Refraction at Plane and Spherical surfaces

PhysicsRay Optics And Optical InstrumentsFor NEET aspirants

Refraction at plane and spherical surfaces is the bending of light as it crosses from one medium into another, governed by Snell's law $n_{1} sin i = n_{2} sin r$ . At a plane surface it makes pools look shallower and shifts objects seen through glass; at a curved surface it forms images given by $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ , the basis of every lens. Refraction at plane and spherical surfaces is tested every year in JEE Main, JEE Advanced and NEET.

On this page1Refractive index2Snell's law3Apparent depth4Glass slab5Curved surface6Flowchart and map7Solved examples

Key Formulas - Quick Reference

★ Must learn $n = \frac{c}{v}$ ; relative index $n_{21} = \frac{n _{2}}{n _{1}} = \frac{v _{1}}{v _{2}} = \frac{λ _{1}}{λ _{2}}$ (frequency unchanged)
★ Must learnSnell's law: $n_{1} sin i = n_{2} sin r$
★ Must learnApparent depth (near the normal): $d^{'} = d \frac{n _{observer}}{n _{object}}$
Layers viewed from air: $d^{'} = \frac{t _{1}}{n _{1}} + \frac{t _{2}}{n _{2}} + \dots$
★ Must learnSlab normal shift: $s = t (1 - \frac{1}{n})$ , towards the observer
Lateral shift: $d = \frac{t sin ( i - r )}{cos r}$
★ Must learnSpherical surface: $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$
Focal lengths of a surface: $f_{1} = - \frac{n _{1} R}{n _{2} - n _{1}}$ , $f_{2} = \frac{n _{2} R}{n _{2} - n _{1}}$
Image speed, plane surface: $v_{I S} = \frac{n _{2}}{n _{1}} v_{O S}$ (relative to the surface)

1. Refraction and Refractive Index

Refraction is the change in direction of light when it passes obliquely from one medium into another. It happens because light travels at different speeds in different media. At normal incidence the ray does not bend, but it still slows down, so this is also refraction.

★ Must learnAbsolute refractive index of a medium:

n = \frac{c}{v} = \frac{speed of light in vacuum}{speed of light in the medium}

. A medium with larger

n

is optically denser (light is slower in it). Relative refractive index of medium 2 with respect to medium 1:

n_{21} = \frac{n _{2}}{n _{1}} = \frac{v _{1}}{v _{2}}

, and

n_{12} = \frac{1}{n _{21}}

Medium	Typical $n$	Speed of light
Vacuum	$1$ (exactly)	$3.00 \times 1 0^{8} m s^{- 1}$
Air	$1.0003 \approx 1$	$\approx 3.00 \times 1 0^{8} m s^{- 1}$
Water	$1.33 \approx \frac{4}{3}$	$2.25 \times 1 0^{8} m s^{- 1}$
Crown glass	$1.5 = \frac{3}{2}$	$2.00 \times 1 0^{8} m s^{- 1}$
Diamond	$2.42$	$1.24 \times 1 0^{8} m s^{- 1}$

Frequency never changes in refraction: it is fixed by the source. Speed and wavelength both scale as $\frac{1}{n}$ : $λ_{medium} = \frac{λ _{vacuum}}{n}$ .
Optical density is not mass density: kerosene is lighter than water but has a larger refractive index ( $1.44$ ).
$n$ depends slightly on wavelength (it is larger for violet than for red). This causes dispersion, treated in the Prism concept.

Key idea

Refraction changes speed and wavelength by the factor

n

; the frequency, and hence the colour, stays the same.

2. Laws of Refraction (Snell's Law)

Laws of refraction:
1. The incident ray, the refracted ray and the normal at the point of incidence lie in one plane.
2. For a given pair of media and a given colour,

\frac{sin i}{sin r}

is constant. In the form used in problems:

n_{1} sin i = n_{2} sin r

, with both angles measured from the normal.

Figure 1: Snell's law in action. Air to glass at

i = 4 5^{\circ}

gives

r = 28. 1^{\circ}

(towards the normal); glass to air at

i = 3 0^{\circ}

gives

r = 48. 6^{\circ}

(away from the normal).

Rarer to denser (

n_{1} < n_{2}

)

Light slows down, $r < i$ : the ray bends towards the normal. Always possible, for every angle of incidence.

Denser to rarer (

n_{1} > n_{2}

)

Light speeds up, $r > i$ : the ray bends away from the normal. Possible only while $i$ is below the critical angle; beyond it, total internal reflection.

2.1 Refraction through parallel layers

Apply Snell's law at every boundary of a stack of parallel layers: $n_{1} sin i_{1} = n_{2} sin i_{2} = n_{3} sin i_{3} = \dots$ The product $n sin θ$ is carried unchanged through the stack.

Figure 2: Across parallel boundaries

n_{1} sin i_{1} = n_{2} sin i_{2} = n_{3} sin i_{3} = \dots

, so the angle in any layer depends only on that layer's

n

. Same outer medium on both sides means the emergent ray is parallel to the incident ray.

Exam Trick

$n sin θ$ is conserved across parallel boundaries. To find the angle in any layer, skip the layers in between: $sin θ_{k} = \frac{n _{1} sin i _{1}}{n _{k}}$ . If the first and last media are the same, the ray leaves parallel to how it entered, whatever lies in between.

Principle of reversibility: a ray retracing its path obeys the same law, so $n_{12} \times n_{21} = 1$ . Applied to parallel layers it also gives the chain rule $n_{12} n_{23} n_{31} = 1$ .

3. Apparent Depth and Normal Shift

An object inside a denser medium, seen from a rarer medium almost along the normal, looks closer to the surface than it is. Rays leaving the object bend away from the normal at the surface, and the eye traces them back to a point above the object.

Figure 3: Looking straight down, the coin appears at

d^{'} = \frac{d}{n}

. Drawn exactly for rays at

1 2^{\circ}

d^{'} / d = 0.737

, close to the paraxial

\frac{3}{4} = 0.750

A ray from $O$ at depth $d$ meets the surface at a point $B$ at horizontal distance $x$ from the normal $A O$ . Angle in the denser medium $i$ , in air $r$ .
Snell's law: $n_{1} sin i = n_{2} sin r$ , where $n_{1}$ is the object's medium and $n_{2}$ the observer's.
Near the normal $sin \approx tan$ , so $n_{1} \frac{x}{d} = n_{2} \frac{x}{d ^{'}}$ .
Hence
$\frac{d}{d ^{'}} = \frac{n _{1}}{n _{2}} \Rightarrow d^{'} = d \frac{n _{2}}{n _{1}}$

Apparent distance $=$ real distance $\times \frac{n _{observer}}{n _{object}}$ , both measured from the surface. Object in water seen from air:

d^{'} = \frac{3 d}{4}

, so the apparent shift is

d - d^{'} = d (1 - \frac{1}{n})

. The same rule works in reverse: a fish sees a bird higher than it is. Valid only for viewing close to the normal (paraxial rays).

Figure 4: Apparent distance

=

real distance

\times \frac{n _{observer}}{n _{object}}

. The bird sees the fish shallower (

36 \times \frac{3}{4} = 27 cm

); the fish sees the bird higher (

36 \times \frac{4}{3} = 48 cm

3.1 Several layers

For an object under layers of thickness $t_{1}, t_{2}, \dots$ and indices $n_{1}, n_{2}, \dots$ , each layer shifts the image on its own. Seen from air:

d^{'} = \frac{t _{1}}{n _{1}} + \frac{t _{2}}{n _{2}} + \dots

If the observer is in a medium of index $n_{obs}$ , multiply the sum by $n_{obs}$ .

Figure 5: For stacked layers seen from air,

d^{'} = \frac{t _{1}}{n _{1}} + \frac{t _{2}}{n _{2}} = \frac{6}{1.5} + \frac{8}{4/3} = 10 cm

(exact ray at

1 0^{\circ}

gives

9.87 cm

Exam Trick

Divide by $n$ looking into the denser medium, multiply looking out of it. Each layer's contribution is thickness over its own index (from air). A $12 cm$ water tank ( $n = \frac{4}{3}$ ) looks $9 cm$ deep; a $6 cm$ glass block ( $n = 1.5$ ) looks $4 cm$ thick.

JEE Advanced

Speed of the image for a plane surface. Differentiating $y = \frac{n _{2}}{n _{1}} x$ (distances from the surface) gives

v_{I S} = \frac{n _{2}}{n _{1}} v_{O S}

where $v_{I S}$ and $v_{O S}$ are the image and object velocities relative to the surface, along the normal. If the surface itself moves (a rising water level), subtract its velocity first, then add it back (Solved Example 7).

Quick Recall: tap to check

A fish is

40 cm

below the water surface (

n = \frac{4}{3}

). Where does a person above see it?

40 \times \frac{3}{4} = 30 cm

below the surface.

Which quantity never changes when light is refracted?

Frequency (and so colour). Speed and wavelength change by the factor

n

A bird is

h

above a pond. Where does a fish see it?

\frac{4 h}{3}

above the surface: looking out of the denser medium, multiply by

n

Key idea

Apparent distance

=

real distance

\times \frac{n _{observer}}{n _{object}}

, measured from the surface, valid near the normal.

4. Refraction Through a Glass Slab

A glass slab has two parallel faces. At the first face $sin i = n sin r$ ; at the second, $n sin r = sin e$ . Therefore $e = i$ : the emergent ray is parallel to the incident ray. The slab produces no deviation, only two kinds of shift.

4.1 Normal shift (object seen through the slab)

Figure 6: Seen through a slab, the object appears shifted towards the observer by

s = t (1 - \frac{1}{n})

. For

n = 1.5

s = \frac{t}{3}

(drawn:

s / t = 0.338

for

9^{\circ}

rays).

First face: the object at distance $x$ appears at $n x$ (seen from inside the glass).
Second face: this image is $n x + t$ inside the glass; seen from air it appears at $x + \frac{t}{n}$ from the second face.
Without the slab the object would be $x + t$ from that face. Shift
$s = (x + t) - (x + \frac{t}{n}) = t (1 - \frac{1}{n})$

Normal shift $s = t (1 - \frac{1}{n})$ , in the direction of the incident light (the object appears closer to the observer). It does not depend on the object's distance from the slab. If the slab sits in a medium of index

n_{s}

, use

s = t (1 - \frac{n _{s}}{n})

. Use this shift when object and observer are in the same medium; when they are in different media use the apparent-depth rule layer by layer.

4.2 Lateral shift (oblique incidence)

Figure 7: Emergent ray parallel to the incident ray, displaced sideways by

d = \frac{t sin ( i - r )}{cos r}

. Here

i = 5 0^{\circ}

r = 30. 7^{\circ}

d = 0.38 t

Inside the slab the ray travels $O B = \frac{t}{cos r}$ . The perpendicular distance between the emergent ray and the original line is $O B sin (i - r)$ :

d = \frac{t sin ( i - r )}{cos r}

It is zero at normal incidence and grows towards $t$ as $i \to 9 0^{\circ}$ .

Normal shift

s

Object viewed through the slab, near the normal. Shift of the image position along the line of sight: $s = t (1 - \frac{1}{n})$ , independent of $i$ .

Lateral shift

d

A single oblique ray. Sideways displacement of the ray: $d = \frac{t sin ( i - r )}{cos r}$ , zero at $i = 0$ .

Quick Recall: tap to check

A glass slab (

n = 1.5

) is

6 cm

thick. How far is an object shifted when seen through it?

6 (1 - \frac{2}{3}) = 2 cm

, towards the observer, wherever the object is.

What is the lateral shift for normal incidence?

Zero:

i = r = 0

gives

sin (i - r) = 0

Key idea

A parallel slab never deviates a ray: it only shifts it sideways, by

\frac{t sin ( i - r )}{cos r}

, and shifts images along the normal, by

t (1 - \frac{1}{n})

JEE Advanced

Variable refractive index. If $n$ changes smoothly with height, $n (y)$ , treat the medium as infinitely many thin layers. $n sin θ$ stays constant along the ray: $n (y) sin θ = n_{0} sin i$ . With $\frac{d y}{d x} = cot θ$ this gives the path

\frac{d y}{d x} = \frac{n ( y ) ^{2} - n _{0}^{2} sin ^{2} i}{n _{0} sin i}

which is integrated to find the trajectory. Mirages and the bending of starlight in the atmosphere are examples.

5. Refraction at a Spherical Surface

When the boundary between two media is part of a sphere (radius $R$ , centre $C$ , pole $P$ ), paraxial rays from a point object meet (or appear to meet) at a single image point. This is the building block of lenses.

Distances $u$ , $v$ , $R$ are measured from the pole, with the New Cartesian convention: positive in the direction of the incident light.
$n_{1}$ is the medium the light comes from, $n_{2}$ the medium it goes into.
$R > 0$ when the centre of curvature lies on the outgoing side (surface convex towards the incoming light); $R < 0$ otherwise.

Figure 8: Sign of

R

\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

. Measure

P C

along the incident light:

R > 0

when

C

lies on the side the light goes into. The same surface changes sign when the light comes from the other side (second face of a lens, light returning after a mirror).

Figure 9: Refraction at a spherical surface. With small angles

α \approx \frac{h}{- u}

β \approx \frac{h}{R}

γ \approx \frac{h}{v}

, Snell's law

n_{1} i = n_{2} r

becomes

\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

In triangle $O A C$ , exterior angle: $i = α + β$ . In triangle $A I C$ : $β = r + γ$ , so $r = β - γ$ .
Snell's law for small angles: $n_{1} i = n_{2} r$ , so $n_{1} (α + β) = n_{2} (β - γ)$ .
Paraxial: $α \approx \frac{h}{- u}$ , $β \approx \frac{h}{R}$ , $γ \approx \frac{h}{v}$ . Substituting and dividing by $h$ : $\frac{n _{1}}{- u} + \frac{n _{2}}{v} = \frac{n _{2} - n _{1}}{R}$ .

Refraction formula for a spherical surface:

\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

Valid for every case (convex or concave surface, real or virtual object) when signs are put in. Lateral magnification

m = \frac{n _{1} v}{n _{2} u}

(see the Magnification concept).

5.1 Focal lengths of a single surface

Second focal length $f_{2}$ (object at infinity, $u \to - \infty$ ): $f_{2} = \frac{n _{2} R}{n _{2} - n _{1}}$ .
First focal length $f_{1}$ (image at infinity): $f_{1} = - \frac{n _{1} R}{n _{2} - n _{1}}$ . Note $\frac{f _{2}}{f _{1}} = - \frac{n _{2}}{n _{1}}$ and $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2}}{f _{2}}$ .
For air to glass with $R = 10 cm$ , $n_{2} = 1.5$ : $f_{2} = 30 cm$ , $f_{1} = - 20 cm$ .

Figure 10:

u = - 30 cm

R = + 10 cm

n_{2} = 1.5

gives

v = + 90 cm

(real, inside the glass) and

m = \frac{n _{1} v}{n _{2} u} = - 2

. Ray through

C

goes undeviated. Heights are drawn twice the horizontal scale.

Figure 11:

\frac{1.5}{v} - \frac{1}{u} = \frac{0.5}{10}

plotted exactly. First focus at

u = f_{1} = - 20 cm

, second focal length

f_{2} = 30 cm

; an object at

u = - 40 cm

gives

v = + 60 cm

Figure 12: Two refractions in a glass sphere (

R = 10 cm

n = 1.5

): paraxial rays focus

5 cm

beyond the far surface. Exact tracing shows the

4 cm

rays crossing 0.7 cm earlier (spherical aberration).

Exam Trick

Put $R = \infty$ and the curved-surface formula becomes apparent depth. $\frac{n _{2}}{v} = \frac{n _{1}}{u}$ gives $v = \frac{n _{2}}{n _{1}} u$ , exactly the plane-surface rule. One formula, two topics: use it with signs for any single refracting surface.

Speed of the image (object moving along the axis): differentiating the formula, $- \frac{n _{2}}{v ^{2}} \frac{d v}{d t} + \frac{n _{1}}{u ^{2}} \frac{d u}{d t} = 0$ , so

\frac{d v}{d t} = \frac{n _{1}}{n _{2}} \frac{v ^{2}}{u ^{2}} \frac{d u}{d t}

The image moves in the same direction as the object (unlike a mirror).

Quick Recall: tap to check

What does the formula

\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

reduce to for a plane surface?

v = \frac{n _{2}}{n _{1}} u

: the apparent-depth rule.

Find the second focal length of an air-glass surface with

R = + 10 cm

n = 1.5

f_{2} = \frac{1.5 \times 10}{0.5} = 30 cm

inside the glass.

When is

R

positive?

When the centre of curvature lies on the side the light goes into (surface bulges towards the incoming light).

Key idea

One surface, one formula:

\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

, with

n_{1}

on the incident side; for several surfaces use it surface by surface.

6. Flowchart and Mind Map

Most refraction questions reduce to one decision: is the surface plane or curved? The flowchart below picks the formula; the mind map collects the whole concept for revision.

Figure 13: Problem-solving flowchart. Decide plane or curved first; for several surfaces (or a surface plus a mirror) apply one formula at a time, each image becoming the next object.

Figure 14: Refraction at plane and spherical surfaces on one page. Revise from this map, then test yourself on the solved examples.

7. Solved Examples

Solved Example 1

Light of wavelength

600 nm

in air enters glass of refractive index

1.5

. Its wavelength and frequency in glass are
(A)

600 nm

5 \times 1 0^{14} Hz

(B)

400 nm

5 \times 1 0^{14} Hz

(C)

400 nm

3.3 \times 1 0^{14} Hz

(D)

900 nm

5 \times 1 0^{14} Hz

Solution:

Frequency is fixed by the source: $f = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{600 \times 1 0 ^{- 9}} = 5 \times 1 0^{14} Hz$ . Wavelength in glass: $\frac{600}{1.5} = 400 nm$ .

Answer: (B).

Solved Example 2

Light passes from medium

a

into medium

b

, making

4 5^{\circ}

with the normal in

a

and

6 0^{\circ}

b

. The speed of light in

b

\frac{c}{3}

. Find the speed in

a

Solution:

$n_{b} = 3$ . Snell's law: $n_{a} sin 4 5^{\circ} = 3 sin 6 0^{\circ}$ , so $n_{a} = \frac{3 \times 3 /2}{1/ 2} = \frac{3 6}{2}$ .

Answer: $v_{a} = \frac{c}{n _{a}} = \frac{2 c}{3 6} = \frac{6}{9} c \approx 0.27 c$ . (Note $n_{a} > n_{b}$ here, consistent with the ray bending away from the normal.)

Solved Example 3

A ray strikes a glass sphere (

n = 3

) at an angle of incidence of

6 0^{\circ}

and leaves after two refractions. Find the angles inside, the angle of emergence and the total deviation.

Solution:

First surface: $sin 6 0^{\circ} = 3 sin r$ gives $sin r = \frac{1}{2}$ , $r = 3 0^{\circ}$ . The chord inside makes equal angles with the two radii (isosceles triangle), so $r^{'} = 3 0^{\circ}$ and $3 sin 3 0^{\circ} = sin e$ gives $e = 6 0^{\circ}$ .

Deviation at each surface $= 6 0^{\circ} - 3 0^{\circ} = 3 0^{\circ}$ , in the same sense.

Answer: $r = r^{'} = 3 0^{\circ}$ , $e = 6 0^{\circ}$ , total deviation $6 0^{\circ}$ .

Solved Example 4

A ray in air strikes a horizontal glass plate (

n = 1.5

) at

4 5^{\circ}

. The plate lies on a layer of water (

n = \frac{4}{3}

) resting on a glass tank bottom with air below. The angle at which the ray finally emerges into the air below is
(A)

4 5^{\circ}

(B)

3 2^{\circ}

(C)

28. 1^{\circ}

(D) it depends on the thicknesses

Solution:

Across parallel boundaries $n sin θ = 1 \times sin 4 5^{\circ}$ throughout (Figure 2). In air again, $sin θ = sin 4 5^{\circ}$ . Thicknesses change only where the ray comes out, not its direction. (In water it travels at $32. 0^{\circ}$ , in glass at $28. 1^{\circ}$ .)

Answer: (A).

Solved Example 5

A bird is

36 cm

above a pond and a fish is

36 cm

below the surface, on the same vertical (

n = \frac{4}{3}

). Find (a) the apparent depth of the fish seen by the bird and (b) the apparent height of the bird seen by the fish.

Solution:

(a) Observer in air, object in water: $d^{'} = 36 \times \frac{1}{4/3} = 27 cm$ below the surface.

(b) Observer in water, object in air: $d^{'} = 36 \times \frac{4/3}{1} = 48 cm$ above the surface (Figure 4).

Answer: (a) $27 cm$ ; (b) $48 cm$ . So the bird judges the fish to be $63 cm$ away, while the fish judges the bird to be $84 cm$ away.

Solved Example 6

A coin lies at the bottom of a beaker containing

8 cm

of water (

n = \frac{4}{3}

) with

6 cm

of oil (

n = 1.5

) floating on it. Seen from directly above, the coin appears to be below the oil surface by
(A)

14 cm

(B)

10 cm

(C)

9.3 cm

(D)

11 cm

Solution:

$d^{'} = \frac{t _{1}}{n _{1}} + \frac{t _{2}}{n _{2}} = \frac{6}{1.5} + \frac{8}{4/3} = 4 + 6 = 10 cm$ (Figure 5).

Answer: (B). The coin appears raised by $4 cm$ .

Solved Example 7

A bird dives vertically at

24 cm s^{- 1}

towards a pond while a fish directly below rises at

12 cm s^{- 1}

. The water level itself rises at

1 cm s^{- 1}

(

n = \frac{4}{3}

). Find (a) the speed of the fish as it appears to the bird and (b) the speed of the bird as it appears to the fish.

Solution:

Take upward as positive and use $v_{I S} = \frac{n _{2}}{n _{1}} v_{O S}$ with velocities relative to the surface.

(a) Fish (object) in water, bird looks from air: $v_{O S} = 12 - 1 = 11$ . $v_{I S} = \frac{1}{4/3} \times 11 = 8.25$ , so $v_{I} = 9.25 cm s^{- 1}$ upward. Relative to the bird (velocity $- 24$ ): $9.25 + 24 = 33.25 cm s^{- 1}$ .

(b) Bird (object) in air, fish looks from water: $v_{O S} = - 24 - 1 = - 25$ . $v_{I S} = \frac{4}{3} (- 25) = - 33.3$ , so $v_{I} = - 32.3 cm s^{- 1}$ . Relative to the fish: $- 32.3 - 12 = - 44.3 cm s^{- 1}$ .

Answer: (a) $\frac{133}{4} = 33.25 cm s^{- 1}$ ; (b) $\frac{133}{3} \approx 44.3 cm s^{- 1}$ (closing speeds).

Solved Example 8

A concave mirror (

R = 40 cm

) lies at the bottom of a tank with

30 cm

of water (

n = \frac{4}{3}

) above it, reflecting side up and axis vertical. A parallel beam falls vertically on the water. Where does the final image appear to an observer above?

Solution:

Vertical rays enter the water undeviated and strike the mirror parallel to its axis, so the mirror forms the image at its focus, $20 cm$ above the mirror, which is $10 cm$ below the water surface. Seen from air: $d^{'} = \frac{10}{4/3} = 7.5 cm$ .

Answer: $7.5 cm$ below the water surface.

Solved Example 9

A glass slab

3 cm

thick (

n = 1.5

) is placed between a point object and a concave mirror of radius

20 cm

. The object is

21 cm

from the mirror. Where is the final image?

Solution:

Shift due to the slab: $3 (1 - \frac{1}{1.5}) = 1 cm$ towards the mirror. For the mirror, the object therefore appears at $21 - 1 = 20 cm$ , which is its centre of curvature. Rays strike the mirror normally, retrace their path through the slab, and the slab undoes its own shift on the way back.

Answer: the image forms on the object itself.

Solved Example 10

A ray strikes a parallel glass slab

10 cm

thick at

6 0^{\circ}

; the angle of refraction is

4 5^{\circ}

. Find the lateral shift.

Solution:

$d = \frac{t sin ( i - r )}{cos r} = \frac{10 sin 1 5 ^{\circ}}{cos 4 5 ^{\circ}} = 102 sin 1 5^{\circ} = 102 \times 0.2588$ .

Answer: $d \approx 3.66 cm$ .

Solved Example 11

A point object in air is

30 cm

in front of the convex end of a long glass rod (

R = 10 cm

n = 1.5

). Locate the image.

Solution:

$u = - 30$ , $R = + 10$ , $n_{1} = 1$ , $n_{2} = 1.5$ : $\frac{1.5}{v} = \frac{0.5}{10} + \frac{1}{- 30} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$ , so $v = + 90 cm$ .

Answer: real image $90 cm$ inside the rod, inverted, $m = \frac{n _{1} v}{n _{2} u} = - 2$ (Figure 10).

Solved Example 12

A parallel beam falls on a glass sphere of radius

10 cm

and

n = 1.5

. Where does it converge?

Solution:

First surface ( $u = - \infty$ , $R = + 10$ ): $\frac{1.5}{v _{1}} = \frac{0.5}{10}$ , $v_{1} = 30 cm$ from the first pole. The rays would meet $10 cm$ beyond the far surface.
Second surface (glass to air, $R = - 10$ ): that point is a virtual object at $u = + 10 cm$ . $\frac{1}{v} - \frac{1.5}{10} = \frac{1 - 1.5}{- 10} = \frac{1}{20}$ , so $\frac{1}{v} = \frac{1}{5}$ .

Answer: $5 cm$ beyond the far surface ( $15 cm$ from the centre), as in Figure 12.

Solved Example 13

A small air bubble is inside a glass sphere (

R = 10 cm

n = 1.5

5 cm

from the centre. Where does it appear when viewed (a) from the side nearer the bubble and (b) from the opposite side?

Solution:

Light goes glass to air: $n_{1} = 1.5$ , $n_{2} = 1$ . Take the direction towards the observer as positive; the centre is behind the surface, so $R = - 10 cm$ .

(a) $u = - 5$ : $\frac{1}{v} = \frac{1 - 1.5}{- 10} + \frac{1.5}{- 5} = 0.05 - 0.3 = - 0.25$ , $v = - 4 cm$ .

(b) $u = - 15$ : $\frac{1}{v} = 0.05 - 0.1 = - 0.05$ , $v = - 20 cm$ .

Answer: (a) $4 cm$ inside the near surface; (b) $20 cm$ inside the far surface, that is, at the opposite surface of the sphere. The same bubble appears in two very different places.

Solved Example 14

A glass rod (

n = 1.5

90 cm

long, has one end ground convex (

R = 10 cm

) and the other end flat and silvered. A point object is at distance

x

in front of the convex end. For which values of

x

does the image form on the object itself?

Solution:

Case 1: rays inside the rod travel parallel to the axis, strike the silvered flat end normally and retrace. Need $v = \infty$ : $0 - \frac{1}{- x} = \frac{0.5}{10}$ , so $x = 20 cm$ .

Case 2: rays converge to a point on the silvered end ( $v = 90 cm$ ). A point on a plane mirror is its own image, so the light returns along paths that lead back to the object. $\frac{1.5}{90} + \frac{1}{x} = \frac{0.5}{10}$ , so $\frac{1}{x} = \frac{1}{20} - \frac{1}{60} = \frac{1}{30}$ , $x = 30 cm$ .

Answer: $x = 20 cm$ or $x = 30 cm$ .

Practice Questions

A ray in water ( $n = \frac{4}{3}$ ) meets the surface with $sin θ_{w} = \frac{3}{5}$ . Find the angle it makes with the normal in air.Answer: $sin θ_{a} = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5}$ , $θ_{a} = sin^{- 1} \frac{4}{5} \approx 5 3^{\circ}$
A coin at the bottom of a $20 cm$ deep water tank ( $n = \frac{4}{3}$ ) is viewed from above. By how much does it appear raised?Answer: $5 cm$ (apparent depth $15 cm$ )
Find the normal shift produced by a slab $15 cm$ thick with $n = 1.5$ in air.Answer: $5 cm$
A point object is $20 cm$ from one face of a glass slab ( $t = 15 cm$ , $n = 1.5$ ); the other face is $10 cm$ from a convex mirror ( $f = 40 cm$ ). How far behind the mirror does the observer (on the object side) see the final image?Answer: Object effectively $40 cm$ from the mirror; $v = + 20 cm$ ; seen through the slab it appears $20 - 5 = 15 cm$ behind the mirror
A point object in air is $5 cm$ in front of a concave glass surface ( $R = - 10 cm$ , $n = 1.5$ ). Locate the image.Answer: $v = - 6 cm$ : virtual, $6 cm$ in front of the surface, in air
In the rod of Solved Example 11 the object ( $u = - 30 cm$ ) moves towards the rod at $2 m s^{- 1}$ . Find the image speed.Answer: $\frac{d v}{d t} = \frac{1}{1.5} \times \frac{9 0 ^{2}}{3 0 ^{2}} \times 2 = 12 m s^{- 1}$ , moving away from the surface (same direction as the object)
The image in Solved Example 11 (inside the rod) meets a plane mirror placed inside the glass $120 cm$ from the convex end. Find the final image after the light returns through the convex end.Answer: Mirror image $150 cm$ from the pole; glass to air: $v = 25 cm$ in front of the rod, real

Common Mistakes to Avoid

Watch out

Inverting the apparent-depth ratio. Real to apparent is $d^{'} = d \frac{n _{observer}}{n _{object}}$ : an object in water seen from air looks shallower.
Saying the frequency changes in refraction. Frequency is fixed; speed and wavelength change.
Using $d^{'} = \frac{d}{n}$ for oblique viewing. It is a paraxial result: valid only near the normal.
Shifting the image away from the observer. A slab shifts it towards the observer, by $t (1 - \frac{1}{n})$ , independent of the object distance.
Forgetting the surrounding medium in the slab shift: use $t (1 - \frac{n _{s}}{n})$ when the slab is in water.
Swapping $n_{1}$ and $n_{2}$ in $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ . $n_{2}$ is always the medium the light enters.
Getting the sign of $R$ wrong. $R > 0$ only when the centre of curvature is on the outgoing side of the surface.
For a sphere or rod, measuring the second object distance from the first pole instead of the second surface.

Frequently Asked Questions

What is refraction of light?

Refraction is the bending of light when it passes obliquely from one medium into another. It happens because light travels at different speeds in different media. Light bends towards the normal when it slows down in a denser medium and away from the normal when it speeds up in a rarer medium, following Snell's law.

Does the frequency of light change on refraction?

No. The frequency is set by the source and stays the same in every medium, which is why colour does not change. Speed and wavelength both decrease by the factor n in a medium of refractive index n; for example 600 nm light becomes 400 nm in glass of index 1.5.

Why does a swimming pool look shallower than it is?

Light from the bottom bends away from the normal as it leaves the water. Your eye traces the rays back in straight lines and they appear to come from a point above the real bottom. Seen from directly above, the apparent depth is the real depth divided by 4/3, about three quarters of it.

What is the difference between normal shift and lateral shift?

Normal shift is the apparent movement of an object seen through a glass slab, t(1 - 1/n) towards the observer, independent of the object distance. Lateral shift is the sideways displacement of an oblique ray passing through the slab, t sin(i - r)/cos r, which is zero at normal incidence.

What is the formula for refraction at a spherical surface?

For a single spherical surface, n2/v - n1/u = (n2 - n1)/R, where n1 is the medium the light comes from, n2 the medium it enters, and u, v and R are measured from the pole with the New Cartesian sign convention. Putting R equal to infinity gives the apparent-depth rule for a plane surface.

Why is the emergent ray from a glass slab parallel to the incident ray?

Both faces of a slab are parallel and the medium on both sides is the same. Snell's law at the first face and at the second face gives sin i = n sin r = sin e, so the angle of emergence equals the angle of incidence. The ray is only displaced sideways, not deviated.

Is refraction at plane and spherical surfaces important for JEE Main and JEE Advanced?

Yes. JEE Main asks apparent depth, slab shift and single-surface image problems almost every year. JEE Advanced combines surfaces with mirrors and slabs, adds image velocity and variable refractive index, and uses spheres and rods with silvered ends. Master the sign convention for n1, n2 and R first.

Which refraction questions are common in NEET?

NEET most often asks about refractive index and speed of light, change of wavelength with frequency unchanged, apparent depth of objects in water, normal shift by a glass slab and simple numericals on the spherical surface formula. Learn apparent depth equals real depth divided by n and slab shift t(1 - 1/n).

Previous year questions on Refraction at Plane and Spherical surfaces

20 questions from past papers, each with a step-by-step solution.

Show all 20 questions

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