From the laws of reflection to the telescope — image formation by mirrors, lenses and prisms
🔬 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. 🔉⇢
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.
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.
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.
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 relation 1/v + 1/u = 1/f between object distance, image distance and focal length, together with the linear magnification m = -v/u.
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.
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.
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 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 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.
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 + ...
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).
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).
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).
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.
The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.
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.
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. 🔉⇢
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. 🔉⇢
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. 🔉⇢
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 →
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 →
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. 🔉⇢
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. 🔉⇢
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 →
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. 🔉⇢
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 →
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. 🔉⇢
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. 🔉⇢
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. 🔉⇢
🔬 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. 🔉⇢
Source: JEE-pattern (NCERT Ch 9)
🔬 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. 🔉⇢
Source: JEE-pattern (NCERT Ch 9)
🔬 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. 🔉⇢
Source: JEE-pattern (NCERT Ch 9)
🔬 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. 🔉⇢
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
These worked examples are taught in full alongside their interactive scene:
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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 |
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.
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
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
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$^2$.]
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)
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
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?
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 $R$?
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
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
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
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.
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
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
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
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
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
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.]
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 $S_1$.]
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)
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
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 $PR$.]
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.]
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?
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?
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?
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 $m$?
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 __________.
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:
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 $n_2$.]
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 ____.
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).]
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)
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
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?
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?
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?
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 :
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?
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 _____.
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 _____.
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 _____ $cm\ s^{-1}$.
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?
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?
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?
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 _______.
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 :
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 :
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 :
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
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 :
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)
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 ___________
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 _____________.
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 $=\frac{4}{3}$)
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 __________.
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
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:
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:
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
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 :
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-
In a reflecting telescope, a secondary mirror is used to:
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 _________.
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 __________ $\mathrm{ms}^{-1}$.
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.
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 ____________ $\mathrm{cm}$.
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 $\left(n_{w}=4 / 3\right)$ ____________.
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) $0^\circ$
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:
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
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:
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.
For a thin symmetric prism made of glass (refractive index 1.5), the ratio of incident angle and minimum deviation will be _______.
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.
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 ?
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 $\_\_\_\_$ .
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
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 $\_\_\_\_$
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 .
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)
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 ________.
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.
Distribution — advanced: 10 · easy: 40 · hard: 20 · medium: 30. Every question carries a source trace; each ends in an SME-verify solution.
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: NCERT Ch 9 (in-text)
Source: NCERT Ch 9 (in-text)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
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Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
Source: JEE-pattern (NCERT Ch 9)
MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.
👁 Observe: Ray-diagram rules for image formation by spherical mirrors, from an IIT faculty.
📚 Teaches: reflection-spherical-mirrors
👁 Observe: Critical-angle derivation and TIR applications from an IIT-PAL lecture.
📚 Teaches: total-internal-reflection
👁 Observe: How optical fibres guide light by repeated total internal reflection.
📚 Teaches: optical-fibres
Full lecture — no clip index.
👁 Observe: Building from single-surface refraction toward the lens maker's formula.
📚 Teaches: refraction-spherical-surface
👁 Observe: Prism refraction and dispersion treated rigorously by IIT faculty.
📚 Teaches: dispersion
👁 Observe: Optical instruments compared and their magnification derived by IIT faculty.
📚 Teaches: telescope
👁 Observe: Foundational geometric-optics concepts from an IIT Madras course.
📚 Teaches: refraction-snells-law
👁 Observe: University-level opening lecture framing rays, reflection and refraction.
📚 Teaches: refraction-snells-law
Full lecture — no clip index.
👁 Observe: Intuition for why the refractive index means light effectively slows in a medium.
📚 Teaches: refractive-index
👁 Observe: Refraction plus dispersion inside raindrops producing the rainbow angle.
📚 Teaches: dispersion
👁 Observe: Fermat's principle as the deeper reason behind refraction and Snell's law.
📚 Teaches: refraction-snells-law
👁 Observe: How R = 2f and the geometry of concave/convex spherical mirrors is set up.
📚 Teaches: reflection-spherical-mirrors
👁 Observe: Full derivation of 1/v + 1/u = 1/f from similar triangles.
📚 Teaches: mirror-formula
👁 Observe: Sign-convention handling while plugging numbers into the mirror formula.
📚 Teaches: mirror-formula
👁 Observe: How bending at an interface follows n1 sin i = n2 sin r.
📚 Teaches: refraction-snells-law
👁 Observe: Applying Snell's law to compute a refraction angle across a boundary.
📚 Teaches: refraction-snells-law
👁 Observe: Refractive index and the laws of refraction explained in Hindi.
📚 Teaches: refractive-index
👁 Observe: Critical angle and the condition for total internal reflection.
📚 Teaches: total-internal-reflection
👁 Observe: Derivation of the single-spherical-surface refraction relation.
📚 Teaches: refraction-spherical-surface
👁 Observe: Derivation of 1/v - 1/u = 1/f for a thin lens.
📚 Teaches: thin-lens-formula
👁 Observe: Thin-lens formula derivation walked through in Hindi.
📚 Teaches: thin-lens-formula
👁 Observe: Ray diagrams and image formation by a converging lens.
📚 Teaches: thin-lens-formula
👁 Observe: Power in dioptres as the reciprocal of focal length in metres.
📚 Teaches: power-of-lens
👁 Observe: Why powers add for lenses placed in contact.
📚 Teaches: combination-of-lenses
👁 Observe: The symmetric-ray condition giving minimum deviation and the prism formula.
📚 Teaches: prism-minimum-deviation
👁 Observe: Minimum deviation and the prism formula explained in Hindi.
📚 Teaches: prism-minimum-deviation
👁 Observe: How a prism splits white light because n varies with wavelength.
📚 Teaches: dispersion
👁 Observe: Dispersion of white light through a prism, in Hindi.
📚 Teaches: dispersion
👁 Observe: Objective-plus-eyepiece geometry and magnification of a compound microscope.
📚 Teaches: microscope
👁 Observe: Compound microscope ray diagram and magnification in Hindi.
📚 Teaches: microscope
👁 Observe: Astronomical refracting telescope in normal adjustment and its magnifying power.
📚 Teaches: telescope
👁 Observe: Why parabolic mirrors avoid spherical aberration, explained in Hindi.
📚 Teaches: reflection-spherical-mirrors
👁 Observe: What distinguishes a virtual image from a real one in mirrors and lenses.
📚 Teaches: reflection-spherical-mirrors
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.
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.
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.
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.
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.
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.
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.
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.
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.
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 $i$.
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.
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.
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.
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 $D=25\,\mathrm{cm}$.
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.
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.
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.
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.
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.)
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?
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.
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.
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).
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.
JEE-pattern (NCERT Ch 9)
| Chapter-mock score | Percentile band | Projected 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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Authoritative & comprehensive JEE Main + Advanced resource · sources traced Tier 1–3 · SME-review state (append ?review=1)