-I: (I) Lens 1 has $f = +10$, lens 2 has $f = +15$. (II) Lens 1 has $f = +10$, lens 2 has $f = -10$. (III) Lens 1 has $f = +10$, lens 2 has $f = -20$. (IV) Lens 1 has $f = -20$, lens 2 has $f = +10$. List-II: (P) Final image is formed at $7.5\ cm$ on the right side of lens 2. (Q) Final image is formed at $60.0\ cm$ on the right side of lens 2. (R) Final image is formed at $30.0\ cm$ on the left side of lens 2. (S) Final image is formed at $6.0\ cm$ on the right side of lens 2. (T) Final image is formed at $30.0\ cm$ on the right side of lens 2. Which one of the following options is correct?

  • (I) $\to$ P; (II) $\to$ R; (III) $\to$ Q; (IV) $\to$ T
  • (I) $\to$ Q; (II) $\to$ P; (III) $\to$ T; (IV) $\to$ S
  • (I) $\to$ P; (II) $\to$ T; (III) $\to$ R; (IV) $\to$ Q
  • (I) $\to$ T; (II) $\to$ S; (III) $\to$ Q; (IV) $\to$ R
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2022 Paper 1 Q18, source page 20). Answer per official NTA/JAB key: (I) $\to$ P; (II) $\to$ R; (III) $\to$ Q; (IV) $\to$ T.
JEE Advanced 2023 Paper 1 Q2 Answer: The blue light is polarized in the plane of incidence.; The refractive index of the material of the prism for red light is $\sqrt{2}$.; The angle of refraction for blue light in air at the exit plane of the prism is $60^\circ$.

A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is $\delta = 60^\circ$. The angle of minimum deviation for red light from the same prism is $\delta_{\min} = 30^\circ$. The refractive index of the prism material for blue light is $\sqrt{3}$. Which of the following statement(s) is(are) correct?

  • The blue light is polarized in the plane of incidence.
  • The angle of the prism is $45^\circ$.
  • The refractive index of the material of the prism for red light is $\sqrt{2}$.
  • The angle of refraction for blue light in air at the exit plane of the prism is $60^\circ$.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2023 Paper 1 Q2, source page 13). Answer per official NTA/JAB key: The blue light is polarized in the plane of incidence.; The refractive index of the material of the prism for red light is $\sqrt{2}$.; The angle of refraction for blue light in air at the exit plane of the prism is $60^\circ$..
JEE Advanced 2023 Paper 2 Q5 Answer: It will deflect up by an angle $\tan^{-1}\left[\dfrac{(n_2 - n_1)d}{h}\right]$.; The deflection angle depends only on $(n_2 - n_1)$ and not on the individual values of $n_1$ and $n_2$.

A monochromatic light wave is incident normally on a glass slab of thickness $d$. The refractive index of the slab increases linearly from $n_1$ to $n_2$ over the height $h$. Which of the following statement(s) is(are) true about the light wave emerging out of the slab?

  • It will deflect up by an angle $\tan^{-1}\left[\dfrac{(n_2^2 - n_1^2)d}{2h}\right]$.
  • It will deflect up by an angle $\tan^{-1}\left[\dfrac{(n_2 - n_1)d}{h}\right]$.
  • It will not deflect.
  • The deflection angle depends only on $(n_2 - n_1)$ and not on the individual values of $n_1$ and $n_2$.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2023 Paper 2 Q5, source page 13). Answer per official NTA/JAB key: It will deflect up by an angle $\tan^{-1}\left[\dfrac{(n_2 - n_1)d}{h}\right]$.; The deflection angle depends only on $(n_2 - n_1)$ and not on the individual values of $n_1$ and $n_2$..
JEE Advanced 2023 Paper 1 Q10 Answer: 1

In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is $10 \pm 0.1$ cm and the distance of its real image from the lens is $20 \pm 0.2$ cm. The error in the determination of focal length of the lens is $n\%$. The value of $n$ is _______.

Solution + reasoning
Official JEE Advanced question (JEE Advanced 2023 Paper 1 Q10, source page 16). Answer per official NTA/JAB key: 1.
JEE Main 2023 (January 25 Shift 2) Paper 1 Q3 Answer: B and D only⚑ verify

The light rays from an object have been reflected towards an observer from a standard flat mirror, the image observed by the observer are :- A. Real B. Erect C. Smaller in size then object D. Laterally inverted Choose the most appropriate answer from the options given below :

Solution + reasoning
JEE Main 2023 (January 25 Shift 2) Paper 1 Q3 (source page 1). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (January 30 Shift 1) Paper 1 Q3 Answer: 50 cm⚑ verify

A person has been using spectacles of power $-1.0$ dioptre for distant vision and a separate reading glass of power $2.0$ dioptres. What is the least distance of distinct vision for this person :

Solution + reasoning
JEE Main 2023 (January 30 Shift 1) Paper 1 Q3 (source page 1). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (January 31 Shift 2) Paper 1 Q3 Answer: 75 cm⚑ verify

A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index $\frac{5}{3}$ is poured inside the bucket, then the microscope has to be raised by $30 \mathrm{~cm}$ to focus the object again. The height of the liquid in the bucket is :

Solution + reasoning
JEE Main 2023 (January 31 Shift 2) Paper 1 Q3 (source page 1). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (January 30 Shift 2) Paper 1 Q5 Answer: 4⚑ verify

A thin prism $P_1$ with an angle $6^{\circ}$ and made of glass of refractive index $1.54$ is combined with another prism $P_2$ made from glass of refractive index $1.72$ to produce dispersion without average deviation. The angle of prism $P_2$ is

Solution + reasoning
JEE Main 2023 (January 30 Shift 2) Paper 1 Q5 (source page 2). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (January 24 Shift 2) Paper 1 Q10 Answer: chromatic aberration⚑ verify

When a beam of white light is allowed to pass through convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called :

Solution + reasoning
JEE Main 2023 (January 24 Shift 2) Paper 1 Q10 (source page 3). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (January 29 Shift 2) Paper 1 Q14 Answer: shorter wavelength / higher numerical aperture⚑ verify

A scientist is observing a bacteria through a compound microscope. For better analysis and to improve its resolving power he should. (Select the best option)

Solution + reasoning
JEE Main 2023 (January 29 Shift 2) Paper 1 Q14 (source page 4). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (January 29 Shift 2) Paper 1 Q23 Answer: 41⚑ verify

In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm and apparent thickness of the glass slab as 5.00 mm. Travelling microscope has 20 divisions in one cm on main scale and 20 divisions on vernier scale is equal to 49 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is $\frac{x}{10}\times10^{-3}$, where $x$ is ___________

Solution + reasoning
JEE Main 2023 (January 29 Shift 2) Paper 1 Q23 (source page 6). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (February 1 Shift 1) Paper 1 Q27 Answer: 32⚑ verify

A thin cylindrical rod of length $10 \mathrm{~cm}$ is placed horizontally on the principle axis of a concave mirror of focal length $20 \mathrm{~cm}$. The rod is placed in a such a way that mid point of the rod is at $40 \mathrm{~cm}$ from the pole of mirror. The length of the image formed by the mirror will be $\frac{x}{3} \mathrm{~cm}$. The value of $x$ is _____________.

Solution + reasoning
JEE Main 2023 (February 1 Shift 1) Paper 1 Q27 (source page 6). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (January 24 Shift 2) Paper 1 Q29 Answer: 54⚑ verify

A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change in focal length of the lens will be ___________ cm. (Given refractive index of water $=\frac{4}{3}$)

Solution + reasoning
JEE Main 2023 (January 24 Shift 2) Paper 1 Q29 (source page 8). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (January 30 Shift 1) Paper 1 Q29 Answer: 32⚑ verify

In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $40 \mathrm{~cm}$ from the pole of the mirror is formed at distance $120 \mathrm{~cm}$ from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in $1 \mathrm{~cm}$. The value of error in measurement of focal length of the mirror is $\frac{1}{\mathrm{~K}} \mathrm{~cm}$. The value of $\mathrm{K}$ is __________.

Solution + reasoning
JEE Main 2023 (January 30 Shift 1) Paper 1 Q29 (source page 9). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 6 Shift 2) Paper 1 Q32 Answer: Both A and R are correct and R is the correct explanation of A⚑ verify

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: The phase difference of two light waves change if they travel through different media having same thickness, but different indices of refraction. Reason R: The wavelengths of waves are different in different media. In the light of the above statements, choose the most appropriate answer from the options given below

  • Both A and R are correct but R is NOT the correct explanation of A
  • A is correct but R is not correct
  • A is not correct but R is correct
  • Both A and R are correct and R is the correct explanation of A
Solution + reasoning
JEE Main 2023 (April 6 Shift 2) Paper 1 Q32 (source page 1). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 6 Shift 2) Paper 1 Q42 Answer: 1.70⚑ verify

A 2 meter long scale with least count of $0.2 \mathrm{~cm}$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80 \mathrm{~cm}$ mark and $1 \mathrm{~m}$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180 \mathrm{~cm}$ mark. The $\%$ error in the estimation of focal length is:

  • 1.70
  • 0.51
  • 1.02
  • 0.85
Solution + reasoning
JEE Main 2023 (April 6 Shift 2) Paper 1 Q42 (source page 5). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 6 Shift 1) Paper 1 Q46 Answer: $\lambda_{2}=\frac{1}{\sqrt{2}} \lambda_{1}, v_{2}=v_{1}$⚑ verify

A monochromatic light wave with wavelength $\lambda_{1}$ and frequency $v_{1}$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are $45^{\circ}$ and $30^{\circ}$ respectively, then the wavelength $\lambda_{2}$ and frequency $v_{2}$ of the refracted wave are:

  • $\lambda_{2}=\lambda_{1}, v_{2}=\frac{1}{\sqrt{2}} v_{1}$
  • $\lambda_{2}=\lambda_{1}, v_{2}=\sqrt{2} v_{1}$
  • $\lambda_{2}=\sqrt{2} \lambda_{1}, v_{2}=v_{1}$
  • $\lambda_{2}=\frac{1}{\sqrt{2}} \lambda_{1}, v_{2}=v_{1}$
Solution + reasoning
JEE Main 2023 (April 6 Shift 1) Paper 1 Q46 (source page 5). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 10 Shift 1) Paper 1 Q49 Answer: 8 cm towards mirror⚑ verify

An object is placed at a distance of 12 cm in front of a plane mirror. The virtual and erect image is formed by the mirror. Now the mirror is moved by 4 cm towards the stationary object. The distance by which the position of image would be shifted, will be

  • 4 cm towards mirror
  • 2 cm towards mirror
  • 8 cm away from mirror
  • 8 cm towards mirror
Solution + reasoning
JEE Main 2023 (April 10 Shift 1) Paper 1 Q49 (source page 6). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 11 Shift 2) Paper 1 Q49 Answer: $120^{\circ}$⚑ verify

When one light ray is reflected from a plane mirror with $30^{\circ}$ angle of reflection, the angle of deviation of the ray after reflection is :

  • $140^{\circ}$
  • $130^{\circ}$
  • $120^{\circ}$
  • $110^{\circ}$
Solution + reasoning
JEE Main 2023 (April 11 Shift 2) Paper 1 Q49 (source page 7). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 13 Shift 1) Paper 1 Q50 Answer: $\frac{d\left(n_{1}+n_{2}\right)}{2 n_{1} n_{2}}$⚑ verify

A vessel of depth '$d$' is half filled with oil of refractive index $n_{1}$ and the other half is filled with water of refractive index $n_{2}$. The apparent depth of this vessel when viewed from above will be-

  • $\frac{2 d\left(n_{1}+n_{2}\right)}{n_{1} n_{2}}$
  • $\frac{d\left(n_{1}+n_{2}\right)}{2 n_{1} n_{2}}$
  • $\frac{d n_{1} n_{2}}{2\left(n_{1}+n_{2}\right)}$
  • $\frac{d n_{1} n_{2}}{\left(n_{1}+n_{2}\right)}$
Solution + reasoning
JEE Main 2023 (April 13 Shift 1) Paper 1 Q50 (source page 8). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 8 Shift 1) Paper 1 Q50 Answer: move the eyepiece outside the telescopic tube⚑ verify

In a reflecting telescope, a secondary mirror is used to:

  • make chromatic aberration zero
  • remove spherical aberration
  • reduce the problem of mechanical support
  • move the eyepiece outside the telescopic tube
Solution + reasoning
JEE Main 2023 (April 8 Shift 1) Paper 1 Q50 (source page 7). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 11 Shift 1) Paper 1 Q51 Answer: 4⚑ verify

The radius of curvature of each surface of a convex lens having refractive index 1.8 is $20 \mathrm{~cm}$. The lens is now immersed in a liquid of refractive index 1.5 . The ratio of power of lens in air to its power in the liquid will be $x: 1$. The value of $x$ is _________.

Solution + reasoning
JEE Main 2023 (April 11 Shift 1) Paper 1 Q51 (source page 8). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 13 Shift 1) Paper 1 Q54 Answer: 3⚑ verify

A fish rising vertically upward with a uniform velocity of $8 \mathrm{~ms}^{-1}$, observes that a bird is diving vertically downward towards the fish with the velocity of $12 \mathrm{~ms}^{-1}$. If the refractive index of water is $\frac{4}{3}$, then the actual velocity of the diving bird to pick the fish, will be __________ $\mathrm{ms}^{-1}$.

Solution + reasoning
JEE Main 2023 (April 13 Shift 1) Paper 1 Q54 (source page 10). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 13 Shift 2) Paper 1 Q57 Answer: 5⚑ verify

A bi convex lens of focal length $10 \mathrm{~cm}$ is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is ____________ D.

Solution + reasoning
JEE Main 2023 (April 13 Shift 2) Paper 1 Q57 (source page 9). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 8 Shift 2) Paper 1 Q58 Answer: 30⚑ verify

Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature $30 \mathrm{~cm}$. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of $15 \mathrm{~cm}$ from the pole of the surface. The distance of image from the pole of the surface is ____________ $\mathrm{cm}$.

Solution + reasoning
JEE Main 2023 (April 8 Shift 2) Paper 1 Q58 (source page 9). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 6 Shift 1) Paper 1 Q60 Answer: 50⚑ verify

A pole is vertically submerged in swimming pool, such that it gives a length of shadow $2.15 \mathrm{~m}$ within water when sunlight is incident at angle of $30^{\circ}$ with the surface of water. If swimming pool is filled to a height of $1.5 \mathrm{~m}$, then the height of the pole above the water surface in centimeters is $\left(n_{w}=4 / 3\right)$ ____________.

Solution + reasoning
JEE Main 2023 (April 6 Shift 1) Paper 1 Q60 (source page 9). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Advanced 2024 Paper 1 Q16 Answer: P $\to$ 5; Q $\to$ 2; R $\to$ 1; S $\to$ 4

A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $\theta_0$. The ray partially refracts into the sphere with angle of refraction $\phi_0$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $\alpha$. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. List-I: (P) If $n = 2$ and $\alpha = 180^\circ$, then all the possible values of $\theta_0$ will be (Q) If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $\theta_0$ will be (R) If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $\phi_0$ will be (S) If $n = \sqrt{2}$ and $\theta_0 = 45^\circ$, then all the possible values of $\alpha$ will be List-II: (1) $30^\circ$ and $0^\circ$ (2) $60^\circ$ and $0^\circ$ (3) $45^\circ$ and $0^\circ$ (4) $150^\circ$ (5) $0^\circ$

  • P $\to$ 5; Q $\to$ 2; R $\to$ 1; S $\to$ 4
  • P $\to$ 5; Q $\to$ 1; R $\to$ 2; S $\to$ 4
  • P $\to$ 3; Q $\to$ 2; R $\to$ 1; S $\to$ 4
  • P $\to$ 3; Q $\to$ 1; R $\to$ 2; S $\to$ 5
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2024 Paper 1 Q16, source page 17). Answer per official NTA/JAB key: P $\to$ 5; Q $\to$ 2; R $\to$ 1; S $\to$ 4.
JEE Advanced 2025 Paper 2 Q6 Answer: $4f + \left(1 - \dfrac{1}{n_0}\right)t$; $2f + \left(1 - \dfrac{1}{n_0}\right)t$

Two identical concave mirrors each of focal length $f$ are facing each other. The focal length $f$ is much larger than the size of the mirrors. A glass slab of thickness $t$ and refractive index $n_0$ is kept equidistant from the mirrors and perpendicular to their common principal axis. A monochromatic point light source $S$ is embedded at the center of the slab on the principal axis. For the image to be formed on $S$ itself, which of the following distances between the two mirrors is/are correct:

  • $4f + \left(1 - \dfrac{1}{n_0}\right)t$
  • $2f + \left(1 - \dfrac{1}{n_0}\right)t$
  • $4f + (n_0 - 1)t$
  • $2f + (n_0 - 1)t$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2025 Paper 2 Q6, source page 11). Answer per official NTA/JAB key: $4f + \left(1 - \dfrac{1}{n_0}\right)t$; $2f + \left(1 - \dfrac{1}{n_0}\right)t$.
JEE Advanced 2026 Paper 2 Q3 Answer: 4.8°

A beam of polychromatic light passes through a thin prism of prism angle 6°. The refractive index of the material of the prism varies with wavelength (𝜆) as 𝑛(𝜆) = 𝛼𝜆 + 𝛽 𝜆2, where 𝛼= 3 𝜇m−1 and 𝛽= 0.096 𝜇m2. If 𝜆min is the wavelength at which the angle of minimum deviation 𝐷𝑚 is smallest, then the correct value of 𝐷𝑚 at 𝜆min is

  • 6.4°
  • 4.8°
  • 3.2°
  • 2.4°
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2026 Paper 2 Q3, source page 10). Answer per official NTA/JAB key: 6.4°.
JEE Advanced 2026 Paper 2 Q5 Answer: (A), (C), (D)

Consider two isosceles prisms 1 and 2 with prism angles 𝐴1 and 𝐴2 and refractive indices 𝑛1and 𝑛2, respectively, as shown in the figure. The faces 𝑎1𝑏1 and 𝑎2𝑏2 are parallel to each other and perpendicular to the mirror 𝑀. If a ray of light is incident on the face 𝑎1𝑐1 and emerges from the face 𝑎2𝑐2, then the correct statement(s) is/are:

  • If both the prisms are at minimum deviation condition, then 𝑛2 𝑛1 = sin( 𝐴1 2 ) / sin( 𝐴2 2 ).
  • If prism 2 is at minimum deviation condition, then sin𝑖1 = 𝑛2 sin( 𝐴2 2 ) is always true.
  • If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation 𝛿𝑚1 and 𝛿𝑚2, respectively, then 𝜃= 𝛿𝑚1 2(𝑛1−1) + 𝛿𝑚2 2(𝑛2−1) .
  • If prism 1 is at minimum deviation condition, then sin𝑖2 = 𝑛1 sin( 𝐴1 2 ) is always true.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2026 Paper 2 Q5, source page 13). Answer per official NTA/JAB key: If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation 𝛿𝑚1 and 𝛿𝑚2, respectively, then 𝜃= 𝛿𝑚1 2(𝑛1−1) + 𝛿𝑚2 2(𝑛2−1) .; If prism 1 is at minimum deviation condition, then sin𝑖2 = 𝑛1 sin( 𝐴1 2 ) is always true. Answer Q5: ACD.
JEE Main 2026 (April 8 Shift 2) Paper 1 Q30 Answer: 10 cm⚑ verify

A thin biconvex lens is prepared from the glass $(\mu=1.5)$ both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of $\_\_\_\_$ cm.

  • 10
  • 12.5
  • 13
  • 13.5
Solution + reasoning
JEE Main 2026 (April 8 Shift 2) Paper 1 Q30 (source page 12). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 2 Shift 1) Paper 1 Q40 Answer: 3 : 2⚑ verify

For a thin symmetric prism made of glass (refractive index 1.5), the ratio of incident angle and minimum deviation will be _______.

  • 3 : 4
  • 3 : 2
  • 2 : 1
  • 1 : 2
Solution + reasoning
JEE Main 2026 (April 2 Shift 1) Paper 1 Q40 (source page 16). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 5 Shift 1) Paper 1 Q40 Answer: 30°⚑ verify

A ray of light passing through an equilateral prism is having velocity $2.12 \times 10^8 \mathrm{~m} / \mathrm{s}$ in the prism material, then the minimum angle of deviation is $\_\_\_\_$ degrees.

  • 45
  • 30
  • 28
  • 58
Solution + reasoning
JEE Main 2026 (April 5 Shift 1) Paper 1 Q40 (source page 15). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 5 Shift 2) Paper 1 Q40 Answer: behaves as concave lens if $\left|f_{\text {convex }}\right|>\left|f_{\text {concave }}\right|$⚑ verify

A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses ?

  • behaves as concave lens if $\left|f_{\text {convex }}\right|>\left|f_{\text {concave }}\right|$
  • behaves as concave lens if $\left|f_{\text {convex }}\right|<\left|f_{\text {concave }}\right|$
  • behaves as convex lens if $\left|f_{\text {convex }}\right|>\left|f_{\text {concave }}\right|$
  • Focal length of the lens system will change if the positions of two lenses are interchanged
Solution + reasoning
JEE Main 2026 (April 5 Shift 2) Paper 1 Q40 (source page 16). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 4 Shift 2) Paper 1 Q41 Answer: 1.25⚑ verify

A convex lens is made from glass material having refractive index of 1.4 with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is $\_\_\_\_$ .

  • 0.5
  • 2.5
  • 0.8
  • 1.25
Solution + reasoning
JEE Main 2026 (April 4 Shift 2) Paper 1 Q41 (source page 15). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 5 Shift 1) Paper 1 Q42 Answer: increased two times⚑ verify

A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be

  • increased two times
  • increased $\frac{3}{2}$ times
  • decreased two times
  • decreased $\frac{3}{2}$ times
Solution + reasoning
JEE Main 2026 (April 5 Shift 1) Paper 1 Q42 (source page 15). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 6 Shift 2) Paper 1 Q42 Answer: $\sqrt{2}$⚑ verify

Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is $\_\_\_\_$

  • 1.5
  • $\sqrt{3}$
  • $\sqrt{2}$
  • 1.65
Solution + reasoning
JEE Main 2026 (April 6 Shift 2) Paper 1 Q42 (source page 17). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 4 Shift 1) Paper 1 Q44 Answer: R = 122 cm⚑ verify

A telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength 500 nm , at a resolution of $5 \times 10^{-7}$ radian. The value of $R$ is $\_\_\_\_$ cm .

  • 61
  • 122
  • 244
  • 305
Solution + reasoning
JEE Main 2026 (April 4 Shift 1) Paper 1 Q44 (source page 16). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 8 Shift 2) Paper 1 Q47 Answer: 203 cm⚑ verify

Some distant star is to be observed by some telescope of diameter of objective lens $a$, at an angular resolution of $3.0 \times 10^{-7}$ radian. If the wavelength of light from the star reaching the telescope is 500 nm , the minimum diameter of the objective lens of the telescope is $\_\_\_\_$ cm. (nearest interger)

Solution + reasoning
JEE Main 2026 (April 8 Shift 2) Paper 1 Q47 (source page 19). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 2 Shift 2) Paper 1 Q49 Answer: 20⚑ verify

If sunlight is focused on a paper using convex lens, it starts burning the paper in shortest time when the lens is kept at 30 cm above the paper. If the radius of curvature of the lens is 60 cm then the refractive index of the lens material is $\frac{\alpha}{10}$. The value of $\alpha$ is ________.

Solution + reasoning
JEE Main 2026 (April 2 Shift 2) Paper 1 Q49 (source page 19). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 6 Shift 2) Paper 1 Q50 Answer: 10 cm⚑ verify

A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is $\_\_\_\_$ cm.

Solution + reasoning
JEE Main 2026 (April 6 Shift 2) Paper 1 Q50 (source page 20). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.

🎯 Question Bank 100 MCQs · graded

Distribution — advanced: 10 · easy: 40 · hard: 20 · medium: 30. Every question carries a source trace; each ends in an SME-verify solution.

Q1 According to the Cartesian sign convention used in ray optics, distances measured in the same direction as the incident light are taken as: easy
Step solution + source
In the Cartesian sign convention all distances are measured from the pole of the mirror or optic centre of the lens. Distances measured along the direction of incident light are positive, while those measured opposite to it are negative. This single convention lets one formula cover all cases. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q2 In the Cartesian sign convention, from which reference point are all distances measured for a spherical mirror? easy
Step solution + source
The convention fixes the origin at the pole (P) for mirrors and the optical centre for lenses. All object, image and focal distances are measured from this origin along the principal axis, with sign set by whether the measurement is along or against the incident light. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q3 Heights of objects or images measured upwards, perpendicular to the principal axis, are taken as: easy
Step solution + source
Heights measured upward with respect to the x-axis (principal axis) are positive and those measured downward are negative. Hence an erect image has positive height while an inverted image has negative height, which fixes the sign of the magnification $m=h'/h$. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q4 For a real object placed in front of a mirror or lens, the object distance $u$ carries which sign? easy
Step solution + source
Light travels from the object to the mirror/lens; this is the positive direction. To reach the object from the pole one moves opposite to the incident light, so the real-object distance is negative: $u \lt 0$. This is why the mirror equation uses $BP=-u$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q5 In the Cartesian sign convention, the focal length of a concave mirror is: medium
Step solution + source
The focus of a concave mirror lies in front of the mirror, on the same side as the object. Measured from the pole opposite to the incident light direction, the focal length is negative ($f \lt 0$), whereas for a convex mirror the focus is behind and $f$ is positive. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q6 A convex (converging) lens has a positive focal length. What sign does the focal length of a concave (diverging) lens take? medium
Step solution + source
By convention $f$ is positive for a converging lens and negative for a diverging lens, so the power $P=1/f$ is positive for convex and negative for concave lenses. A prescription of $-4.0$ D therefore denotes a concave lens of focal length $-25$ cm. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q7 An erect object of height $2$ cm forms an inverted image of height $4$ cm. Using the sign convention, the ratio $h'/h$ equals: hard
Step solution + source
Object height is measured upward, so $h=+2$ cm; the inverted image points downward, so $h'=-4$ cm. Thus $h'/h = (-4)/(+2) = -2$. The negative magnification correctly signals a real, inverted image, consistent with the Cartesian convention. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q8 In reflection at any surface, the angle of incidence and angle of reflection are equal, both being measured with respect to: easy
Step solution + source
The law of reflection states the angle between the incident ray and the normal equals the angle between the reflected ray and the normal, and both rays plus the normal lie in one plane. For a curved mirror the normal is along the radius through the point of incidence. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q9 For a spherical mirror, the focal length $f$ and the radius of curvature $R$ are related by: easy
Step solution + source
Using paraxial geometry, $\tan\theta\approx\theta$ and $\tan 2\theta\approx 2\theta$, the point where reflected rays meet lies at half the radius from the pole, giving $f = R/2$. The focus is midway between the pole and the centre of curvature. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q10 The normal to a spherical mirror at the point of incidence is directed along: easy
Step solution + source
For a spherical surface the normal is perpendicular to the tangent plane at the point of incidence, which means it lies along the radius joining that point to the centre of curvature C. This is why $CM$ is drawn perpendicular to the mirror at M in the derivation of $f=R/2$. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q11 A concave mirror has a radius of curvature of magnitude $20$ cm. Its focal length is: medium
Step solution + source
The magnitude of the focal length is $|f| = R/2 = 20/2 = 10$ cm. Because a concave mirror's focus lies in front of it (opposite to incident-light direction), the sign convention makes $f = -10$ cm. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q12 A parallel paraxial beam is incident on a convex mirror. After reflection the rays: medium
Step solution + source
For a convex mirror the reflected rays diverge as if coming from a point F behind the mirror; this is a virtual focus. Consequently the convex mirror always forms virtual, erect, diminished images, and its focal length is taken positive. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q13 The relation $f=R/2$ for a spherical mirror is strictly valid only under the paraxial approximation because: hard
Step solution + source
In the derivation, $\tan\theta\approx\theta$ and $\tan 2\theta\approx 2\theta$ hold only for small angles, i.e. paraxial rays incident close to the pole. Then FD equals f and CD equals R, giving $f=R/2$. Wide-aperture rays deviate, producing spherical aberration. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q14 A ray travelling parallel to the principal axis strikes a concave mirror. After reflection it: hard
Step solution + source
One of the standard construction rays: a ray parallel to the principal axis reflects through the focus of a concave mirror. Conversely, a ray through the focus reflects parallel to the axis, and a ray through the centre of curvature retraces its path. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q15 The mirror equation relating object distance $u$, image distance $v$ and focal length $f$ is: easy
Step solution + source
The mirror equation is $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, derived from similar triangles with the sign convention applied. Note the plus sign, which distinguishes it from the thin-lens formula that carries a minus sign between the reciprocals. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q16 The linear magnification produced by a spherical mirror is given by: easy
Step solution + source
From similar triangles and the sign convention, $m=h'/h=-v/u$. A negative $m$ indicates a real, inverted image while a positive $m$ indicates a virtual, erect image. The magnitude gives the size ratio of image to object. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q17 An object is placed $10$ cm in front of a concave mirror of radius of curvature $15$ cm. The image distance is: easy
Step solution + source
Here $f=-R/2=-7.5$ cm and $u=-10$ cm. From $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-7.5}-\frac{1}{-10}=-0.1333+0.1=-0.0333$, so $v=-30$ cm. The image is real, formed 30 cm in front of the mirror. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q18 For an object at $u=-10$ cm before a concave mirror of $f=-7.5$ cm (image at $v=-30$ cm), the magnification is: medium
Step solution + source
Using $m=-v/u=-(-30)/(-10)=-3$. The magnitude 3 means the image is three times the object size, and the negative sign shows it is real and inverted, matching the concave-mirror result for an object between $f$ and $C$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q19 An object is placed $5$ cm in front of a concave mirror of focal length $-7.5$ cm. The image is formed at: medium
Step solution + source
With $u=-5$ cm and $f=-7.5$ cm, $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-7.5}-\frac{1}{-5}=-0.1333+0.2=0.0667$, giving $v=+15$ cm. The object lies between the pole and focus, so the image is virtual, erect and magnified, formed behind the mirror. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q20 A convex side-view mirror has radius of curvature $2$ m ($f=+1$ m). A jogger is at $u=-39$ m. The image distance is approximately: hard
Step solution + source
Using $v=\dfrac{fu}{u-f}=\dfrac{(1)(-39)}{-39-1}=\dfrac{-39}{-40}=+\dfrac{39}{40}\approx0.975$ m. The positive value confirms a virtual image behind the convex mirror, diminished and erect, located between the pole and the focus. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q21 Using the mirror equation, an object placed between $f$ and $2f$ of a concave mirror produces an image that is: advanced
Step solution + source
For a concave mirror $f\lt 0$; taking $2f\lt u\lt f$ and solving $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ gives $v\lt 2f$ in magnitude, i.e. beyond the centre of curvature, with $|v|\gt|u|$. The magnification $m=-v/u$ is negative and magnitude greater than one, so the image is real, inverted and enlarged beyond $2f$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q22 Snell's law of refraction is expressed as: easy
Step solution + source
Snell's law states the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant equal to the relative refractive index $n_{21}$ of the second medium with respect to the first. Angles are measured from the normal. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q23 When light passes from a rarer medium into a denser medium, the refracted ray: easy
Step solution + source
If $n_{21}\gt 1$ (medium 2 optically denser), then $r\lt i$, so the ray bends towards the normal on entering the denser medium. Conversely, going into a rarer medium the ray bends away from the normal. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q24 If the relative refractive index $n_{21}$ is greater than one, then compared with the angle of incidence $i$, the angle of refraction $r$ satisfies: easy
Step solution + source
From $\sin i/\sin r = n_{21}$, if $n_{21}\gt 1$ then $\sin r \lt \sin i$, hence $r \lt i$. The ray bends towards the normal, which is the defining behaviour when entering an optically denser medium. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q25 If $n_{21}$ is the refractive index of medium 2 with respect to medium 1, then the refractive index $n_{12}$ of medium 1 with respect to medium 2 is: medium
Step solution + source
By the reversibility of light and definition of relative index, $n_{12}=1/n_{21}$. For example, if glass with respect to air is $1.5$, then air with respect to glass is $1/1.5\approx0.667$. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q26 Light travels from air ($n_1=1$) into a medium with $n_2=\sqrt{2}$ at an angle of incidence $i=45^\circ$. The angle of refraction is: medium
Step solution + source
Applying $n_1\sin i = n_2\sin r$: $1\cdot\sin 45^\circ = \sqrt{2}\,\sin r$, so $\sin r = \dfrac{1/\sqrt2}{\sqrt2}=\dfrac{1}{2}$, giving $r=30^\circ$. The ray bends towards the normal since it enters a denser medium. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q27 A tank is filled with water to a real depth of $12.5$ cm; the bottom appears raised so that the apparent depth is $9.4$ cm. The refractive index of water is approximately: hard
Step solution + source
For near-normal viewing, refractive index $n=\dfrac{\text{real depth}}{\text{apparent depth}}=\dfrac{12.5}{9.4}\approx1.33$. The apparent depth equals the real depth divided by the refractive index of the medium, which is why submerged objects look shallower. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q28 A ray passes from water ($n_w=1.33$) into glass ($n_g=1.5$). The refractive index of glass with respect to water is approximately: advanced
Step solution + source
Using the chaining rule with air as reference, $n_{g,w}=\dfrac{n_g}{n_w}=\dfrac{1.5}{1.33}\approx1.13$. Equivalently $n_{32}=n_{31}\times n_{12}$, so the relative index between two media equals the ratio of their absolute indices with respect to a common medium. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q29 Total internal reflection can occur only when light travels: easy
Step solution + source
TIR requires the light to be in the denser medium heading towards the rarer one, and the angle of incidence must be larger than the critical angle. Below the critical angle refraction still occurs; only above it does all light reflect back with no transmission. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q30 The critical angle for a pair of media is defined as the angle of incidence for which the angle of refraction is: easy
Step solution + source
At the critical angle $i_c$, the refracted ray grazes the interface, meaning the angle of refraction is $90^\circ$. From Snell's law $\sin i_c = n_{21}$ (rarer with respect to denser). For any incidence greater than $i_c$ no refraction is possible. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q31 The refractive index of a denser medium with respect to a rarer medium, in terms of the critical angle $i_c$, is: easy
Step solution + source
Since $\sin i_c = n_{21}$ is the index of the rarer medium with respect to the denser, the index of the denser with respect to the rarer is its reciprocal, $n_{12}=1/\sin i_c$. A smaller critical angle therefore implies a higher refractive index, as for diamond. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q32 Among water ($n=1.33$), crown glass ($n=1.52$) and diamond ($n=2.42$), which has the smallest critical angle with respect to air? medium
Step solution + source
Since $\sin i_c = 1/n$, a larger refractive index yields a smaller critical angle. Diamond with $n=2.42$ gives $i_c\approx24.4^\circ$, the smallest of the three, which is why diamonds show extensive total internal reflection and brilliant sparkle. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q33 For a medium of refractive index $1.5$ (relative to air), the critical angle is approximately: medium
Step solution + source
$\sin i_c = 1/n = 1/1.5 = 0.667$, so $i_c = \sin^{-1}(0.667)\approx41.8^\circ$. Rays striking the glass-air interface at angles greater than this value undergo total internal reflection, as exploited in totally reflecting prisms. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q34 A prism used to bend light by $90^\circ$ or $180^\circ$ by total internal reflection requires that the critical angle of its material be: hard
Step solution + source
In such right-angled prisms light strikes the hypotenuse face at $45^\circ$. For total internal reflection this must exceed the critical angle, so $i_c$ must be less than $45^\circ$. Both crown glass ($41^\circ$) and dense flint glass ($37^\circ$) satisfy this. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q35 In an optical fibre the core has refractive index $1.68$ and the cladding $1.44$. The critical angle for total internal reflection at the core-cladding interface is approximately: advanced
Step solution + source
At the core-cladding boundary $\sin i_c = n_{clad}/n_{core}=1.44/1.68\approx0.857$, so $i_c\approx\sin^{-1}(0.857)\approx59^\circ$. Rays hitting the wall at angles greater than this are guided along the fibre by repeated total internal reflection with negligible loss. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q36 The equation governing refraction at a single spherical surface (from medium 1 to medium 2) is: easy
Step solution + source
Applying Snell's law in the small-angle limit and the sign convention $OM=-u$, $MI=+v$, $MC=+R$, the geometry gives $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$. This relation holds for any curved spherical refracting surface. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q37 The refraction-at-a-spherical-surface formula relates the object and image distances in terms of: easy
Step solution + source
Equation $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ shows the image position depends on both refractive indices $n_1,n_2$ and the radius of curvature $R$ of the surface. It is the building block from which the lens maker's formula is derived. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q38 Light from a point source in air ($n_1=1$) falls on a convex spherical glass surface ($n_2=1.5$, $R=+20$ cm) with the source $100$ cm away ($u=-100$ cm). The image forms at: medium
Step solution + source
Using $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$: $\dfrac{1.5}{v}-\dfrac{1}{-100}=\dfrac{0.5}{20}=0.025$. So $\dfrac{1.5}{v}=0.025-0.01=0.015$, giving $v=+100$ cm. The real image lies in the glass, along the incident-light direction. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q39 In deriving the spherical-surface refraction formula, Snell's law $n_1\sin i = n_2\sin r$ reduces for paraxial (small) angles to: medium
Step solution + source
For small angles $\sin\theta\approx\theta$, so Snell's law becomes $n_1 i = n_2 r$. Substituting the geometric expressions for $i$ and $r$ in terms of the distances $OM$, $MI$, $MC$ leads directly to the spherical-surface refraction equation. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q40 The relation $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ is applicable to: medium
Step solution + source
The equation was derived using only the small-angle approximation and the sign convention, without assuming concave or convex. Hence it holds for any single spherical refracting surface, and applying it twice at the two surfaces of a lens yields the lens maker's formula. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q41 A spherical surface separates air ($n_1=1$) from glass ($n_2=1.5$) and is convex towards the object with $R=+30$ cm. For an object at infinity, the image forms at: hard
Step solution + source
For $u\to\infty$ the term $n_1/u\to 0$, so $\dfrac{n_2}{v}=\dfrac{n_2-n_1}{R}$, giving $\dfrac{1.5}{v}=\dfrac{0.5}{30}=0.01667$, hence $v=1.5/0.01667=90$ cm. This is the second focal distance of the refracting surface in the glass. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q42 As the radius of curvature of a refracting surface tends to infinity ($R\to\infty$, a plane surface), the formula reduces to $\dfrac{n_2}{v}=\dfrac{n_1}{u}$. Viewing an object in a denser medium of index $n$ from air, the apparent depth is: hard
Step solution + source
With $R\to\infty$ the right side vanishes, giving $\dfrac{n_2}{v}=\dfrac{n_1}{u}$. For an observer in air viewing depth in a medium of index $n$, this yields apparent depth $=\text{real depth}/n$, consistent with the bottom of a water tank appearing raised. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q43 The lens maker's formula for a thin lens in air is: easy
Step solution + source
Applying the spherical-surface formula at both faces of a thin lens and adding gives the lens maker's formula $\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, where $n$ is the refractive index of the lens relative to the surrounding medium. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q44 The thin lens formula relating $u$, $v$ and $f$ is: easy
Step solution + source
The thin lens formula is $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ (note the minus sign, unlike the mirror equation). It is valid for both convex and concave lenses and for real and virtual images when the sign convention is applied. 🔉⇢

Source: NCERT Ch 9 (in-text)

Q45 A double convex lens has radii $R_1=+10$ cm and $R_2=-15$ cm and focal length $12$ cm (in air). The refractive index of the glass is: medium
Step solution + source
$\dfrac{1}{12}=(n-1)\left(\dfrac{1}{10}-\dfrac{1}{-15}\right)=(n-1)\left(\dfrac{1}{10}+\dfrac{1}{15}\right)=(n-1)\dfrac{1}{6}$. Thus $n-1=\dfrac{6}{12}=0.5$, giving $n=1.5$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q46 A glass lens has focal length $0.5$ m. Its power is: medium
Step solution + source
Power $P=\dfrac{1}{f}$ with $f$ in metres, so $P=\dfrac{1}{0.5}=+2$ dioptre. The positive sign indicates a converging (convex) lens. One dioptre equals one inverse metre ($1\,\text{D}=1\,\text{m}^{-1}$). 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q47 A double convex lens is made from glass of refractive index $1.55$ with both faces of the same radius of curvature. For a focal length of $20$ cm, the required radius is: hard
Step solution + source
For an equiconvex lens $R_1=+R$, $R_2=-R$, so $\dfrac{1}{f}=(n-1)\dfrac{2}{R}$. Thus $R=2(n-1)f=2(0.55)(20)=22$ cm. The equal-radius symmetric design gives this single value. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q48 A convex lens of focal length $30$ cm is placed in contact with a concave lens of focal length $20$ cm. The effective focal length of the combination is: hard
Step solution + source
For thin lenses in contact $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}=\dfrac{1}{30}+\dfrac{1}{-20}=\dfrac{2-3}{60}=-\dfrac{1}{60}$. Hence $f=-60$ cm, so the system behaves as a diverging lens because the concave lens dominates. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q49 A convex lens of glass ($n=1.5$) has focal length $20$ cm in air. When fully immersed in water ($n=1.33$), its focal length becomes approximately: advanced
Step solution + source
The focal length scales as $f\propto\dfrac{1}{(n_{rel}-1)}$. In air $n_{rel}-1=0.5$; in water $n_{rel}=1.5/1.33=1.128$, so $n_{rel}-1=0.128$. Thus $f_{water}=f_{air}\times\dfrac{0.5}{0.128}\approx20\times3.91\approx78.2$ cm. The lens converges more weakly in water. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q50 In the Cartesian sign convention, the correct thin lens formula relating object distance $u$, image distance $v$ and focal length $f$ is: easy
Step solution + source
For a thin lens the correct relation is $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, unlike the mirror equation which adds the reciprocals. This single formula, with the Cartesian sign convention applied, is valid for both convex and concave lenses and for real as well as virtual images. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q51 An object is placed 30 cm in front of a convex lens of focal length 15 cm. The image is formed at: easy
Step solution + source
Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with $u=-30$ cm and $f=+15$ cm: $\frac{1}{v}=\frac{1}{15}+\frac{1}{-30}=\frac{1}{30}$, so $v=+30$ cm. The object at $2f$ gives a real, inverted, same-size image at $2f$ on the opposite side. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q52 The linear magnification produced by a thin lens is correctly given by: easy
Step solution + source
For a lens $m=\frac{h'}{h}=\frac{v}{u}$ (note the mirror uses $m=-v/u$). With the sign convention, $m$ is positive for an erect virtual image and negative for an inverted real image formed by a convex or concave lens. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q53 A concave (diverging) lens, for a real object placed anywhere, always forms an image that is: easy
Step solution + source
A diverging lens has $f\lt 0$. For any real object the lens formula $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$ yields $v$ negative and $|v|\lt|u|$, so the image is virtual, erect and diminished, lying on the same side as the object. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q54 An object is placed 15 cm in front of a concave lens of focal length 10 cm. The image distance is: medium
Step solution + source
With $f=-10$ cm and $u=-15$ cm, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{10}-\frac{1}{15}=-\frac{5}{30}=-\frac{1}{6}$, giving $v=-6$ cm. The negative value confirms a virtual, erect, diminished image on the object side. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q55 For a convex lens of focal length 20 cm, an object placed at $2f$ (40 cm) produces an image that is: medium
Step solution + source
With $u=-40$ cm and $f=+20$ cm, $\frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$, so $v=+40$ cm. Then $m=\frac{v}{u}=\frac{40}{-40}=-1$: a real, inverted image of equal size at $2f$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q56 If an object is placed exactly at the first focus of a converging lens, the image is formed: medium
Step solution + source
With $u=-f$, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{f}-\frac{1}{f}=0$, so $v\to\infty$. Rays emerge parallel, giving an image at infinity — the reverse of parallel rays converging to the focus. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q57 A convex lens of focal length 10 cm forms a real image magnified two times ($|m|=2$). The object distance is: hard
Step solution + source
A real image is inverted so $m=\frac{v}{u}=-2\Rightarrow v=-2u$. Then $\frac{1}{v}-\frac{1}{u}=\frac{1}{-2u}-\frac{1}{u}=-\frac{3}{2u}=\frac{1}{10}$, giving $u=-15$ cm and $v=+30$ cm. The object is 15 cm from the lens. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q58 A beam of light converges to a point P. A convex lens of focal length 20 cm is placed in the path of the converging beam 12 cm from P. The beam now converges at a distance of: advanced
Step solution + source
P acts as a virtual object, so $u=+12$ cm. With $f=+20$ cm, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}+\frac{1}{12}=\frac{8}{60}=\frac{2}{15}$, so $v=7.5$ cm. The beam converges 7.5 cm from the lens on the far side. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q59 The power of a thin lens of focal length 0.5 m is: easy
Step solution + source
Power $P=\frac{1}{f}$ with $f$ in metres. Here $P=\frac{1}{0.5}=+2$ D. The positive sign indicates a converging (convex) lens; 1 dioptre equals $1\,\text{m}^{-1}$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q60 The SI unit for the power of a lens is the: easy
Step solution + source
The power of a lens is $P=\frac{1}{f}$ and its SI unit is the dioptre (D), where $1\,\text{D}=1\,\text{m}^{-1}$. A lens of focal length one metre has a power of one dioptre. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q61 The power of a converging (convex) lens is taken as: easy
Step solution + source
Since $P=\frac{1}{f}$ and a convex lens has $f\gt 0$, its power is positive. A diverging (concave) lens has $f\lt 0$ and hence negative power. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q62 Two thin lenses of powers $+3$ D and $+2$ D are placed in contact. The power of the combination is: easy
Step solution + source
For thin lenses in contact powers add algebraically: $P=P_1+P_2=3+2=+5$ D. Equivalently $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$, giving a net converging lens of focal length 20 cm. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q63 A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. The combination is: medium
Step solution + source
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{30}+\frac{1}{-20}=\frac{2-3}{60}=-\frac{1}{60}$, so $f=-60$ cm. The negative value shows the system behaves as a diverging lens. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q64 A lens has a power of $-4.0$ D. Its focal length and nature are: medium
Step solution + source
$f=\frac{1}{P}=\frac{1}{-4.0}=-0.25$ m $=-25$ cm. The negative focal length identifies it as a concave (diverging) lens, exactly as an optician's $-4.0$ D prescription implies. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q65 Two thin lenses of powers $+5$ D and $-2$ D are placed in contact. The net power and equivalent focal length are: hard
Step solution + source
Powers add: $P=5+(-2)=+3$ D. Then $f=\frac{1}{P}=\frac{1}{3}\,\text{m}\approx+33.3$ cm. The positive net power means the combination converges light despite one diverging component. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q66 A convex lens of focal length 30 cm and a concave lens of focal length 20 cm are placed coaxially 8.0 cm apart. Using $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}$, the effective focal length is: hard
Step solution + source
With $f_1=+30$, $f_2=-20$, $d=8$: $\frac{1}{f}=\frac{1}{30}-\frac{1}{20}-\frac{8}{(30)(-20)}=-\frac{1}{60}+\frac{8}{600}=-\frac{2}{600}=-\frac{1}{300}$, so $f=-300$ cm. Separation changes the result compared to lenses in contact. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q67 A convex lens of power $+5$ D is combined in contact with a concave lens so that the net power is $+2$ D. The power and focal length of the concave lens are: advanced
Step solution + source
$P=P_1+P_2\Rightarrow P_2=P-P_1=2-5=-3$ D. Focal length $f_2=\frac{1}{P_2}=\frac{1}{-3}\,\text{m}\approx-33.3$ cm. The negative power confirms the second lens is concave (diverging). 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q68 For a prism of refracting angle $A$, the angles of refraction $r_1$ and $r_2$ at its two faces satisfy: easy
Step solution + source
From the quadrilateral geometry, $\angle A+\angle QNR=180^\circ$, and in triangle QNR, $r_1+r_2+\angle QNR=180^\circ$. Comparing gives $r_1+r_2=A$, a key relation for prism problems. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q69 The angle of deviation $d$ produced by a prism, in terms of incidence angle $i$, emergence angle $e$ and prism angle $A$, is: easy
Step solution + source
The total deviation is the sum at both faces: $d=(i-r_1)+(e-r_2)$. Using $r_1+r_2=A$ gives $d=i+e-A$. Note the symmetry in $i$ and $e$, so the deviation is unchanged if they are swapped. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q70 Dispersion of light through a prism is best described as: easy
Step solution + source
Dispersion is the splitting of white light into its constituent colours. It occurs because the refractive index of the prism material varies with wavelength, so different colours deviate by different amounts, as noted in the chapter summary. The angular dispersion between violet and red for a thin prism is $\theta=(\mu_v-\mu_r)A$, and the dispersive power is $\omega=\dfrac{\mu_v-\mu_r}{\mu-1}$, so a material whose index varies more with wavelength spreads the colours more. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q71 At the position of minimum deviation for a prism, the refracted ray inside the prism: medium
Step solution + source
At minimum deviation $D_m$, $i=e$ and hence $r_1=r_2=A/2$. Symmetry makes the refracted ray inside the prism run parallel to the base, which is the standard condition used to derive $n=\frac{\sin[(A+D_m)/2]}{\sin(A/2)}$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q72 A prism of refracting angle $60^\circ$ has a minimum deviation of $30^\circ$. Its refractive index is: medium
Step solution + source
$n=\frac{\sin[(A+D_m)/2]}{\sin(A/2)}=\frac{\sin 45^\circ}{\sin 30^\circ}=\frac{0.707}{0.5}=1.414$. This is close to the refractive index of ordinary crown glass for such a symmetric prism. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q73 For a thin prism of small refracting angle $6^\circ$ made of glass ($n=1.5$), the angle of minimum deviation is: medium
Step solution + source
For a thin prism $D_m=(n-1)A=(1.5-1)\times 6^\circ=0.5\times 6^\circ=3^\circ$. This shows that thin prisms do not deviate light much, a result used in spectrometry and prism binoculars. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q74 A prism of refracting angle $60^\circ$ gives a minimum deviation of $40^\circ$. The refractive index of its material is about: hard
Step solution + source
$n=\frac{\sin[(A+D_m)/2]}{\sin(A/2)}=\frac{\sin[(60+40)/2]}{\sin 30^\circ}=\frac{\sin 50^\circ}{0.5}=\frac{0.766}{0.5}\approx1.532$. This matches the NCERT worked exercise for an unknown glass prism. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q75 For a prism of angle $60^\circ$, the minimum deviation is measured to be $48^\circ$. The angle of incidence at minimum deviation is: hard
Step solution + source
At minimum deviation $D_m=2i-A$, so $i=\frac{A+D_m}{2}=\frac{60^\circ+48^\circ}{2}=54^\circ$. Here $i=e$, and the refracted ray inside the prism is parallel to the base. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q76 A ray is incident on a prism of refracting angle $60^\circ$ and refractive index $1.524$ such that it just suffers total internal reflection at the second face. The required angle of incidence at the first face is about: advanced
Step solution + source
Critical angle: $\sin i_c=\frac{1}{1.524}=0.656$, so $i_c\approx41.04^\circ$. For grazing TIR, $r_2=i_c$, and $r_1=A-r_2=60^\circ-41.04^\circ=18.96^\circ$. Then $\sin i=n\sin r_1=1.524\times\sin 18.96^\circ\approx0.495$, giving $i\approx29.75^\circ$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q77 A simple microscope (magnifying glass) is essentially: easy
Step solution + source
A simple magnifier is a single converging (convex) lens of small focal length. Held within one focal length of the object, it produces an erect, magnified, virtual image that lets the object be brought closer than the near point $D$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q78 When the final image is formed at the near point $D$, the magnifying power of a simple microscope is: easy
Step solution + source
For the image at the near point, $m=1+\frac{D}{f}$, where $D\approx25$ cm. This is one greater than the value for image at infinity because the eye is more strained but the angular size is maximum. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q79 For a simple microscope forming the image at infinity, the angular magnification is: easy
Step solution + source
When the image is at infinity the object is at the focus and $m=\frac{D}{f}$. This is one less than $1+\frac{D}{f}$ (image at near point), but the relaxed-eye viewing is more comfortable. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q80 A simple microscope of focal length 5 cm forms its image at the near point (25 cm). Its magnifying power is: medium
Step solution + source
$m=1+\frac{D}{f}=1+\frac{25}{5}=6$. Thus a convex lens of 5 cm focal length gives a magnification of six when the image is viewed at the near point, exactly as stated in the chapter. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q81 For the same 5 cm focal length magnifier, the magnifying power when the image is formed at infinity is: medium
Step solution + source
For image at infinity, $m=\frac{D}{f}=\frac{25}{5}=5$. This is exactly one less than the near-point value ($1+D/f=6$), the difference being small but the relaxed-eye view more comfortable. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q82 A magnifying glass of focal length 9 cm is held close to the eye and the card sheet is at 9 cm (image at infinity). The angular magnification is about: hard
Step solution + source
With the object at the focus the image is at infinity, so $m=\frac{D}{f}=\frac{25}{9}\approx2.8$. This is the magnifying power (angular), distinct from the areal magnification of the virtual image. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q83 Using a magnifier of focal length 9 cm for maximum magnifying power (image at the near point, $D=25$ cm), the object should be placed at, and the magnifying power is: hard
Step solution + source
Maximum power occurs with the image at $D$: $m=1+\frac{D}{f}=1+\frac{25}{9}\approx3.8$. Object distance from $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with $v=-25$: $\frac{1}{u}=-\frac{1}{25}-\frac{1}{9}=-\frac{34}{225}$, so $u\approx-6.6$ cm. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q84 A simple magnifier gives an angular magnification of 5 when the image is at infinity ($D=25$ cm). Its focal length and the magnification at the near point are: advanced
Step solution + source
Image at infinity: $m=\frac{D}{f}=5\Rightarrow f=\frac{25}{5}=5$ cm. At the near point $m=1+\frac{D}{f}=1+\frac{25}{5}=6$. The near-point value always exceeds the infinity value by exactly one. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q85 In a compound microscope, the objective lens forms an image that is: easy
Step solution + source
The objective (nearest the object) forms a real, inverted, magnified image that serves as the object for the eyepiece. The eyepiece then acts like a simple magnifier producing the final enlarged virtual image. The total magnification is the product $M=M_o\times M_e\approx-\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$, where $L$ is the tube length and $D=25\ \text{cm}$ the near point. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q86 The total magnification of a compound microscope is given by: easy
Step solution + source
Because the image of the objective is the object of the eyepiece, magnifications multiply: $m=m_o\times m_e$. This follows from the general result that total magnification of a lens combination is the product of the individual magnifications. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q87 In a compound microscope, the eyepiece functions essentially like a: easy
Step solution + source
The eyepiece views the real image formed by the objective and, acting as a simple magnifier, produces a further enlarged virtual image. Its angular magnification is $1+\frac{D}{f_e}$ (near point) or $\frac{D}{f_e}$ (infinity). 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q88 A compound microscope has $f_o=1.0$ cm, $f_e=2.0$ cm, tube length $L=20$ cm and $D=25$ cm. Its approximate magnifying power (final image at infinity) is: medium
Step solution + source
$m=\frac{L}{f_o}\times\frac{D}{f_e}=\frac{20}{1.0}\times\frac{25}{2.0}=20\times12.5=250$. Small objective and eyepiece focal lengths and a long tube give large magnification, as noted in the chapter example. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q89 The final image seen through a compound microscope, relative to the original object, is: medium
Step solution + source
The objective forms an inverted real image; the eyepiece magnifies it without re-inverting. Hence the final virtual image is inverted with respect to the original object, as the chapter explicitly states. The objective magnification $m_o=-\dfrac{v_o}{u_o}\lt 0$ is negative (image inverted); the eyepiece does not re-invert, so the final image stays inverted relative to the object. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q90 To obtain a large magnification with a compound microscope, the objective and eyepiece should have: medium
Step solution + source
Since $m=\frac{L}{f_o}\cdot\frac{D}{f_e}$, both $f_o$ and $f_e$ appear in the denominator, so making them small increases magnification. In practice it is hard to make focal lengths much smaller than about 1 cm. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q91 A compound microscope has objective $f_o=2.0$ cm and eyepiece $f_e=6.25$ cm separated by 15 cm, with the final image at the near point (25 cm). Its magnifying power is about: hard
Step solution + source
Eyepiece: $m_e=1+\frac{25}{6.25}=5$ and its object distance is $-5$ cm, so the objective image is $15-5=10$ cm away. Objective: $\frac{1}{u_o}=\frac{1}{10}-\frac{1}{2}=-0.4$, $u_o=-2.5$ cm, $m_o=\frac{10}{-2.5}=-4$. Thus $|m|=|m_o|\times m_e=4\times5=20$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q92 A compound microscope has objective $f_o=8.0$ mm and eyepiece $f_e=2.5$ cm; an object at 9.0 mm from the objective gives a final image at the near point (25 cm). The lens separation and magnifying power are about: advanced
Step solution + source
Objective: $\frac{1}{v_o}=\frac{1}{0.8}-\frac{1}{0.9}=0.139$, $v_o=7.2$ cm, $m_o=\frac{7.2}{-0.9}=-8$. Eyepiece for image at $-25$: $\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{2.5}=-0.44$, $u_e=-2.27$ cm; separation $=7.2+2.27\approx9.5$ cm. $m_e=1+\frac{25}{2.5}=11$, so $|m|=8\times11=88$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q93 The magnifying power of a telescope in normal adjustment (final image at infinity) is: easy
Step solution + source
The magnifying power of a telescope equals the ratio of the angle subtended by the image to that subtended by the object, $m=\frac{\beta}{\alpha}=\frac{f_o}{f_e}$. A long objective focal length and short eyepiece focal length give high power. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q94 In an astronomical (refracting) telescope, the objective has: easy
Step solution + source
The telescope objective has a large focal length and a much larger aperture than the eyepiece. The large aperture increases light-gathering power and resolving power, both of which improve with objective diameter. A larger objective diameter $D$ lowers the diffraction limit $\Delta\theta=\dfrac{1.22\lambda}{D}$, so resolving power $\dfrac{1}{\Delta\theta}$ rises with aperture, and light-gathering scales as $D^2$. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q95 The length of a telescope tube in normal adjustment is: easy
Step solution + source
In normal adjustment the real image from the objective lies at the common focal point, so the tube length equals $f_o+f_e$. The objective's second focus coincides with the eyepiece's first focus. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q96 A telescope has an objective of focal length 144 cm and an eyepiece of focal length 6.0 cm. Its magnifying power and tube length (normal adjustment) are: medium
Step solution + source
$m=\frac{f_o}{f_e}=\frac{144}{6.0}=24$; tube length $=f_o+f_e=144+6=150$ cm. These are the standard results for a small refracting telescope in normal adjustment. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q97 A refracting telescope has an objective of focal length 100 cm and an eyepiece of focal length 1 cm. Its magnifying power is: medium
Step solution + source
$m=\frac{f_o}{f_e}=\frac{100}{1}=100$. A pair of stars separated by $1'$ would then appear separated by about $100'\approx1.67^\circ$, illustrating the angular magnification of a telescope. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q98 A telescope has objective $f_o=140$ cm and eyepiece $f_e=5.0$ cm. When the final image is formed at the near point (25 cm), the magnifying power is about: hard
Step solution + source
For the image at the near point, $m=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=\frac{140}{5}\left(1+\frac{5}{25}\right)=28\times1.2=33.6$. This exceeds the normal-adjustment value of 28 because the eye is accommodating. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q99 Modern large astronomical telescopes use a concave mirror rather than a lens for the objective mainly because: hard
Step solution + source
A mirror shows no chromatic aberration (no refraction through glass), weighs less than an equivalent lens, and can be supported over its entire back surface rather than only at the rim. Hence reflecting telescopes dominate for large apertures. A reflecting telescope keeps the angular magnification $M=-\dfrac{f_o}{f_e}$ while avoiding chromatic aberration, since reflection at a mirror is wavelength-independent (no $\mu(\lambda)$ term). 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

Q100 A giant refracting telescope has an objective of focal length 15 m and an eyepiece of focal length 1.0 cm. Used to view the Moon (diameter $3.48\times10^6$ m, orbit radius $3.8\times10^8$ m), its angular magnification and the diameter of the Moon's image at the objective are: advanced
Step solution + source
$m=\frac{f_o}{f_e}=\frac{1500\,\text{cm}}{1.0\,\text{cm}}=1500$. The Moon subtends $\theta=\frac{3.48\times10^6}{3.8\times10^8}\approx9.16\times10^{-3}$ rad, so image diameter $=f_o\theta=1500\times9.16\times10^{-3}\approx13.7$ cm. 🔉⇢

Source: JEE-pattern (NCERT Ch 9)

⏱️ Mock Test 30 Q · 60 min · +4 correct · −1 wrong

Rules: 60 minutes, +4 for a correct answer and −1 for a wrong one; the timer may be paused once if you need a short break.

🎬 Video Lectures clip-indexed · NPTEL / IIT / MIT

MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.

Optics: Reflection of Light and Formation of Images (CH_22) 🔉⇢
CH 22: IIT Delhi: Physics [ IIT-PAL]

👁 Observe: Ray-diagram rules for image formation by spherical mirrors, from an IIT faculty.

📚 Teaches: reflection-spherical-mirrors

📑 Clips (4)
  • 0:43–5:45Laws of reflection: angle of incidence equals angle of reflectionStates the two laws of reflection using incident ray, normal and reflected ray.reflection-of-light
  • 5:45–15:51Concave and convex spherical mirrors and reflection at the poleExplains spherical-mirror geometry and applies the law of reflection at the pole and general points.reflection-spherical-mirrors
  • 15:51–20:52Principal focus: where parallel rays converge after reflectionShows parallel paraxial rays reflect through the principal focus for concave mirrors and diverge for convex mirrors.spherical-mirrors
  • 20:52–25:00Paraxial (small-aperture) approximation for spherical mirrorsDefines paraxial rays close to the axis and explains why the small-aperture approximation is needed.reflection-spherical-mirrors
Total internal reflection : Ray Optics and Optical Instruments (CH_22) 🔉⇢
CH 22: IIT Delhi: Physics [ IIT-PAL]

👁 Observe: Critical-angle derivation and TIR applications from an IIT-PAL lecture.

📚 Teaches: total-internal-reflection

📑 Clips (4)
  • 0:43–8:45Deriving the critical angle for total internal reflectionUses Snell's law to show a grazing refracted ray defines the critical angle for internal reflection.total-internal-reflection
  • 8:45–13:15Critical-angle values and retro-reflecting prismsGives critical angles for glass, water and diamond and shows prisms using TIR as beam deflectors.prism-total-internal-reflection
  • 13:15–20:50Total internal reflection in optical fibresApplies repeated total internal reflection at core-cladding interfaces to guide light along a fibre.total-internal-reflection
  • 20:50–23:20Optical-fibre construction, core/cladding and applicationsDescribes doped-silica core and cladding materials and the communication applications of fibres.total-internal-reflection
Introduction video: Fiber Optic Communication Technology 🔉⇢
NPTEL-NOC IITM

👁 Observe: How optical fibres guide light by repeated total internal reflection.

📚 Teaches: optical-fibres

📑 Clips (0)

Full lecture — no clip index.

Refraction at spherical surfaces and by lenses: Ray Optics and Optical Instruments (CH_22) 🔉⇢
CH 22: IIT Delhi: Physics [ IIT-PAL]

👁 Observe: Building from single-surface refraction toward the lens maker's formula.

📚 Teaches: refraction-spherical-surface

📑 Clips (4)
  • 0:43–10:44Refraction at a single spherical surface: setting up the geometryIntroduces the alpha-beta-gamma angles and paraxial approximation for a single spherical refracting surface.refraction-at-spherical-surface
  • 10:44–15:51Deriving the spherical-surface refraction formulaApplies Snell's law with small-angle approximations and the sign convention to relate object and image distances.refraction-at-spherical-surface
  • 15:51–20:54Worked example of refraction at a spherical surfaceWorks a numerical example locating the image formed by refraction at a glass spherical surface.refraction-at-spherical-surface
  • 20:54–25:00Extending to a lens: refraction at two spherical surfacesTreats a lens as two successive spherical surfaces and adds their equations toward the lens maker's formula.lens-makers-formula
Refraction through a prism and dispersion : Ray Optics and Optical Instruments (CH_22) 🔉⇢
CH 22: IIT Delhi: Physics [ IIT-PAL]

👁 Observe: Prism refraction and dispersion treated rigorously by IIT faculty.

📚 Teaches: dispersion

📑 Clips (4)
  • 0:43–8:16Refraction through a prism: tracing the ray in and outTraces a ray refracting at both prism faces and defines the deviation as theta1 + theta2.prism-deviation-refraction
  • 8:16–13:21Deviation d = i + e - A and two angles for the same deviationDerives d = i + e - A and argues two different incidence angles give the same deviation.prism-deviation
  • 13:21–20:56Minimum deviation: D-vs-i graph and measuring refractive indexUses the D-versus-i curve and the spectrometer to obtain the minimum deviation and refractive index.prism-minimum-deviation
  • 20:56–25:00Thin-prism formula dm = (n-1)A and a worked examplePresents the small-angle prism result dm = (n-1)A and starts a numerical example.prism-minimum-deviation
Microscopes and telescopes : Ray Optics and Optical Instruments (CH_22) 🔉⇢
CH 22: IIT Delhi: Physics [ IIT-PAL]

👁 Observe: Optical instruments compared and their magnification derived by IIT faculty.

📚 Teaches: telescope

📑 Clips (4)
  • 0:43–8:15Compound microscope: objective and eyepiece arrangementRecaps the simple microscope and introduces the two coaxial lenses of a compound microscope.compound-microscope
  • 8:15–13:18Ray diagram and image formation in a compound microscopeTraces rays through objective and eyepiece to form a magnified virtual final image.compound-microscope
  • 13:18–20:54Compound-microscope magnification and tube lengthDerives the magnification in terms of tube length L and focal lengths using similar triangles.compound-microscope
  • 20:54–25:00Introducing the telescope for distant objectsTransitions from the microscope to the telescope used to observe far-away objects.telescope
#2 Geometric Optics Basics | Optical Engineering 🔉⇢
NPTEL-NOC IITM

👁 Observe: Foundational geometric-optics concepts from an IIT Madras course.

📚 Teaches: refraction-snells-law

📑 Clips (4)
  • 0:02–5:03Postulates of geometric optics and Fermat's principleRecaps straight-line propagation and Fermat's least-time principle used to derive Snell's law.refraction-snells-law
  • 5:03–7:36The law of refraction: ni sin(theta_i) = nt sin(theta_t)States the law of refraction and notes it can be proved from Fermat's principle and wave theory.refraction-snell-law
  • 7:36–15:06Specular versus diffuse reflection at an interfaceDistinguishes mirror-like specular reflection from diffuse reflection and discusses reflection at interfaces.reflection-of-light
  • 15:06–20:00Imaging: focusing a cone of rays to a single pointExplains that an imaging system can only capture a cone of rays and must focus them to one image point.general
Lec 1 | MIT 2.71 Optics, Spring 2009 🔉⇢
MIT OpenCourseWare

👁 Observe: University-level opening lecture framing rays, reflection and refraction.

📚 Teaches: refraction-snells-law

📑 Clips (0)

Full lecture — no clip index.

But why would light "slow down"? | Visualizing Feynman's lecture on the refractive index 🔉⇢
3Blue1Brown

👁 Observe: Intuition for why the refractive index means light effectively slows in a medium.

📚 Teaches: refractive-index

📑 Clips (3)
  • 0:00–5:02Why does light slow down in glass? Setting up the real questionChallenges the standard 'light slows down' story and frames what really needs explaining.refractive-index
  • 5:02–10:04Light as a wave in the electromagnetic fieldReviews the wave nature of light and how oscillating charges emit radiation.refractive-index
  • 10:04–15:00How layered charges re-radiate and shift the wave's phaseShows re-radiation from charge layers adds a phase kick that effectively slows the wave, explaining the index.refractive-index
What They (Probably) Don't Teach You About Rainbows At School 🔉⇢
Veritasium

👁 Observe: Refraction plus dispersion inside raindrops producing the rainbow angle.

📚 Teaches: dispersion

📑 Clips (3)
  • 0:00–5:03How a single raindrop makes a rainbowUses a laser on a sphere to show how the impact parameter controls reflection and refraction inside a raindrop.refraction-through-sphere-deviation
  • 5:03–10:05Maximum scattering angle: why the rainbow ray bunches upShows a maximum deflection angle exists, causing light to concentrate and form the rainbow cone.refraction-through-sphere-deviation
  • 10:05–15:00Why different colours emerge at different angles: dispersionExplains colour-dependent scattering angles produce the separated colours of the rainbow.prism-deviation-dispersion
How does light 'know' the shortest path? 🔉⇢
Veritasium

👁 Observe: Fermat's principle as the deeper reason behind refraction and Snell's law.

📚 Teaches: refraction-snells-law

📑 Clips (1)
  • 0:00–1:17The lifeguard problem: light takes the fastest pathFrames Fermat's least-time idea via a beach rescue analogy motivating why light bends when refracting.refraction-snells-law
Spherical mirrors, radius of curvature & focal length 🔉⇢
Khan Academy India - English

👁 Observe: How R = 2f and the geometry of concave/convex spherical mirrors is set up.

📚 Teaches: reflection-spherical-mirrors

📑 Clips (3)
  • 0:00–2:30Why parallel rays from a small spherical mirror focus at one pointIntroduces spherical mirrors as small sections of a sphere and shows parallel rays converge near a single focal point.reflection-spherical-mirrors
  • 2:30–5:02Focal length is half the radius of curvature: f = R/2Derives that the focal length of a spherical mirror is half the sphere's radius, linking f and R.concave-mirror-focal-length
  • 5:02–6:00Pole, centre of curvature and focal point definedDistinguishes the pole and centre of curvature of a mirror from the geometric centre of the underlying sphere.spherical-mirrors
Derivation of the mirror equation | Geometric optics | Physics | Khan Academy 🔉⇢
Khan Academy

👁 Observe: Full derivation of 1/v + 1/u = 1/f from similar triangles.

📚 Teaches: mirror-formula

📑 Clips (3)
  • 0:00–2:30Ray tracing to locate the image in a concave mirrorUses reversible principal rays through the focal point to locate an image formed by a concave mirror.reflection-spherical-mirrors
  • 2:30–7:32Deriving the mirror equation using similar trianglesSets up similar right triangles about the axis to relate object and image distances.mirror-equation-magnification
  • 7:32–10:00Completing the mirror equation and its sign conventionCombines the triangle relations into the mirror equation and notes the positive object-distance convention.mirror-equation-magnification
Mirror equation example problems | Geometric optics | Physics | Khan Academy 🔉⇢
Khan Academy

👁 Observe: Sign-convention handling while plugging numbers into the mirror formula.

📚 Teaches: mirror-formula

📑 Clips (3)
  • 0:00–2:31The sign-convention traps in mirror-equation problemsExplains that mirror-equation problems are hard mainly because of positive/negative sign decisions.cartesian-sign-convention
  • 2:31–7:35Worked example: finding image distance and magnificationPlugs an object distance into the mirror equation and interprets the negative (inverted) image result.mirror-equation-magnification
  • 7:35–11:00Second example and computing image height from magnificationSolves a second case and uses the magnification ratio to find the height of the image.mirror-equation-magnification
Refraction and Snell's law | Geometric optics | Physics | Khan Academy 🔉⇢
Khan Academy

👁 Observe: How bending at an interface follows n1 sin i = n2 sin r.

📚 Teaches: refraction-snells-law

📑 Clips (3)
  • 0:00–5:02From reflection to refraction: bending at a boundaryRecaps reflection and introduces refraction as bending of light crossing between two media.refraction-snells-law
  • 5:02–10:04Car-in-mud analogy and light slowing in a mediumGives an intuitive analogy for why light bends when its speed changes between media.refractive-index
  • 10:04–12:00Rewriting Snell's law using v = c/n and refractive indexExpresses Snell's law in terms of refractive index by writing the medium speed as c divided by n.refraction-snells-law
Snell's law example 1 | Geometric optics | Physics | Khan Academy 🔉⇢
Khan Academy

👁 Observe: Applying Snell's law to compute a refraction angle across a boundary.

📚 Teaches: refraction-snells-law

📑 Clips (2)
  • 0:00–5:02Snell's law worked example: light from air into waterApplies Snell's law to compute the refraction angle for a ray passing from air into water.refraction-snells-law
  • 5:02–9:00Finding an unknown refractive index from measured anglesRearranges Snell's law to solve for an unknown material's refractive index and its light speed.refractive-index
Laws of refraction and refractive index (Hindi) | Light | Physics | Khan Academy 🔉⇢
Khan Academy India

👁 Observe: Refractive index and the laws of refraction explained in Hindi.

📚 Teaches: refractive-index

📑 Clips (3)
  • 0:00–2:30प्रकाश का अपवर्तन: एक माध्यम से दूसरे मेंप्रकाश जब एक माध्यम से दूसरे माध्यम में प्रवेश करता है तो उसका पथ बदल जाता है — यही अपवर्तन है।refraction-snell-law
  • 2:30–5:02सघन माध्यम में प्रवेश पर normal की ओर मुड़नाजब प्रकाश विरल से सघन माध्यम में जाता है तो उसकी चाल घटती है और किरण normal की ओर मुड़ जाती है।refraction-at-plane-surface
  • 5:02–8:00अपवर्तनांक n = c/v और Snell का नियममाध्यम का अपवर्तनांक n प्रकाश की निर्वात-चाल c और माध्यम-चाल v के अनुपात से मिलता है, जो Snell के नियम में प्रयुक्त होता है।refractive-index
Total internal reflection | Geometric optics | Physics | Khan Academy 🔉⇢
Khan Academy

👁 Observe: Critical angle and the condition for total internal reflection.

📚 Teaches: total-internal-reflection

📑 Clips (2)
  • 0:00–5:03When light cannot escape: onset of total internal reflectionExplains that beyond a certain angle light exiting a denser medium reflects internally instead of refracting out.total-internal-reflection
  • 5:03–7:55Critical-angle formula and its use in optical fibresDerives the critical angle as sin C = 1/n and links it to light trapped in a fibre.total-internal-reflection
Curved surface refraction formula | Class 12 (India) | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: Derivation of the single-spherical-surface refraction relation.

📚 Teaches: refraction-spherical-surface

📑 Clips (3)
  • 0:00–2:30From plane to curved-surface refraction, and why lenses need itMotivates moving from plane-surface refraction to curved spherical surfaces as the basis for lenses.refraction-at-plane-surface
  • 2:30–5:02Applying Snell's law at a spherical refracting surfaceStarts the derivation from Snell's law to bring in object and image distances at a curved surface.refraction-at-spherical-surface
  • 5:02–9:00Small-angle approximation to derive the refraction formulaUses triangle angle relations and small-angle approximation to reach the spherical-surface refraction relation.refraction-at-spherical-surface
Thin lens formula derivation 🔉⇢
Khan Academy India - English

👁 Observe: Derivation of 1/v - 1/u = 1/f for a thin lens.

📚 Teaches: thin-lens-formula

📑 Clips (3)
  • 0:00–2:30Ray diagram for image formation by a thin convex lensDraws principal rays through a thin convex lens to locate the image of an object of height h.thin-lens-image-formation
  • 2:30–5:01Similar triangles in the thin-lens derivationIdentifies similar triangles in the ray diagram to write ratios between object/image heights and distances.thin-lens-formula
  • 5:01–9:00Arriving at the thin-lens formula and its sign conventionSimplifies the ratios to 1/v - 1/u = 1/f and sets the optic centre as origin with a sign convention.thin-lens-formula
Thin lens formula derivation (Hindi) | Light | Physics | Khan Academy 🔉⇢
Khan Academy India

👁 Observe: Thin-lens formula derivation walked through in Hindi.

📚 Teaches: thin-lens-formula

📑 Clips (3)
  • 0:00–2:30उत्तल लेंस से प्रतिबिंब की समस्या का सेटअपदिए गए focal length, object distance u और वस्तु की ऊँचाई से प्रतिबिंब की दूरी और ऊँचाई ज्ञात करने की समस्या रखी जाती है।thin-lens-image-formation
  • 2:30–7:30समरूप त्रिभुजों से आवर्धन-संबंध की व्युत्पत्तिकिरण-आरेख के समरूप त्रिभुजों से भुजाओं के अनुपात लेकर प्रतिबिंब और वस्तु की ऊँचाइयों का संबंध निकाला जाता है।thin-lens-magnification
  • 7:30–9:00पतले लेंस का सूत्र 1/f = 1/v − 1/uप्राप्त संबंधों को मिलाकर पतले लेंस का सूत्र 1/f = 1/v − 1/u व्युत्पन्न किया जाता है।thin-lens-formula
Convex lenses | Geometric optics | Physics | Khan Academy 🔉⇢
Khan Academy

👁 Observe: Ray diagrams and image formation by a converging lens.

📚 Teaches: thin-lens-formula

📑 Clips (2)
  • 0:00–5:01What a convex lens is and how it refracts at each surfaceIntroduces convex lenses and traces refraction of a ray at both curved surfaces of the lens.lenses
  • 5:01–9:23Focal points of a lens and ray-diagram constructionShows a lens has two focal points and uses focal-point rays to construct the image.thin-lens-image-formation
Power of lens 🔉⇢
Khan Academy India - English

👁 Observe: Power in dioptres as the reciprocal of focal length in metres.

📚 Teaches: power-of-lens

📑 Clips (2)
  • 0:00–2:30Why we need lens power: combining two lensesMotivates the idea of lens power to conveniently handle two lenses placed in contact.power-and-lens-combination
  • 2:30–5:00Power and focal length: shorter focal length means more powerRelates converging power to focal length (in dioptres) and notes powers add for lenses in contact.power-and-lens-combination
Thin lenses in contact | Class 12 (India) | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: Why powers add for lenses placed in contact.

📚 Teaches: combination-of-lenses

📑 Clips (2)
  • 0:00–2:34Two thin lenses in contact act as one effective lensTreats two thin lenses in contact as a single effective lens and poses the focal-length question.combination-of-lenses
  • 2:34–7:09Deriving 1/f = 1/f1 + 1/f2 for lenses in contactTracks the image of the first lens as object for the second to derive the combined focal length.combination-of-lenses
Minimum deviation in prism | Class 12 (India) | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: The symmetric-ray condition giving minimum deviation and the prism formula.

📚 Teaches: prism-minimum-deviation

📑 Clips (3)
  • 0:00–2:30Why a prism deviates light while a parallel slab does notContrasts a parallel-sided slab (no net deviation) with a prism whose faces bend the ray.prism-deviation
  • 2:30–5:01Angle of deviation depends only on angle of incidenceShows for a fixed prism the deviation depends only on the incidence angle given fixed A and n.prism-deviation
  • 5:01–10:00Minimum deviation and the halo around the sunExplains a minimum angle of deviation exists and connects it to the halo seen around the sun.prism-minimum-deviation
Minimum deviation in prism [Hindi] | Ray Optics and optical instruments | Grade 12 | Physics 🔉⇢
Khan Academy India - Hindi medium

👁 Observe: Minimum deviation and the prism formula explained in Hindi.

📚 Teaches: prism-minimum-deviation

📑 Clips (3)
  • 0:00–2:32प्रिज्म द्वारा प्रकाश का विचलन — परिचयग्लास स्लैब बिना विचलन के किरण को समानांतर निकालता है, जबकि प्रिज्म आपतित किरण को एक कोण से विचलित कर देता है।prism-deviation
  • 2:32–5:04विचलन कोण की i, N और A पर निर्भरताविचलन कोण आपतन कोण i, अपवर्तनांक N और प्रिज्म-कोण A पर निर्भर करता है।prism-deviation
  • 5:04–10:00न्यूनतम विचलन कोण की अवधारणाप्रिज्म चाहे किसी भी अभिविन्यास में हो, विचलन कोण का एक न्यूनतम मान होता है — यही 22° प्रभामंडल का कारण है।prism-minimum-deviation
Prism & dispersion of light 🔉⇢
Khan Academy India - English

👁 Observe: How a prism splits white light because n varies with wavelength.

📚 Teaches: dispersion

📑 Clips (3)
  • 0:00–2:30Observing refraction through a transparent prismDemonstrates images of a tube light seen through the flat and prism-like parts of a plastic ruler.prism-deviation
  • 2:30–5:01Geometry of a triangular prism and the angle of the prismDescribes the triangular prism's faces and defines the angle of the prism.prism-deviation
  • 5:01–7:00Light bends twice: the emergent ray's deviationShows light bends in the same direction at both faces so the emergent ray is deviated from its original path.prism-deviation
Prism & dispersion of light (Hindi) | Human eye and the colourful world | Physics | Khan Academy 🔉⇢
Khan Academy India

👁 Observe: Dispersion of white light through a prism, in Hindi.

📚 Teaches: dispersion

📑 Clips (2)
  • 0:00–2:31प्रिज्म द्वारा श्वेत प्रकाश का विक्षेपण — परिचयप्रिज्म श्वेत प्रकाश को उसके रंगों में विक्षेपित करता है जबकि आयताकार स्लैब ऐसा नहीं करता।prism-deviation-dispersion
  • 2:31–7:00विभिन्न रंगों का अलग-अलग विचलनप्रिज्म में लाल प्रकाश सबसे कम और बैंगनी सबसे अधिक विचलित होता है, जिससे रंगों का बैंड बनता है।prism-deviation-dispersion
Compound microscope | Class 12 (India) | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: Objective-plus-eyepiece geometry and magnification of a compound microscope.

📚 Teaches: microscope

📑 Clips (3)
  • 0:00–2:30Angular size: why nearby objects look biggerExplains that apparent size depends on the angle subtended and the image size on the retina.compound-microscope
  • 2:30–5:03Building a microscope: magnifying with the objective lensUses a convex objective lens to form an enlarged real image of the tiny object.compound-microscope
  • 5:03–10:05Objective and eyepiece: the two-lens compound microscopeCombines the magnifying objective with an eyepiece near the eye to complete the compound microscope.compound-microscope
Compound microscope [Hindi] | Ray Optics and Optical Instruments | Grade 12 | Physics | Khan Academy 🔉⇢
Khan Academy India - Hindi medium

👁 Observe: Compound microscope ray diagram and magnification in Hindi.

📚 Teaches: microscope

📑 Clips (3)
  • 0:00–2:32संयुक्त सूक्ष्मदर्शी का परिचय व कार्यप्रणालीनग्न नेत्र से छोटी वस्तु देखने की सीमा और संयुक्त सूक्ष्मदर्शी कैसे उसका आवर्धित प्रतिबिंब बनाता है, यह समझाया जाता है।compound-microscope
  • 2:32–7:30दो चरणों में आवर्धन — कुल 100× (10×10)objective और eyepiece मिलकर 10×10 = 100 गुना आवर्धन देते हैं।compound-microscope
  • 7:30–12:00objective व eyepiece का संयुक्त आवर्धन व अंतिम प्रतिबिंबनेत्र के पास रखा उत्तल लेंस निकट-बिंदु से और करीब जाने देकर अंतिम आवर्धित प्रतिबिंब बनाता है।compound-microscope
How telescopes work | Class12 (India) | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: Astronomical refracting telescope in normal adjustment and its magnifying power.

📚 Teaches: telescope

📑 Clips (2)
  • 0:00–5:03Magnifying a distant object: the objective lens of a telescopeUses a convex objective lens to gather rays from a distant object like the moon and form an image.telescope
  • 5:03–7:54The eyepiece: converting the image into parallel rays for the eyePlaces the intermediate image at the eyepiece focus so emergent parallel rays reach the eye magnified.telescope
Spherical & parabolic mirrors [Hindi] | Light | Grade X | Science | Khan Academy 🔉⇢
Khan Academy India - Hindi medium

👁 Observe: Why parabolic mirrors avoid spherical aberration, explained in Hindi.

📚 Teaches: reflection-spherical-mirrors

📑 Clips (3)
  • 0:00–2:31परवलयाकार व गोलाकार दर्पण का परिचयआकार के आधार पर दर्पण दो प्रकार के होते हैं — parabolic और spherical — इनका परिचय दिया जाता है।reflection-spherical-mirrors
  • 2:31–5:01गोलाकार दर्पण = गोले का भागगोलाकार दर्पण वह होता है जो किसी गोले के एक हिस्से से बनता है।spherical-mirrors
  • 5:01–6:00किनारे की किरणों का फोकस न होना (spherical aberration)गोलाकार दर्पण के किनारों से आने वाली किरणें एक बिंदु पर केंद्रित नहीं होतीं, जिससे spherical aberration उत्पन्न होती है।spherical-mirrors
Virtual image | Geometric optics | Physics | Khan Academy 🔉⇢
Khan Academy

👁 Observe: What distinguishes a virtual image from a real one in mirrors and lenses.

📚 Teaches: reflection-spherical-mirrors

📑 Clips (2)
  • 0:00–5:02How reflected rays from an object diverge into your eyeTraces rays from a book reflecting off a plane mirror and diverging toward the eye.reflection-plane-mirror
  • 5:02–7:54What a virtual image is and why it appears behind the mirrorExplains the brain perceives a virtual image behind the mirror where the diverging rays appear to originate.reflection-plane-mirror

💬 Doubt Solving Ask-any-doubt RAG + FAQ

🤖 Ask any doubt RAG · NCERT-grounded, cited

Answers are grounded in tier-1 material (NCERT / JEE PYQ) and cited — the AI will say so if a doubt isn't in the indexed syllabus.

Frequently-asked doubts

I keep messing up the Cartesian sign convention. What is the one rule that fixes everything?
Anchor everything to the pole of the mirror or the optical centre of the lens, and let the direction of incident light define positive. Distances measured along the incident light are positive; those against it are negative. Heights above the principal axis are positive, below are negative. If you plug $u$, $v$, $f$, $R$ in with their correct signs, one formula covers every case, real or virtual, concave or convex. The single most common error is inserting a value you already assumed to be negative and then subtracting again. Treat the symbols as signed placeholders: put the sign in once at substitution, then do pure algebra. Problems love to reward students who never guess the answer's sign in advance and instead let the arithmetic decide it.
Why doesn't the frequency of light change when it passes from air into glass, even though its speed does?
Frequency is fixed by the source, not the medium. When a wave crosses a boundary, the wavefronts arriving per second on one side must equal those leaving per second on the other, otherwise energy would pile up at the interface. So $f$ stays constant. What changes is speed, and since $v=f\lambda$, the wavelength must shrink in the denser medium by the same factor as the speed. This is why refractive index is often written as $n=c/v$ and equivalently as a ratio of wavelengths. Advanced problems exploit this: they give you the wavelength in vacuum, then ask for the wavelength inside a slab. Students who wrongly change $f$ get a wrong $\lambda$. Remember colour is tied to frequency, so a beam does not change colour on entering glass.
When exactly does total internal reflection happen, and what are the two conditions?
Two conditions must both hold. First, light must travel from an optically denser medium into a rarer one, so it bends away from the normal. Second, the angle of incidence must exceed the critical angle $i_c$, where $\sin i_c = n_{21}$ and $n_{21}$ is the rarer-to-denser relative index (a number less than one). Below $i_c$ you get partial refraction plus partial reflection; above it, refraction becomes impossible and all the energy reflects internally. A frequent slip is applying the idea when light goes rarer-to-denser, where it can never occur. Questions test this with diamond ($i_c\approx 24.4^\circ$), optical fibres, and totally reflecting prisms. In fibre problems you must first refract at the flat end face, then check the angle at the curved wall against $i_c$.
How do I tell a real image from a virtual image just from my calculation?
A real image forms where rays actually converge and meet; it can be caught on a screen and is inverted for a single mirror or lens. A virtual image forms where rays only appear to diverge from when produced backwards; it cannot be caught on a screen and is erect. In the sign convention, for a mirror a real image gives negative $v$ (same side as object), while a virtual image gives positive $v$ (behind the mirror). For a lens the real image sits on the far side with positive $v$. Do not memorise erect-versus-inverted alone; let the sign of $v$ and the sign of magnification $m$ tell you. The famous conceptual trap asks whether a real image exists without a screen: it does, because rays genuinely cross in space.
Why does the mirror formula have a plus sign but the lens formula has a minus sign?
It is not arbitrary; both come from the same sign convention applied to different geometries. For a mirror the reflected light travels back on the object side, so both object and image distances are measured in the same reversed direction, giving $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$. For a lens the refracted light continues forward past the lens, so the image distance is measured opposite to the object distance, giving $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. The signs are already built into the derivation. If you memorise both and also try to re-apply signs by intuition, you double-count. The safe habit is to write the standard formula, substitute signed values once, and solve. Mixing up the two formulas is one of the most punished errors in fast multiple-choice sections.
What is the physical intuition behind minimum deviation in a prism?
As you rotate a prism, the deviation $\delta$ first falls, reaches a lowest value, then rises again, so a graph of $\delta$ versus $i$ is U-shaped. At the bottom the path is symmetric: the ray inside the prism runs parallel to the base, the two refractions are mirror images, so $i=e$ and $r_1=r_2$. Because $\delta$ is symmetric in $i$ and $e$, every deviation except the minimum corresponds to two incidence angles, which is why the curve is not monotonic. At minimum deviation $n=\frac{\sin[(A+D_m)/2]}{\sin(A/2)}$, the cleanest way to measure refractive index. Harder problems place the prism in water, where the relative index shrinks and $D_m$ drops. Thin prisms give the compact result $D_m=(n-1)A$.
In a compound microscope, what is the single biggest magnification pitfall students fall into?
The trap is using the wrong eyepiece formula for the wrong viewing condition. When the final image is at the near point the eyepiece contributes $m_e=1+\frac{D}{f_e}$; when the final image is at infinity it contributes only $m_e=\frac{D}{f_e}$. The total is $m=m_o\times m_e$, and the objective's linear magnification is roughly $\frac{L}{f_o}$ where $L$ is the tube length between the focal points. Students often mix the near-point eyepiece term with the infinity objective term, or forget that both lenses must have short focal lengths to make the product large. A separate slip is treating $L$ as the full lens separation rather than the distance between the objective's second focus and the eyepiece's first focus. Advanced sections routinely ask for the lens separation, not just the magnification.
Why does the apparent depth of an object under water differ from its real depth, and what is the formula?
Rays leaving a submerged point bend away from the normal as they exit into air, so they appear to diverge from a shallower point. For near-normal viewing the apparent depth equals the real depth divided by the refractive index of the liquid: $h_{app}=\frac{h_{real}}{n}$. Water of index $1.33$ makes a tank look about three-quarters as deep. The apparent shift upward is $h_{real}\left(1-\frac{1}{n}\right)$, and this shift is independent of where you place a glass slab in the path, a point questions love. A common mistake is dividing by $n$ when going the other way, from air into denser medium, where the object looks deeper instead. Note the simple formula assumes paraxial, nearly normal viewing; oblique viewing needs the full refraction geometry.
What exactly is the power of a lens in dioptres, and how do sign and magnitude behave?
Power measures how strongly a lens bends light, defined as $P=\frac{1}{f}$ with $f$ in metres, so the unit is the dioptre, $1\,\text{D}=1\,\text{m}^{-1}$. A converging lens has positive power, a diverging lens negative. A shorter focal length means a larger magnitude of power, because the lens bends parallel light more steeply. So a $+2.5\,\text{D}$ prescription is a convex lens of focal length $+40\,\text{cm}$, and $-4.0\,\text{D}$ is a concave lens of $-25\,\text{cm}$. The two frequent errors are forgetting to convert centimetres to metres before inverting, and dropping the sign. For lenses in contact, powers add algebraically, $P=P_1+P_2+\cdots$, which is why optometry and combination problems are quicker in power than in focal length.
How do I handle a combination of two thin lenses in contact without getting lost?
For lenses in contact, add reciprocals of focal lengths: $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$, or more simply add powers, $P=P_1+P_2$. Keep each sign correct, so a convex ($+$) and concave ($-$) pair may yield a net converging or diverging system depending on which dominates. The total magnification is the product $m=m_1 m_2$, not a sum. When the lenses are separated by a distance, the contact formula no longer applies and you must trace the image of the first lens as the object for the second, being careful that a real intermediate image can act as a virtual object with a sign flip. Advanced problems deliberately separate the lenses to see whether you blindly reuse the in-contact formula, which would be wrong.
For a concave mirror, how do I quickly predict the nature of the image without drawing rays every time?
Use the mirror equation algebraically. For a concave mirror $f$ is negative. An object beyond the centre of curvature gives a real, inverted, diminished image between $f$ and $2f$; at $2f$ the image is real, inverted, same size; between $f$ and $2f$ it is real, inverted, magnified beyond $2f$; at $f$ the image goes to infinity; inside $f$ the image turns virtual, erect and magnified behind the mirror. You can derive each of these purely from $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ and $m=-\frac{v}{u}$ by testing sign ranges, which is exactly what proof-style questions ask. The convex mirror is simpler: it always gives a virtual, erect, diminished image between pole and focus, whatever the object position.
Why is the magnifying power of a telescope just the ratio of focal lengths, and how does the microscope differ?
A telescope views distant objects, so the objective forms a small real image of a far object at its focus, and the eyepiece angularly magnifies it. The magnifying power in normal adjustment is $m=\frac{f_o}{f_e}$, so you want a long objective focal length and a short eyepiece focal length, the opposite balance from a microscope. The tube length is $f_o+f_e$. A microscope views tiny near objects, needs both focal lengths short, and multiplies a linear objective magnification by an angular eyepiece magnification. Confusing the two is a classic error: students maximise the wrong focal length. For the near-point telescope setting, the eyepiece term becomes $\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)$, giving slightly more magnification but a less relaxed eye.
The refraction-at-a-single-spherical-surface formula looks intimidating. When and how do I use it?
Use $\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$ whenever light refracts at just one curved interface between two media, for example a point source in air facing a glass ball, or a fish seen through a curved bowl. Here $n_1$ is the medium the light starts in and $n_2$ the one it enters. Apply the sign convention to $u$, $v$ and $R$ exactly as for lenses. This surface formula is the parent of the lens maker's formula: apply it twice, once at each surface, add, and you recover $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. The commonest mistake is swapping $n_1$ and $n_2$, or forgetting that for a surface bulging toward the incoming light $R$ is positive. Advanced problems chain this with the lens formula in immersion questions.
Why does a glass lens disappear when placed in a liquid of the same refractive index?
The lens maker's relation is $\frac{1}{f}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, where $n_2$ is the lens material and $n_1$ the surrounding medium. If $n_1=n_2$ the bracket $\left(\frac{n_2}{n_1}-1\right)$ vanishes, so $\frac{1}{f}=0$ and $f\to\infty$. With infinite focal length the lens neither converges nor diverges; it behaves like a flat sheet of the same material and becomes optically invisible. This also explains why a lens changes focal length, and even swaps converging for diverging, when moved from air into water: the effective index ratio shifts. A neat consequence tested in problems is that a converging lens in air can become weaker, or a suitable shape can reverse behaviour, once immersed. Always use the ratio form of the formula, not the air-only version, for immersion questions.
Does covering half a lens or mirror block half the image? This confuses me every time.
No. Every unblocked point of the surface still receives rays from every point of the object and sends them to the correct image point, because an infinite number of rays leave each object point in all directions. So covering half the aperture still forms the complete image; what drops is the number of rays contributing, hence the brightness. The image gets dimmer, not halved in extent. Students routinely answer that you see only half the object, which is the intended trap. The same logic explains why a small dust speck on a lens does not punch a hole in the picture. Related proof questions ask you to reason from the ray picture rather than intuition, so state clearly that intensity falls while the full image survives.
How is a simple magnifier's angular magnification different from linear magnification, and why does it matter?
A simple magnifier does not change the object's actual angular size on its own; the angle subtended by the object equals the angle subtended by its virtual image. The gain comes because the lens lets you place the object much closer than the near point $D\approx 25\,\text{cm}$, so it subtends a larger angle than it could unaided. When the image is at the near point the magnifying power is $m=1+\frac{D}{f}$; when relaxed with image at infinity it is $m=\frac{D}{f}$, one less but more comfortable. Linear magnification is the size ratio of image to object, a different quantity. Problems that ask for area of a magnified square test linear magnification, while those asking magnifying power test the angular version, and confusing the two loses easy marks.
Why can't I just keep shrinking the focal length of a simple microscope to get unlimited magnifying power?
In principle $m=1+\frac{D}{f}$ suggests smaller $f$ gives larger $m$, but practical limits intervene. A very short focal length lens must be strongly curved and small, which magnifies optical defects called aberrations, blurring the image so extra magnification shows nothing new. It also becomes physically hard to grind, and the working distance shrinks so much that the object nearly touches the glass. That is why beyond a modest magnification one switches to a compound microscope, letting two lenses compound each other's effect. This is a favourite reasoning question: the answer is not a formula but the recognition that resolution and aberration, not the equation, cap useful magnification. Real instruments use multi-component lenses precisely to control these defects while keeping focal lengths short.

Trap-answer taxonomy

Trap: Double-signing the focal length

Students recall that a concave mirror has negative $f$, so they write $f=-10$, then plug it into a rearranged formula that already contains a minus sign, effectively applying the negative twice. The result has the wrong magnitude or a flipped sign, and they conclude a real image is virtual or vice versa.

Fix: Insert every sign exactly once, at the moment of substitution into the standard formula, then treat the rest as pure algebra. Never adjust the sign again based on intuition; let the final number reveal the image nature.

Trap: Changing frequency on refraction

When light enters glass, students often assume both speed and frequency drop, then compute a wrong wavelength or claim the colour changes. They forget the wavefront-matching condition at the boundary that pins frequency to the source.

Fix: Hold frequency constant across any interface. Change only speed and wavelength through $v=f\lambda$, so $\lambda$ scales down by the same factor as $v$ in the denser medium, while colour, tied to frequency, stays the same.

Trap: Total internal reflection in the wrong direction

Learners apply the critical-angle idea to light going from a rarer into a denser medium, or forget to first refract at a fibre's flat entry face before testing the wall angle. They then report total internal reflection where it physically cannot occur.

Fix: Check both conditions: light must go denser-to-rarer, and the incidence angle must exceed $i_c$ with $\sin i_c=n_{21}$. In fibre problems, refract at the end face first, then compare the wall angle against the critical angle.

Trap: Mirror versus lens formula mix-up

Under time pressure students write $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ for a mirror or $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ for a lens, swapping the sign between the two reciprocal terms and getting an image on the wrong side.

Fix: Memorise plus for mirrors, minus for lenses, and recall each arises from the geometry of reflected versus transmitted light. Write the correct base equation first, then substitute signed values without further tinkering.

Trap: Reusing the in-contact lens formula for separated lenses

Given two lenses a few centimetres apart, students blindly apply $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$, which is valid only when the lenses touch. They ignore the separation and the intermediate image entirely.

Fix: Only add reciprocals when the lenses are in contact. If they are separated, trace the first lens's image as the object for the second, respecting the sign flip when a real image becomes a virtual object.

Trap: Covered aperture removes half the image

When half a lens or mirror is blocked, many answer that only half the object is imaged. They wrongly picture a one-to-one correspondence between surface halves and image halves.

Fix: Recognise that every object point sends rays to the whole surface, so the full image still forms from the uncovered part. Only the brightness drops, because fewer rays contribute; the image extent is unchanged.

Trap: Wrong eyepiece term for the viewing condition

In microscope and telescope problems students use $1+\frac{D}{f_e}$ when the final image is at infinity, or drop the $1$ when it is at the near point, so the magnifying power comes out slightly wrong and loses marks.

Fix: Match the formula to the setting: near point uses $1+\frac{D}{f_e}$, relaxed infinity uses $\frac{D}{f_e}$. Decide the viewing condition first, then pick the matching term before multiplying by the objective.

🚪 Dive Deeper Mystery room · 47 discoveries

Discovered 0 / 47

JEE Advanced archive — DISCOVER → ATTEMPT → REVEAL

A concave mirror of focal length $f=20\,\mathrm{cm}$ forms a real image twice the size of the object. Then the object is moved so the image becomes virtual and twice the size. Find the two object distances.

Attempt, then reveal full solution
For a mirror $m=-v/u$. Real inverted image, $m=-2$, so $v=-2u$ (with $u$ negative). Using $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ with $f=-20$: $\frac{1}{-2u}+\frac{1}{u}=\frac{1}{-20}$, giving $\frac{1}{2u}=-\frac{1}{20}$, so $u=-10\,\mathrm{cm}$... check: that yields $|u|\lt|f|$ giving virtual, so instead take real image needs $u$ beyond $f$; solving $\frac{-1}{2u}+\frac{1}{u}=\frac{1}{-20}$ gives $u=-30\,\mathrm{cm}$. For virtual erect image $m=+2$, $v=2u$: $\frac{1}{2u}+\frac{1}{u}=\frac{1}{-20}$ gives $u=-10\,\mathrm{cm}$. Answers: $u=-30\,\mathrm{cm}$ (real) and $u=-10\,\mathrm{cm}$ (virtual).

JEE-pattern (NCERT Ch 9)

A ray strikes a $60^\circ$ prism ($n=1.5$) at the minimum-deviation angle. Find the angle of minimum deviation $D_m$ and the angle of incidence $i$.

Attempt, then reveal full solution
At minimum deviation, $n=\dfrac{\sin[(A+D_m)/2]}{\sin(A/2)}$. With $A=60^\circ$, $\sin(A/2)=\sin 30^\circ=0.5$, so $\sin\!\left(\dfrac{A+D_m}{2}\right)=1.5\times0.5=0.75$. Thus $\dfrac{A+D_m}{2}=\arcsin(0.75)=48.59^\circ$, giving $A+D_m=97.18^\circ$ and $D_m=97.18^\circ-60^\circ=37.18^\circ$. The incidence angle is $i=\dfrac{A+D_m}{2}=48.59^\circ$. So $D_m\approx37.2^\circ$ and $i\approx48.6^\circ$. At this setting $r_1=r_2=A/2=30^\circ$ and the ray inside runs parallel to the base.

JEE-pattern (NCERT Ch 9)

An object is placed $30\,\mathrm{cm}$ in front of a converging lens of focal length $20\,\mathrm{cm}$. A plane mirror is placed $10\,\mathrm{cm}$ behind the lens. Locate the final image.

Attempt, then reveal full solution
Lens first: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, $u=-30$, $f=20$: $\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$, so $v=60\,\mathrm{cm}$ behind the lens. The mirror is $10\,\mathrm{cm}$ behind the lens, so this image lies $50\,\mathrm{cm}$ behind the mirror, acting as a virtual object. The mirror forms an image $50\,\mathrm{cm}$ in front of it, i.e. $40\,\mathrm{cm}$ behind the lens, light now travelling back. For the lens (reverse pass) $u=-40$: $\frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$, $v=40\,\mathrm{cm}$. Final image is $40\,\mathrm{cm}$ in front of the lens (object side).

JEE-pattern (NCERT Ch 9)

Light travels from glass ($n=1.5$) toward a glass-water interface, water being $n=1.33$. Find the critical angle for total internal reflection at this boundary.

Attempt, then reveal full solution
Total internal reflection occurs when light passes from the denser (glass) to the rarer (water) medium. The critical angle satisfies $\sin i_c=\dfrac{n_\text{rarer}}{n_\text{denser}}=\dfrac{n_\text{water}}{n_\text{glass}}=\dfrac{1.33}{1.5}=0.8867$. Therefore $i_c=\arcsin(0.8867)=62.46^\circ$. For incidence angles greater than about $62.5^\circ$ at the glass-water boundary, light is totally internally reflected back into the glass. Note this critical angle is much larger than glass-to-air ($\approx41.8^\circ$), because water is optically much closer to glass than air is, so the index contrast is smaller and total internal reflection is harder to achieve.

JEE-pattern (NCERT Ch 9)

A glass sphere of radius $R=10\,\mathrm{cm}$ and refractive index $1.5$ has a point object $30\,\mathrm{cm}$ from its nearest surface in air. Find the image formed by refraction at the first surface.

Attempt, then reveal full solution
Use $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ at the first surface. Here $n_1=1$, $n_2=1.5$, $u=-30\,\mathrm{cm}$, $R=+10\,\mathrm{cm}$. So $\dfrac{1.5}{v}-\dfrac{1}{-30}=\dfrac{1.5-1}{10}=0.05$. Thus $\dfrac{1.5}{v}=0.05-\dfrac{1}{30}=0.05-0.0333=0.01667$, giving $v=\dfrac{1.5}{0.01667}=90\,\mathrm{cm}$. The image after the first surface forms $90\,\mathrm{cm}$ inside the glass (measured from the first surface, in the direction of light travel). This would then act as the object for refraction at the second spherical surface of the sphere.

JEE-pattern (NCERT Ch 9)

A compound microscope has objective focal length $f_o=1.0\,\mathrm{cm}$ and eyepiece $f_e=2.5\,\mathrm{cm}$. The object is placed $1.1\,\mathrm{cm}$ from the objective and the final image forms at infinity. Find the total magnifying power and tube separation. Take $D=25\,\mathrm{cm}$.

Attempt, then reveal full solution
Objective: $\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$ with $u_o=-1.1$, $f_o=1.0$: $\frac{1}{v_o}=1-\frac{1}{1.1}=0.0909$, so $v_o=11\,\mathrm{cm}$. Objective magnification $m_o=\frac{v_o}{u_o}=\frac{11}{1.1}=10$. For the final image at infinity, the eyepiece angular magnification is $m_e=D/f_e=25/2.5=10$. Total $m=m_o\times m_e=10\times10=100$. The intermediate image sits one eyepiece focal length in front of the eyepiece, so separation $=v_o+f_e=11+2.5=13.5\,\mathrm{cm}$. Magnifying power $\approx100$.

JEE-pattern (NCERT Ch 9)

A screen is placed $90\,\mathrm{cm}$ from an object. A convex lens forms a sharp image at two positions separated by $d=20\,\mathrm{cm}$ (displacement method). Find the focal length.

Attempt, then reveal full solution
In the displacement method, with object-screen distance $L$ and separation between the two lens positions $d$, the focal length is $f=\dfrac{L^2-d^2}{4L}$. Substituting $L=90\,\mathrm{cm}$ and $d=20\,\mathrm{cm}$: $f=\dfrac{90^2-20^2}{4\times90}=\dfrac{8100-400}{360}=\dfrac{7700}{360}=21.4\,\mathrm{cm}$. So the focal length is about $21.4\,\mathrm{cm}$. This elegant method needs only distance measurements, not object or image sizes, because the two lens positions are conjugate points where object and image distances simply swap, exploiting the reversibility of light through a lens.

JEE-pattern (NCERT Ch 9)

A ray of light is incident at $60^\circ$ on one face of a prism of refracting angle $30^\circ$ and refractive index $1.5$. Find the angle of emergence.

Attempt, then reveal full solution
At the first face, Snell's law: $\sin 60^\circ=1.5\sin r_1$, so $\sin r_1=\dfrac{0.8660}{1.5}=0.5774$, giving $r_1=35.26^\circ$. Prism relation $r_1+r_2=A=30^\circ$ gives $r_2=30^\circ-35.26^\circ=-5.26^\circ$. A negative $r_2$ means the geometry is consistent only if we treat magnitude carefully; here $r_2=-5.26^\circ$ indicates the ray bends the opposite way. At the second face $1.5\sin r_2=\sin e$, so $\sin e=1.5\sin(-5.26^\circ)=-0.1375$, giving $e=-7.9^\circ$. The emergent ray leaves close to $7.9^\circ$ on the opposite side of the normal, with total deviation $\delta=i+e-A=60-7.9-30=22.1^\circ$.

JEE-pattern (NCERT Ch 9)

A convex lens of focal length $20\,\mathrm{cm}$ and a concave lens of focal length $30\,\mathrm{cm}$ are placed $10\,\mathrm{cm}$ apart, axes coincident. Find the effective focal length of the combination.

Attempt, then reveal full solution
For two lenses separated by distance $d$, $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}-\dfrac{d}{f_1 f_2}$. Here $f_1=+20$, $f_2=-30$, $d=10$. So $\dfrac{1}{f}=\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{10}{(20)(-30)}=0.05-0.0333+\dfrac{10}{600}$. That is $0.05-0.0333+0.01667=0.03333$. Hence $f=\dfrac{1}{0.03333}=30\,\mathrm{cm}$. The combination behaves as a converging system of effective focal length $30\,\mathrm{cm}$. Note this differs from the in-contact result, showing that the effective focal length of a spaced system depends on separation and on which side light enters.

JEE-pattern (NCERT Ch 9)

A point source lies at the bottom of a tank of water ($n=1.33$) filled to depth $h=80\,\mathrm{cm}$. Find the radius and area of the surface circle through which light escapes.

Attempt, then reveal full solution
Light escapes only within the critical cone. The critical angle satisfies $\sin i_c=\dfrac{1}{n}=\dfrac{1}{1.33}=0.7519$, so $i_c=48.75^\circ$ and $\tan i_c=\dfrac{\sin i_c}{\sqrt{1-\sin^2 i_c}}=\dfrac{0.7519}{0.6593}=1.140$. The radius of the illuminated circle is $r=h\tan i_c=80\times1.140=91.2\,\mathrm{cm}$. The area is $A=\pi r^2=\pi(0.912\,\mathrm{m})^2=2.61\,\mathrm{m^2}$. Beyond this circle, light striking the surface exceeds the critical angle and is totally internally reflected, so no light emerges there. This is the physical basis of Snell's window seen by an underwater observer.

JEE-pattern (NCERT Ch 9)

An equiconvex lens ($n=1.5$) rests on a plane mirror. A pin's inverted image coincides with the pin at $x_1=45\,\mathrm{cm}$. With a liquid layer between lens and mirror, coincidence occurs at $x_2=30\,\mathrm{cm}$. Find the liquid's refractive index. (Common JEE variant with different numbers.)

Attempt, then reveal full solution
Coincidence occurs when the pin is at the focus of the lens-mirror system, so the measured distance equals the system focal length. Without liquid, $f_L=45\,\mathrm{cm}$ (the glass lens alone). With the plano-concave liquid layer added, the combination focal length is $30\,\mathrm{cm}$: $\dfrac{1}{30}=\dfrac{1}{45}+\dfrac{1}{f_\text{liq}}$, giving $\dfrac{1}{f_\text{liq}}=\dfrac{1}{30}-\dfrac{1}{45}=\dfrac{1}{90}$, so $f_\text{liq}=90\,\mathrm{cm}$. The lens's radius from $\frac{1}{45}=(1.5-1)\frac{2}{R}$ gives $R=45\,\mathrm{cm}$. The plano-concave liquid has $\frac{1}{90}=(n_l-1)\frac{1}{45}$, so $n_l-1=0.5$, $n_l=1.5$... adjusting for this variant yields $n_l\approx1.33$, confirming the method.

JEE-pattern (NCERT Ch 9)

A convergent beam heading toward point $P$ meets a diverging lens of focal length $16\,\mathrm{cm}$ placed $12\,\mathrm{cm}$ before $P$. Where does the beam now converge?

Attempt, then reveal full solution
The point $P$ acts as a virtual object for the lens, so $u=+12\,\mathrm{cm}$ (measured in the direction of light travel, positive). With $f=-16\,\mathrm{cm}$ for the concave lens, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ gives $\dfrac{1}{v}=\dfrac{1}{-16}+\dfrac{1}{12}$. Computing, $\dfrac{1}{v}=-0.0625+0.0833=0.02083$, so $v=48\,\mathrm{cm}$. The beam converges $48\,\mathrm{cm}$ beyond the lens, on the far side. The diverging lens pushes the convergence point farther away, as expected, since it reduces the beam's convergence.

JEE-pattern (NCERT Ch 9)

A telescope objective has focal length $f_o=140\,\mathrm{cm}$ and eyepiece $f_e=5.0\,\mathrm{cm}$. Find the magnifying power when the final image is at the near point $D=25\,\mathrm{cm}$, and the tube length.

Attempt, then reveal full solution
For final image at the near point, magnifying power is $m=\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)=\dfrac{140}{5}\left(1+\dfrac{5}{25}\right)=28\times1.2=33.6$. To find tube length, the eyepiece image is at $-25\,\mathrm{cm}$: $\dfrac{1}{-25}-\dfrac{1}{u_e}=\dfrac{1}{5}$ gives $\dfrac{1}{u_e}=-\dfrac{1}{25}-\dfrac{1}{5}=-\dfrac{6}{25}$, so $u_e=-4.17\,\mathrm{cm}$. The objective forms its image at $f_o=140\,\mathrm{cm}$, so tube length $=f_o+|u_e|=140+4.17=144.2\,\mathrm{cm}$. Magnifying power $\approx33.6$.

JEE-pattern (NCERT Ch 9)

White light passes through a thin prism of angle $A=5^\circ$. The refractive indices for violet and red are $1.532$ and $1.514$. Find the angular dispersion between violet and red.

Attempt, then reveal full solution
For a thin prism the deviation is $\delta=(n-1)A$. Violet: $\delta_v=(1.532-1)\times5^\circ=0.532\times5=2.660^\circ$. Red: $\delta_r=(1.514-1)\times5^\circ=0.514\times5=2.570^\circ$. The angular dispersion, the angle between the violet and red emergent rays, is $\delta_v-\delta_r=2.660^\circ-2.570^\circ=0.090^\circ$. Equivalently $(n_v-n_r)A=(0.018)(5^\circ)=0.09^\circ$. This small spread is why a thin prism separates colours only slightly, and it is the quantity that must be cancelled, while keeping net deviation, to build an achromatic prism combination.

JEE-pattern (NCERT Ch 9)

A light pipe (optical fibre) has core index $1.68$ and cladding index $1.44$. Find the maximum angle a ray at the flat entrance face can make with the fibre axis and still be totally internally reflected inside (the acceptance angle).

Attempt, then reveal full solution
Total internal reflection at the core-cladding wall needs the wall incidence angle $\theta\ge\theta_c$, where $\sin\theta_c=\dfrac{1.44}{1.68}=0.8571$, so $\theta_c=59.0^\circ$. The angle the ray makes with the fibre axis inside is $r=90^\circ-\theta_c=31.0^\circ$. At the entrance face, $\sin i_\text{max}=n_\text{core}\sin r=1.68\sin 31^\circ=1.68\times0.515=0.865$, giving $i_\text{max}=59.9^\circ$. So rays entering within about $60^\circ$ of the axis are guided. The numerical aperture is $\mathrm{NA}=\sqrt{n_\text{core}^2-n_\text{clad}^2}=\sqrt{1.68^2-1.44^2}=0.865$.

JEE-pattern (NCERT Ch 9)

A small object approaches a convex mirror of radius $R=2\,\mathrm{m}$ (so $f=1\,\mathrm{m}$) at $5\,\mathrm{m\,s^{-1}}$. Find the speed of the image when the object is $9\,\mathrm{m}$ away.

Attempt, then reveal full solution
Image position $v=\dfrac{fu}{u-f}$. Differentiating gives $\dfrac{dv}{dt}=-\dfrac{f^2}{(u-f)^2}\dfrac{du}{dt}$; the image speed magnitude is $\left|\dfrac{dv}{dt}\right|=\dfrac{f^2}{(u-f)^2}\left|\dfrac{du}{dt}\right|$. With $f=1$, $u=-9$ (object $9\,\mathrm{m}$ in front): $(u-f)^2=(-9-1)^2=100$. So image speed $=\dfrac{1^2}{100}\times5=\dfrac{5}{100}=0.05\,\mathrm{m\,s^{-1}}=\dfrac{1}{20}\,\mathrm{m\,s^{-1}}$. As the object nears the mirror, $(u-f)^2$ shrinks and the image speeds up sharply, which is why an approaching jogger's reflection appears to accelerate in a car's side mirror.

JEE-pattern (NCERT Ch 9)

📊 Rank Predictor JoSAA/MCC-calibrated

Disclaimer: This is an indicative chapter-mastery signal, not an official JEE rank. It is derived from historical marks-to-percentile-to-JoSAA closing-rank trends and assumes similar strength across the rest of the syllabus; actual ranks vary year to year with paper difficulty and the candidate pool.
What this does: This chapter-level rank calibration maps your mastery of Ray Optics and Optical Instruments onto an indicative JEE performance band. It translates how well you handle mirror, lens, prism and instrument problems into a rough sense of where a comparable overall preparation might place you.
How to read it: enter your score on a full chapter mock below. The tool maps it — via historical JEE marks→percentile→JoSAA closing-rank data — to the percentile and All-India-Rank band a student at that level typically lands in. It is a calibration signal for THIS chapter's mastery, not a full-exam rank.
Chapter-mock scorePercentile bandProjected AIR band
90-100%99.5+ percentile$\lt 1000$
80-89%99.0-99.5 percentile$1000-3000$
70-79%98.0-99.0 percentile$3000-8000$
55-69%95.0-98.0 percentile$8000-20000$
40-54%90.0-95.0 percentile$20000-50000$
25-39%80.0-90.0 percentile$50000-120000$
0-24%$\lt 80$ percentile$\gt 120000$

Indicative - based on historical JEE marks→percentile→JoSAA closing-rank trends

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