1 Ray Optics

Topics Covered

1. Plane Reflection
2. Laws of Reflection
3. Reflection at Spherical Surfaces
4. Mirror Formula
5. Refraction
6. Laws of Refraction
7. Apparent Shift
8. Total Internal Reflection
9. Prism
10. Grazing Incidence and Grazing Emergence
11. Dispersion
12. Combination of Prisms
13. Reflection at Spherical Surfaces
14. Thin Lenses
15. Focus and Focal Length
16. Image Formation
17. Refraction Rules for Diverging Lens
18. Len’s Maker Formula
19. Sign Convention
20. Thin Lenses Continued
21. Optical Power
22. Lenses in Contact
23. Lens with One Surface Silvered

IIT-JEE Syllabus: optics: rectilinear propagation of light; reflection and refraction at plane and spherical
surfaces; total internal reflection; deviation and dispersion of light by a prism; thin lenses; combinations
of mirrors and thin lenses; magnification.

Ray Optics
The branch of optics that ignores the finiteness of the wavelength of light and, therefore, assumes
that light travels in perfectly straight lines is called geometrical optics or ray optics. Traditionally, light
and its propagation have been satisfactorily represented by rays. Rays are thin straight lines (or line
segments) with arrowheads to show their direction of propagation.

Ray Optics 143

 First, we must understand that the rays are just a way of modelling light and do not exist in reality. An
equally valid representation of light and its propagation is through wavefronts. We will study more about
wavefronts later.
Second, we are able to see things because light rays coming from them fall into our eyes. If there were
no light emanating from an ‘object’, it would not be visible to us.
An image of a point on the object is formed.
1. Either when rays emanating from that point finally converge to meet at another point;
2. Or when these rays ‘appear’ to converge to meet at a point.
In the first case, we say that the image is real; in the second case, the image is virtual.

Reflection
Reflection takes place when rays of light return to the same medium after striking the surface of
another medium. The wavelength and velocity of the light do not change upon reflection.
N
incident
reflected
ray
ray i r i r
i r

     Fig. 1.1 Fig. 1.2

Some Definitions
1. Angle of incidence, i: it is the angle, which the incident ray makes with the normal to the surface at
the point of incidence.
2. Angle of reflection, r: it is the angle that the reflected ray makes with the normal at the point of
incidence.
3.  ngle of deviation, δ: it is the angle through which the incident ray turns to become the reflected
A
ray.

Laws of Reflection
(Obeyed at every reflecting surface)
1. The angle of incidence is equal to the angle of reflection, i.e., i = r.
2. The incident ray, reflected ray and the normal lie in one plane.

Reflection at Plane Surfaces

1. The image is virtual and erect. N
2. The image is of the same size as the object (thus, magnification
= 1). Q
P
3. The distance of the image from the mirror, v, is equal to the i r
distance of the object from the mirror, u.
A B
4. While the image is erect, it is laterally inverted.
O
5. Angle of deviation for a single reflection in a plane mirror,
  QOB  BOR  ( 90 o  r )  ( 90 o  i )
o
 180  2 i or
o
180  2 r R
N'
Fig. 1.3

144 Ray Optics

When a plane mirror is rotated through an angle θ, the reflected ray rotates through an angle 2θ.
When two plane mirrors are kept facing each other at an angle θ degrees and an object is placed
between them, then multiple images are formed due to successive reflections.
360
If m  , then the number of images thus formed is given by the following rules,

1. If m is an odd integer, and the object is not on the bisector of θ, then N = m.
2. If m is an odd integer, and the object is on the bisector of θ, then N = m – 1.
3. If m is an even integer, then N = m – 1.
4. If m is not an integer (has a fractional part), then N = Int (m), or the integer part of m.

Example 1: Two plane mirrors are inclined to each other such that a ray of light incident on the first
mirror and parallel to the second is reflected from the second mirror and parallel to the
first mirror. Determine the angle between the two mirrors.
Solution: Let θ be the angle between the two mirrors OM1
and OM2. The incident ray AB is parallel to mirror M1
OM2 and strikes the mirror OM1 at an angle of D
incidence ‘a’. It is reflected along BC; the angle
B 
of reflection is a. A

From Fig. 1.4, we have,   N1
∠M1BA = ∠OBC = ∠M1OM2 = θ N2
Similarly, for reflection at mirror OM2,
  
we have ∠M2CD = ∠BCO = ∠M2OM1 = θ O M2
C
Now in ∠OBC, 3θ = 180°,
∴ θ = 60° Fig. 1.4

Example 2: Find the inclination of two plane mirrors so that it will give three images of a single object.
360
n 1
Solution: 
360
3 1

  90

Reflection on Spherical Surfaces

Spherical mirrors are of two kinds: convex and concave. Each of these kinds is a part of a larger sphere.
If the reflecting surface is facing outside, then the mirror is a convex mirror. If the reflecting surface is
facing inside the sphere (or imaginary sphere), then the mirror is a concave mirror.
Convex Mirror Concave Mirror

F C C F

Fig. 1.5 Fig. 1.6

Ray Optics 145

 Frequently Used Terms in Spherical Mirrors
Centre of curvature: it is the centre of the sphere of which the mirror is a part.
Radius of curvature: it is the radius of the sphere of which the mirror is a part.
Pole: it is the geometrical centre of the circular cross-section of the spherical reflecting surface.
Principal axis: it is the straight line joining the centre of curvature to the pole.
Focus: when a narrow beam of rays of light, parallel to the principal axis and close to it, is incident on
the surface of a mirror, the reflected beam is found to converge to or appear to diverge from a point on
the principal axis. This point is the focus.
Focal length: it is the distance between pole and the focus.
Sign Convention: it is always advised to use the new
Incident ray
cartesian sign convention.
1. All distances are measured from the pole ‘P’.
+ve
2. Distances measured along the direction of incident rays
are taken as positive.
3. Distances measured along a direction opposite to the P
incident rays are taken as negative.
4. Distances above the principal axis are positive. –ve
5. Distances below the principal axis are negative.

Fig. 1.7

Angles, when measured from the normal, in anticlockwise direction are positive, while in clockwise
direction are negative.

Locating Images–Concave Mirror

The following rules can be followed while tracing images for a concave mirror.
1. A ray parallel to the principal axis passes through the focus after reflection.
2. Conversely, a ray passing through the focus becomes parallel to the principal axis after reflection.
3. A ray passing through the centre of curvature retraces its path.

3
Object 1

C F

Real 2
image
Fig. 1.8

The following table summarises the various cases of image formation for a concave mirror. Here, P is
the pole, F is the focus, and C is the centre of curvature.

146 Ray Optics

 Object Position Image Position Image Size Nature of Image
Between P and F Behind the mirror Enlarged Virtual
At F At infinity Highly enlarged Real
Between F and C Beyond C Enlarged Real
At C At C Same size as object Real
Beyond C Between F and C Diminished Real
At infinity At F Highly diminished Real
Table 1.1

Locating Images–Convex Mirror

The following rules can be followed while tracing images for a convex mirror.
1. A ray parallel to the principal axis after reflection would seem to emerge from the focus.
2. Conversely, a ray directed towards the focus becomes parallel to the principal axis after reflection.
3. A ray directed towards the centre of curvature retraces its path after reflection.

3
2

Object Virtual F C
image

Fig. 1.9

Mirror Formula
Consider an object AB whose image is A’B’. The ray diagram is shown below. An incident ray AD parallel
to the principal axis passes through the focus F after reflection. The ray AP reflects along PA’ after
reflection (obeying the laws of reflection such that ∠i = ∠r). This is shown in Fig. 1.10.
Since ∆ABC and ∆A’B’C are similar, we have
AB CB A
 ... (i) D
AB CB
Also, ∆ABP and ∆A’B’P are similar. Hence, we have
AB PB B' i
 r P
... (ii) B C F
AB
  PB
From equations (i) and (ii)
A'
CB PB
 ... (iii)
CB PB
Thus,
u
PB  PC PB v
 ... (iv)
PC  PB PB Fig. 1.10 Real image formed by a mirror

Ray Optics 147

 According to new Cartesian sign conventions,
Distance of the object = PB = –u, PC = –R
Distance of the image = PB’ = –v
Thus, equation (iv) becomes,
u  R u
 or uR + vR = 2uv
R  v v
u  R u
On dividing by uvR we get, 
R  v v
Since R = 2F,

1 1 2
Therefore,   . It represents the required mirror formula.
v u R

Magnification
The ratio of the size of the image (I) to the size of the object (O) is defined as the magnification, m.
I v
m 
O u

Notes

The magnification is negative, if the image is real (and inverted); it is positive when the image is
virtual.

Example 1: A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror with radius of
curvature 36 cm. At what distance from the mirror should the screen be placed in order to
receive a sharp image? Describe the nature of the image.
R
Solution: Given, u = –27 cm, f = = –18 cm
2
Thus, using mirror formula
1 1 1 1 1 1
    
v u f v 18 27
 v  54 cm

Since v is negative; the image is real and

v
magnification: m   2
u
The image is twice as large as the object.
Example 2: A
 square wire of side 3 cm is placed 25 cm from a concave mirror of focal length 10 cm.
What is the area enclosed by the image of the wire? (The centre of the wire is on the axis
of the mirror, with its two sides normal to the axis.)
Solution: Given, u = –25 cm, f = –10 cm

148 Ray Optics

 Thus, using mirror formula
1 1 1 1 1 1
    
v u f v 10 25
2
 v  16 cm
3
Hence, magnification:
v 50 2
m   ( 25) 
u 3 3

2
Thus, we get the image of a square of side  3  2 cm.
3
Area of the image is 2 × 2 = 4 cm2.

Example 3: A
 rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm
in such a way that the end closer to the pole is 20 cm away from it. Find the length of the
image.
Solution: Here, R = 2f = 20 cm. Thus, the end of the rod B is at the centre of curvature of the mirror,
and so its image B’ will also be there.

A B

B’ A’ F

Fig. 1.11
Now, we find the image of A. For A we have,
u = – 30 cm and f = – 10 cm. Thus,
1 1 1
 
u v f

Putting the values and solving, we get,

v = – 15 cm.
Thus, the image is formed 15 cm from the pole, and so, the length of the image is
20 – 15 = 5 cm.
Example 4: A
 n object is placed in front of a concave mirror with radius of curvature 15 cm. If the size
of the image is three times that of the object, then find the distance of the object from the
mirror.
Solution: Given that magnification,
v
m  3  v   3u
u

Case I: v = 3 u
1 1 1
   u   10 cm
3 u u 7.5

Case II: v = –3u

1 1 1
   u   5 cm
3 u u 7.5

Ray Optics 149

 The speed of light in vacuum is known to be a constant equal to 3 × 10 8 m/s. This speed, however,
depends on the medium in which light is travelling. Any medium is optically denser than vacuum, and
the speed of light reduces in such a medium. In other words, the speed of light reduces, when it travels
from a rarer to denser medium. The speed, on the other hand, increases, when light travels from a
denser to rarer medium.
To represent this phenomenon diagrammatically, we show the light ray deviates from its original path as
soon as it enters another medium. If it enters a medium in which its speed reduces (a denser medium),
it bends towards the normal to the interface. Conversely, if its speed increases on entering a medium (a
rarer medium), the ray is shown to move away from the normal. This representation is consistent with
observations.
We must therefore be able to define a property of the medium, which determines the speed of light.
This property is called the refractive index of the medium and is defined as,
Speed of light in vacuum c
m  
Speed of light in medium vm

Since vm is less than c, the refractive index of a medium is normally greater than 1.

Notes

Considering light to be a wave, a change in speed would imply a corresponding change in the
frequency and/or the wavelength. The frequency, which is source dependent, does not change. So
it is the wavelength that changes. We say that when light enters a denser medium, its wavelength
v
reduces (since   ), and when it enters a rarer medium, its wavelength increases.
f

Laws of Refraction
1. The incident ray, the refracted ray, and the normal to the point of incidence lie in the same plane.
2.  nell’s law: for any two media, the ratio of the sine of the angle of incidence to the sine of the angle
S
of refraction is a constant for a light of a given frequency.
i
μ1
i
µ2 μ1
r
μ2
r
Fig. 1.12            Fig. 1.13

sin i 2
 or,  1 sin i   2 sin r
sin r 1

When we say that the refractive index of glass is 1.5, we mean that, this is the refractive index with
respect to vacuum or air. In general, refractive index of medium 2 wrt medium 1,
2 v1
2  
1
1 v2

150 Ray Optics

Example: Refractive index of glass with respect to water is 1.125. If the absolute refractive index of
glass is 1.5, find the absolute refractive index of water.
g
Solution: w
 g  1.125 
w
g 1.5 4
 w   
1.125 1.125 3

Angle of Deviation:   i  r

Notes

1. It is important to note at this stage that the refractive index of a medium also depends on the
frequency of light. Light of higher frequency undergoes a greater reduction of velocity than light
of lower frequency on entering a denser medium. For example, violet colour light, which has
the highest frequency in the visible spectrum, will bend the most on entering the glass, and
red colour light, which has the least frequency in the visible spectrum, will bend the least. In
other words, violet light will undergo the maximum deviation in the glass.
2. Light incident at an interface between two mediums also undergoes partial reflection. While
studying refraction, we may ignore the reflected ray, but it is important to remember that there
is also a reflected ray.

Apparent Shift
As the bending of light at the interface of two different media takes place, the image formation due to
refraction creates an illusion of the shifting of the object position.
Consider an object O in medium 1. After refraction, the ray at Medium 2
the interface bends. The bent ray, when it falls on our eyes is rarer (2)
perceived as coming from I.
For nearly normal incident rays, θ 1 and θ2 will be very small. 2
B
tan 1  sin 1 , similarly tan2  sin2 A
2 1 denser ( 1 )
Medium 1
AB I
sin 1 2  1
  1 2   OB  2
sin2 1 AB 1
O
BI
Fig. 1.14

1 Real depth

2 Apparent depth

Ray Optics 151

 Notes

The apparent shift can be positive or negative. A positive shift is when the object appears to be
farther than its real position. A negative shift is when the object appears nearer than the real
position. Is there a simple way to determine both the magnitude and sign of the shift? Here is a
simple algorithm that can be used for this purpose.
1. Apparent shift is always related to an observer. The observer is in a medium, say of refractive
index µ 1.
2. The object being observed is in another medium, say of refractive index µ 2 .
3. Suppose y1 is the distance of the interface between the two mediums from the observer.
4. Suppose y2 is the distance of the interface between the two mediums from the object.
5. Then the apparent shift,
 y2  y2
 y  y1    (y 1  y 2)   y2
 2  2
 1  1

 4
Example: A vessel contains water     up to a height 10 cm and a small coin at the bottom. By
 3
how much will the image of the coin appear to be shifted, when viewed from the top? Now,
if the coin is suspended 10 cm above the water surface and is somehow viewed from the
bottom of the vessel (inside), then by how much will the image appear to have shifted with
respect to the object?
I
Solution:

O
Fig. 1.15 Fig. 1.16

Case I: Coin inside water, viewed from the top (eye outside water): here, medium 1 is air
and medium 2 is water.
y2 10
y   y2   10  3.3 cm
1
2 3/4

Case II: Coin outside, viewed from the bottom (eye inside water): here, medium 1 is water,
and medium 2 is air.
y2 10
y   y2   10  3.3 cm
1
2 3/4

152 Ray Optics

Total Internal Reflection
Consider a ray travelling from a denser to rarer N4
N1
medium. The angle of refraction is greater than the
Rarer N2 N3
angle of incidence. So, as the angle of incidence medium A B 90° D
increases, so does the angle of refraction until we Denser Totally
find that there is a critical angle of incidence, ic, for medium C i r reflected
ray
which the refracted ray grazes along the interface
between the two mediums or the angle of refraction
is 90°. For angles of incidence greater than the critical
angle ic, the incident ray does not emerge out into the O
rarer medium; instead, it gets reflected back into the Total internal reflection
denser medium, as if the interface between the two Fig. 1.17
mediums has turned into a mirror. This phenomenon
is called total internal reflection.

Critical Angle
From Snell’s law, r = 90° Medium 2

 1 sin i C   2 sin 90 o
ic Medium 1
1
2
 i C  sin ( )
1
Fig. 1.18

Example 1: A ray of light from a denser medium strikes a rarer medium at an angle of incidence i.
The reflected and refracted rays make an angle of with each other. The angles of
reflection and refraction are r and r’, respectively. What is the critical angle for total internal
reflection?

i r

r'

Fig. 1.19
Solution: Let the denser medium be medium 1 and the rarer medium be medium 2. Then,

1
 2 
i C  sin  
 1 
 
2 sin i sin r
Also,  1 sin i   2 sin r '   
1 sin r ' sin r '

Since the reflected and refracted rays make 90° with each other,
o o
r  r '  90  r '  90  r

2 sin r sin r
Therefore,    tan r
1 sin r ' o
sin (90  r )

Hence, i C  sin  1 ( tan r )

Ray Optics 153

 Example 2: A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). What is the
condition for a light ray incident normally on the face AB to be totally reflected to reach
the face BC?
B A


Fig. 1.20
Solution: It can be easily seen using geometry that the angle of incidence on the face AC will be θ.
If the critical angle for the glass water interface is ic, then,
1 4 / 3 1  8
i C  sin    sin  
 1.5  9 B A

For total internal reflection, 
  iC

1 8
   sin   C
9
Fig. 1.21
Example 3: Monochromatic light is incident on a plane interface AB between two media of refractive
indices n1 and n2 (n2 > n1) at an angle of incidence as shown. This angle is infinitesimally
greater than the critical angle for the two media, so that total internal reflection takes place.
Now, if a transparent slab DEFG of uniform thickness and refractive index n3 is introduced
on the interface, show that for any value of n3, all light will ultimately be reflected back into
the medium II. Consider separately the cases: (i) n3 < n1, and (ii) n3 > n1.
Medium I
(n1)
D
E
Medium III
(n3)
G F
A B
Medium II
 (n2 )

Fig. 1.22
Solution: Since    C for the mediums I and II,
n1
sin 
n2

Case I: n 3 < n 1

n3 n1
n3  n1  
n2 n2

154 Ray Optics

 n3
Therefore, sin 
n2
 hus, θ is greater than the critical angle for the mediums I and III. Hence, the light ray will
T
undergo total internal reflection.
Case II: n 3 > n 1

Using Snell’s law

n 2 sin  n 3 sin r
n2
 sin r  sin (n1)
n3

(n3) r
Also, since    C for mediums
A B
I and II, we can write, (n2)

n2 n2
sin  sin C
n3 n3
Fig. 1.23
n2 n2 n1
 sin  .
n3 n3 n2
n1
 sin r 
n3
n1
But, if θ C ' be the critical angle for mediums I and III, then sin C '  . So, we have
n3
sin r  sin C '  r  C '

 herefore the ray will undergo total internal reflection at the interface between mediums
T
III and I, and will ultimately emerge out into medium II.

Prism
A prism is a portion of a certain optical medium bounded by two inclined faces. The angle A between
the inclined faces is known as the angle of prism.
A

D

B C
i e
y r1 r2
nt ra emergent ray
ide
inc i E

Fig. 1.24

Angle of deviation (δ) = Angle between incident ray and emergent ray (produced).
Let i = incident angle, e = Emergent angle and r 1 and r2 are angles of refraction.
In the quadrilateral ABEC, angles ABE and ECA are right angles. Therefore,

Ray Optics 155

 A   BEC  180  ... (i)

In triangle BEC, r 1  r 2   BEC  180  ... (ii)

Comparing (i) and (ii), we get A = r1 + r2

Angle of deviation   (i  r 1 )  (e  r 2 )  (i  e)  A

Clearly, the angle of deviation δ is a function of the angle of incidence i. Experimentally, the variation of
δ vs i looks like the graph shown below.
We can make the following observations from the graph.
1. For a specific angle of incidence, i, if the angles of 
emergence and deviation are e and δ, respectively, then
due to the reversibility principle, the ray will retrace the
path in the reverse direction, if the angle of incidence is e.
For the angle of incidence e, the angles of emergence
and deviation will be i and δ, respectively.
2. Starting from a low value of i, as we increase i, both e
and δ decrease. We get the minimum δ, when i = e.
3. When δ is minimum, and i = e, the refracted ray inside
the prism makes equal angles with the sides of the
i/e i=e e/i i
prism.
Thus, for minimum deviation, Fig. 1.25

A
i = e and therefore, r= r=
1 2
2
 m  ( i  e)  A  2 i  A

Am
 i
2
Using Snell’s law,
Am 
sin  
sin i  2 
   
sin r A
sin
2

where, µ is the refractive index of the prism material.

For a very thin prism, both A and δ m are small so that
Am

 2
A
2

  m  (   1) A

156 Ray Optics

Example: A ray of light undergoes a deviation of 30° when incident on an equilateral prism of
refractive index 2 . What is the angle made by the ray inside the prism with the base
of the prism?

Solution: Since,   2 and A  60,

60°
Am 
sin  
 2  30°
  
A
sin
2

 60 o   
m
sin   Fig. 1.26
 2 
Or, 2  
o
sin 30

 60 o   
 sin 
m
 1
 2  2
 
o
60   m o
  45
2
o
  m  30

Thus, the given deviation is the minimum deviation. Therefore, the refracted ray inside the prism makes
equal angles with the sides, or in other words, is parallel to the base of the prism. Thus, angle made by
the ray with the base is zero.

Grazing Incidence and Grazing Emergence A

A
If a light ray is an incident on a prism ABC along OB, i.e., grazing
along face AB, then it emerges out of the prism in a direction CR,
i.e., grazing along AC.
P r r Q
Thus, i = i ' = 90° for grazing incidence and grazing emergence.
T S
From Fig. 1.27, in APTQ ( – 2r)
π – 2r + A = π B C
⇒ A = 2r
Thus, for obtaining an emergent ray from a prism A ≤ 2r O R
Fig. 1.27

Maximum Deviation by Prism

Maximum deviation δmax occurs, when the angle of incidence on the face 1 i
C r 2
of the prism is 90°. Here, ray has ‘grazing incidence’ on the face of the
prism, as shown in Fig. 1.28. Angle of emergence is ‘i’.
∴ δmax = d1 + d2 = (90 – θc) + (i – r) B C
Since, the angle in the glass is critical angle θc when the angle in air is 90°.
O
Fig. 1.28

Ray Optics 157

 Dispersion Produced by a Prism
When a heterochromatic (consisting of many wavelengths) light ray falls on a prism, it is broken into its
component colours. Such type of phenomenon is known as dispersion.

Angle of Dispersion
The difference in angles of deviation of any two rays is called the
angle of dispersion for those rays, e.g., for violet and red, the angle of
dispersion, R V
VR  V   R   V  R  A
Red
Where, A = Angle of prism
Violet
Dispersive Power
Dispersive power is the ratio of the dispersion between any two
Fig. 1.29
colours to the deviation suffered by the same prism by the mean ray.
The dispersive power is denoted by ‘ω’.
VR  V  R 
  
Y  Y  1
Here, mean ray is a yellow coloured ray.
R  V
 Y 
2

Do You Know?

Dispersive power ‘ω’ is independent of the angle of the prism, therefore, it does not depend on the
size of the prism.

Combination of Prism
Dispersion without Deviation
Two prisms, one of crown glass and the other of flint glass, may be combined in such a way that the
deviation suffered by the mean ray is the same for both the prisms.
Here, two prisms (one of crown and other flint glass) are combined together such that their refracting
angles are opposite to each other.
A ray of white light is incident on prism and is
L
dispersed. Here, R, V, and Y represent red, violet,
and yellow-ray, respectively. R
A Y
For such a combination, V
ight
 1   2  0 or
A


Y'  1  White l O
A '  Y  1
A'
P
∴ The total dispersion produced is,
   1  2   1  1  2  Fig. 1.30

158 Ray Optics

Deviation without Dispersion (Or Achromatic Combination of Prism)

t
te ligh
i
A Wh

a y
ite r
Wh

A'
Fig. 1.31

In this case,

 1  2  0 or
A


V'  R' 
A '  V  R 

The total deviation produced is

 1 1 
   1  2  1   
 1 2 

Example 1: A prism of crown glass refracting angle of 5° and mean refractive index = 1.51 is combined
with one flint glass prism of refractive index = 1.65 to produce no deviation. Find the angle
of flint glass and net dispersion.
Given,
v  1.523, R  1.513 (For crown glass)
v'  1.665, R'  1.645 (For flint glass)
Solution: Let A' be the angle of flint glass prism.
A = 5° and µ = 1.51 for crown glass prism.
     1 A   1.51  1  5  2.55

Deviation produced by flint glass.

    ' 1 A '   1.65  1 A '  0.65 A ' ; for no deviation  '   or 0.65 A ' = 2.55

2.55
A'   3.92
0.65
Net dispersion
 v  
 R  A  v'  R' A '

  1.523  1.513  5   1.655  1.645  3.92  0.284

Ray Optics 159

 Example 2: A
 light ray is incident perpendicular to one face of 90° prism and is totally internally
reflected at the glass air interface. If angle of reflection is 45°, find the condition for
refractive index.

45°

Fig. 1.32
Solution: For i = c, r = 90°
For given Fig. 1.32, 45 > c
So, c < 45°
sin c < sin 45°
1
 sinc 
2

1
n=
sin c

Hence, n > 2

Refraction at Spherical Surfaces

A
i
r
  
O P C I

Fig. 1.33
Snell’s law is followed for refraction at spherical surfaces also. In Fig. 1.33, O is the object placed in a
medium of refractive index, say µ 1. Two rays coming from the object meet to form the image I. Let us
say that the refractive index of the second medium is µ 2. Then,  1 sin i   2 sin r .
Here, we assume that the rays are paraxial or close to the principal axis. Under this assumption, the
angles i and r are small. So, we can write,
1 i  2 r

From plane geometry,

i  

and r   

So,  1 (   )   2 (    )

  1    2   ( 2   1 ) 
   

160 Ray Optics

 AP AP AP
Since, the angles are small, we can write   ,  and   . Substituting, we get,
OP PC PI
1 2  2  1
   ... (i)
OP PI PC

Sign Convention
We will adopt the same sign convention as we had for spherical mirrors, (i.e.,) the Cartesian system.
Then, OP = – u, PI = v and PC = R.
Hence, we can rewrite (i) as,
2 1  2  1
 
v u R

Example 1: A narrow beam of light in air falls on a transparent solid sphere of radius R and is brought
to focus on the diametrically opposite surface. What is the refractive index of the material
of the sphere?
Solution: Here, u  ,  1  1, v  2 R.
2 1 2 1
So,  
2R  R
2 1 2 1
 
2R  R

 2  2 Fig. 1.34

Example 2: In a thin spherical fish bowl of radius 10 cm filled with water of refractive index (4/3), there
is a small fish at a distance of 4 cm from the centre, as shown. Where will the fish appear
to be if seen from (i) E, (ii) F? (Neglect the thickness of the glass.)

10 cm
F E

4 cm

Fig. 1.35
Solution: (i) Fish viewed from E: u  6 cm ,  1  4 / 3 ,  2  1 , R  10 cm

1  2  1 2
 
v u R
1 4/3 14/3
 
v
6 10
1 1 2 17
   
v 30 9 90
90
 v  cm
17

Ray Optics 161

 Notes

The image is again virtual; the fish appears farther than what it is.

(ii) Fish viewed from F: u  14 cm ,  1  4 / 3 ,  2  1 , R  10 cm

2 1  2  1
 
v u R

1 4/3 14/3
 
v 14 10

1 1 2 13
   
v 30 21 210

210
 v  cm
13

Notes

The image is again virtual; the fish appears farther than what it is.

Example 3: Fig. 1.36 shows an irregular block of material of refractive index 2 . A ray of light strikes
the face AB as shown. After refraction, it is incident on a spherical surface CD of radius
of curvature 0.4 m, and enters a medium of refractive index 1.514 to meet PQ at E.
Find the distance OE.
B C
45°

P O
Q
E
=1 = 2  = 1.514
60°
A D

Fig. 1.36
Solution: Let us find the angle of refraction for the face AB.
 1 sin i   2 sin r B C

o
45°
1  sin 45  2  sin r

O
P Q
1 E
 sin r 
2 =1 = 2  = 1.514
 r  30
o 60°
D
A
Fig. 1.37

162 Ray Optics

 If we apply simple geometry, we will find that r = 30° means that the refracted ray is
parallel to the axis POQ inside the block. For the spherical surface, this implies that
the object is at infinity. Note that after refraction through the spherical surface, the ray
converges towards the axis because the medium index of 1.514 is greater than 2. If the
medium were air, the ray would have diverged after refraction.
Here, u   ,  1  2 ,  2  1.514 , R  0.4 m.

2 1  2  1
So,    
v u R

1.514 2 1.514  2 1.514  1.414

   ∴ V = 4 × 1.514 = 6.056 M
v  0 .4 0.4

Example 4: A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P
is kept at a distance of (mR) from it. Find the value of m for which a ray from P will emerge
parallel to the table, as shown.

mR R
Fig. 1.38

Solution: The first surface of refraction is a plane vertical surface.

u   m R ,  1  1 ,  2  1.5 , R  

2
1  2  1
 
v
u R

1.5 1 1 .5  1
  0
v  mR 

 v   1.5 m R

 o, the image from the first surface is virtual. This image becomes the virtual object for
S
the second surface. Note that this virtual object is 1.5 mR + R = (1.5 m + 1) R behind the
second surface. So, u   ( 1.5 m  1) R ,  1  1.5 ,  2  1 , R   R and v  .

2
1  2  1
 
v
u R

1 1.5 1  1.5
 
  ( 1.5 m  1) R R

1.5 0.5
 
( 1.5 m  1) R R

4
 m
3

Ray Optics 163

 Thin Lenses
A thin lens is normally made of a transparent material, is circular in shape, and has two spherical
surfaces.
The surfaces may be both convex, both concave, or one convex and one concave. One of the surfaces
can also be plane. Normally, unless mentioned otherwise, we will consider lenses with both surfaces
spherical.

P
C2 P C1 C1 C2

Fig. 1.39    

Fig. 1.40
Each of the surfaces will have a centre of curvature and hence, a radius of curvature. We will call the
centres of curvature C1 and C2 and the corresponding radii R1 and R2. The line joining C1 and C2 is called the
principal axis. The point P at the centre of the lens lying on the principal axis is called the optical centre.
Note: the difference in the position of C1 and C2 between Fig. 1.39 and Fig. 1.40 is shown here. C1 and C2
are the centres of curvature of the first and the second surfaces, respectively, hence, the difference.

Focus and Focal Length

When a paraxial beam of light, travelling parallel to the principal axis, strikes the lens, and goes through,
it either converges to a point on the axis, or the rays diverge away from the axis. In the second case, the
diverging rays, when extended back, would seem to meet at a point on the axis.
In either case, the point where the rays converge or seem to meet when extended back is called the
focus of the lens. The first lens is called a converging or a convex lens, and the second is called a
diverging or a concave lens.

F2 F2

Fig. 1.41 Fig. 1.42

Please note that this focus, as shown in Fig. 1.41 and Fig. 1.42, above, is called the second focus. The
distance of the second focus from the optical centre is called the second focal length f2.
Since the lens has two faces or sides, rays parallel to the axis can also travel from the right and strike
the second surface first. When this happens, the corresponding foci are called the first foci and the
focal lengths are called the first focal lengths f1.

F1 F1

   
Fig. 1.43 Fig. 1.44

164 Ray Optics

 Notes

The first and second focal lengths will be different only if the medium on one side of the lens is
different from the medium on the other side. If the mediums on both sides are the same, e.g., air,
then f1 = f2 = f and the lens has a single focal length.

Image Formation
Rules of Refraction for a Converging (convex ) Lens
1. An incident ray travelling parallel to the principal axis of a converging lens will refract through the
lens and pass through the focal point on the opposite side of the lens.
2. An incident ray travelling through the focal point on the way to the lens will refract through the lens
and travel parallel to the principal axis on the other side of the lens.
3. An incident ray which passes through the centre of the lens will continue in the same direction that
it had when it entered the lens.
Based on these rules, we show the nature and position of the image at various positions.
Case 1: Object located beyond 2F

Object Object

2F F F 2F 2F F F 2F

Image Image

 
Fig. 1.45              Fig. 1.46

Case 2: Object located at 2F Case 3: Object located between F and 2F

Object
Object

2F F F 2F
2F F F 2F

Image
Image

Fig. 1.47              Fig. 1.48

Ray Optics 165

 Case 4: Object located at F Case 5: Object located in front of F

Virtual
image

Object Object

2F F F 2F
2F F F 2F

Fig. 1.49        Fig. 1.50

Refraction Rules for a Diverging Lens

1. Any incident ray travelling parallel to the principal axis of a diverging lens will refract through the
lens and travel in line with the focal point (i.e., in a direction such that its extension will pass
through the focal point).
2. Any incident ray travelling towards the focal point on the way to the lens will refract through the
lens and travel parallel to the principal axis.
3. An incident ray, which passes through the centre of the lens will continue in the same direction that
it had when it entered the lens. Based on these rules, the image formation by a diverging lens is as
shown below.

2F F F 2F 2F F F 2F 2F F F 2F

  Fig. 1.51           Fig. 1.52 Fig. 1.53

The above diagrams show that in each case, for a diverging lens, the image is,
1. Located behind the lens
2. A virtual image
3. An upright image
4. Reduced in size (i.e., smaller than the object)

Lens Maker’s Formula

Consider a point object O kept on the principal axis of a converging lens. The image I is formed on the
other side of the lens.

166 Ray Optics

 O' O I

Fig. 1.54
Let the medium on either side of the lens be the same and of refractive index µ 1, and the refractive
index of the lens be µ 2. Then for the first surface, we can write:

2 1 2  1
   ... (i)
v' u R1

Here, v ′ is the image distance from the optical centre. In Fig. 1.54, this image has been shown as O′
and acts as a virtual object for the second surface. So, for the second surface, we have:

1 2 1  2
   ... (ii)
v v' R2

Adding (i) and (ii) we get

1 1  1 1 
1 (  )  (  2   1)   
v u  R1 R2 
 

1 1 2  1 1 
Or,  (  1)    ... (iii)
v u 1  R1 R2 
 
When the object is very far from the lens, the image is formed close to the focus. So, for u  , we have
v = f. Hence,
1 2  1 1 
(  1)     ... (iv)
f 1  R1 R2 
 

This is called the lens maker’s formula because it is used by lens makers to decide the curvatures to
get a desired focal length, or ‘power.’
If we combine (iii) and (iv) we will get the lens formula.
1 1 1
 
v u f

Sign Convention
We will use the same sign convention as we had followed in the case of spherical mirrors. The optical
centre is taken as the origin and the principal axis as the x-axis. The positive direction is taken along
the direction of the incident ray.

Magnification
Magnification is defined in the same manner as for spherical mirrors.
I v
m
= =
O u

Ray Optics 167

 Notes

1. Focal length of a converging lens is always positive.

As an example, consider a converging lens of µ = 1.5 and radii of curvature 30 cm and
60 cm, respectively.
A B

C2 C1

Fig. 1.55
Case I:
The light from the object is incident on surface A,

1  2   1 1 
 1   
f  1   R1 R2 
   
 1 1  1
 ( 1.5  1)   
 30 60  40
 f  40 cm

Case II:
The light from the object is incident on surface B,

1  2   1 1 
 1   
f  1   R2 R1 
   
 1 1  1
 ( 1.5  1)   
 60 30  40
 f  40 cm

2. Focal length of a diverging lens is always negative.

A B

C1 C2

Fig. 1.56
Let us determine the focal length of a diverging lens with
  1.5 , R 1  30 cm , R 2  60 cm
No matter which way we consider, we will get,

1  2   1 1 
 1   
f  1   R1 R2 
   
 1 1  1
 ( 1.5  1)   
 30 60  40
 f   40 cm

168 Ray Optics

Example: A convex lens made of a material of refractive index 1.5 has a focal length of 0.3 m in air.
What is its focal length in water of refractive index 4/3?
1 1.5  1 1 
Solution: In air, (  1)     ... (i)
fA 1  R1 R2 
 

1 1.5  1 1 
In water, (  1)    ... (ii)
fW 4/3  R1 R2 
 
Therefore,
fW
=4
fA

∴ fW = 4f4 = 1.2 m

Lens Continued
In this session, the concluding session for ray optics, we will continue our study of lenses. More
specifically, we will look at what happens when we combine lenses or lens with mirrors.
Let us first ask ourselves, what happens if we cut a lens into two halves, as shown below?

Fig. 1.57
If we use only one half, do we get only half the image? We must understand here that although only two
rays are sufficient to trace the image of an object, and we often show only two rays, in reality, there are
infinitely many rays that create the image. Look at Fig. 1.58 and Fig. 1.59 below to convince yourself that
even with half the lens, we get the full image.

Fig. 1.58

Fig. 1.59
The only difference: when we use half the lens, the other half is absent and therefore, plays no role in
the image formation. This means, as compared to the full lens, the number of rays contributing to image
creation becomes half. Hence, the intensity of the image becomes half the intensity which would be
there if there was a full lens.
What happens if we use both halves together but are separated by a small distance? Do we get only one
image as we get for the full lens? Let us try and answer this by tracing the image(s) of a point object.

Ray Optics 169

 I1

I2
Fig. 1.60
We see that two images are formed in this case. Why? There are actually two principal axes, one each
for each of the two halves. For the upper half, the object lies below the principal axis. For the lower half,
it lies above the principal axis.
Let us consider the following examples.
Example 1: A point object is placed at a distance of 0.3 m from a convex lens (focal length 0.2 m) cut
into two halves, each of which is separated by 0.0005 m, as shown. Find the position of
the image(s). How many images are formed? What is the distance between them?
Solution: Even though there are two halves of the lens, they have the same focal length = 0.2 m.
Hence,
1 1 1
 
u f
v
1 1 1
 
v 0.3 0.2
1 1 1
  
v 0.2 0.3
32 1
 
0.6 0.6
 v  0.6 m
The image distance is the same for both images. To find the distance between them, we will
use plane geometry (similarity of triangles).
I1 I2 u  v

L 1 L2 u

u  v 0.3  0.6
 I1 I2   L 1 L2   0.001  0.003 m
u 0.3

I1
L1

0.0005 m

L2
I2

Fig. 1.61

170 Ray Optics

Example 2:  thin plano-convex lens of focal length f is split into two halves: one of the halves is shifted
A
along the principal axis. The separation between the object and image planes is 1.8 m.
 he magnification of the image formed by one of the half lenses is 2. Find the focal length
T
of the lens and the separation between the two halves.

1.8 m

Fig. 1.62
Solution: How many images are formed? Let us find out by tracing the image(s).

I1

I2

Fig. 1.63
 here will be two images. Since the problem states only one image plane, both these are
T
formed on the same plane, i.e., at the same position on the optical axis. The magnification
2 can be only due to the first half lens, as verified from Fig. 1.63. If the object distance for
this half is x, then the image distance is (1.8 – x). Thus, magnification,
v ( 1.8  x )
m  2
u x
 x  0.6 m
Therefore, u   0.6 m , v  1.2 m. Using the lens equation,
1 1 1
 
u f
v

1 1 1 1 1 3
     
f 1.2 0.6 1.2 0.6 1.2
 f  0.4 m

This is also the focal length for the second half lens. If the object distance for this half is y,
then the image distance is (1.8 – y). So,
1 1 1
 
u f
v
1 1 1 1.8 1
    
1.8  y  y 0.4 y ( 1.8  y ) 0.4

Ray Optics 171

 2
 1.8 y  y  0.72
2
y  1.8 y  0.72  0 or ,
2
y  0.6 y  1.2 y  0.72  0

 ( y  0.6)( y  1.2)  0

 y  0.6 m , 1.2 m

 ince the object and image positions are fixed, the two values of y give us the positions of
S
the two half lenses with respect to the object. Note that the magnification for one value of
y corresponds to 2, while for the other value of y corresponds to 0.5. Clearly, the distance
between the two half lenses is (1.2 – 0.6) = 0.6 m.

Fig. 1.64

Optical Power
The optical power of a lens is given by
1
P=
f

where, f is the focal length in metres. Hence, the unit of power is m–1, which is known as the diopter. Since,
the focal length of a converging lens is positive, its power is positive. For the same reason, the power of a
diverging lens is negative. This power is the same as used by optometrists for making spectacles.

Thin Lenses in Contact

Two lenses are in contact, as shown in Fig. 1.64. Suppose their focal lengths are f1 and f2. Suppose an
object is placed at a distance u, from the first lens, so that the image due to it is formed at a distance
v’ away. This image becomes the object for the second lens, and let v be the final image distance. Then,
for the first lens,
1 1 1
 
v ' u f1

For the second lens,

1 1 1
 
v v ' f2

Combining, we get,
1 1 1 1
  
v u f1 f2

Therefore, if the focal length of the combination is f, then,

1 1 1
 
f f1 f2
In effect, we say that the power of the combination is equal to the sum of the powers of both lenses.

172 Ray Optics

Convex Lens with One Surface Silvered
Let us consider a convex lens with one surface silvered. Let the focal length of the lens be fL, and that
R2
of the concave mirror be fM (the mirrored surface will act as a concave mirror). Then, f M = .
2

Fig. 1.65
Such a lens will act as a concave mirror. The combined power is given by
P  PL  PM  PL

1 1
where PL  and PM  
fL fM
The effective focal length of the combination is given by
1
f  
P

Example 3: The radius of curvature of the convex face of a plano-convex lens is 12 cm and its refractive
index is 1.5.
1. Find the focal length of this lens. The plane surface of the lens is now silvered.
2. At what distance from the lens will parallel rays incident on the convex face converge?
3. Find the image distance, when a point object is placed on the axis 20 cm from the lens.
Solution: 1. Focal length of the lens,
1 1 1 1 1 1
 (   1)(  )  ( 1.5  1)(  ) 
fL R1 R2 12  24

 f L  24 cm

2. Focal length of the silvered lens,
P  PL  PM  PL  2 PL  PM

1 1 1
 2  
0.24  0.12
 f   0.12 m   12 cm

Now,
1 1 1 1 1 1
    
v f u 12 20 30

 v  30 cm

Example 4: The convex surface of a thin concavo-convex lens of glass with refractive index 1.5 has
a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The
convex side is silvered and placed on a horizontal surface.

Ray Optics 173

 1.  here should a pin be placed on the optic axis such that its image is formed at the
W
same place?
2. If the concave part is filled with water of refractive index (4/3), find the distance
through which the pin should be moved, so that the image of the pin again coincides
with the pin?
Solution: 1. Combined power,
P  PL  PM  PL  2 PL  PM

1 1 1 1
PL   ( 1.5  1) (  ) D
fL 0.60 0.20 0.60

1 1 1
PM     D
fM 0.20 / 2 0.1

Fig. 1.66
Therefore,
2 1 2
P  2 PL  PM    D
0 . 60 0 . 1 0 . 15

1
 f    7.5 cm
P
 hus, the combination acts as a concave mirror of focal length 7.5 cm. For the image
T
to coincide with the object, the object must be placed at the centre of curvature.
Hence, u  R  2 f  15 cm .
2. W
 hen the concave part is filled with water, the water forms another lens (plano-
convex).
 o, a ray of light from the object would undergo two refractions at each of the lenses
S
and one reflection in the mirror. Hence,
P  PW  PL  PM  PL  PW  2 PL  PM  2 PW

1 4 1 1 1
PW   (  1) (  ) D
f W
3  0 . 60 1 . 80

Therefore,
2 1 2 26
P  2 PL  PM  2 PW     D
0.60 0.1 1.80 1.80

1 180 cm
 f     6.92 cm
P 26
The pin needs to be placed at a distance of 13.84 cm from the lens, and hence, the
distance by which it needs to be displaced = 15 – 13.84 = 1.16 cm.

174 Ray Optics

 Practice Exercise

Multiple Choice Questions (Single Choice (C) the object is virtual and the image is real
Correct) (D) the object and the image are real
Level I 8. If the angle of incidence is doubled for an
incident ray, then the angle of reflection ____.
1. Reflection of light can be explained by ____.
(A) is halved (B) is doubled
(A) wave nature
(C) is same (D) none of these
(B) particle nature
(C) both wave and particle nature 9. In the previous question, what can be said
(D) none of these about the initial angle of incidence which is
doubled?
2. If a ray of light is incident on a mirror, making
angle 30° with the mirror, then the angle of (A) i  90 (B) i  60
reflection is ____. (C) i  45 (D) None of these
(A) 30° (B) 60° 10. The focal length of a convex mirror is 7 cm.
(C) 150° (D) 45° An object is placed 8 cm from it. Magnification
of the image is ____.
3. If you are moving with a speed of 0.5 m/s
(A) 1.5 (B) 0.47
towards a mirror, the speed of image as seen
(C) 1 (D) 0.9
by you is ____.
(A) 0.5 (B) 1 11. Concave mirrors can be used to create fire
(C) 0 (D) 1.5 from sunlight because ____.
(A) s un rays converge at focus, so the
4. When the object is placed in-between focus
intensity of ray is high
and centre of curvature of a mirror, then the
(B) sun rays converge at centre of curvature
image is formed ____.
of the mirror
(A) between C and F
(C) sun rays do not converge but have enough
(B) beyond C and is diminished
intensity to cause fire
(C) beyond C and is enlarged
(D) none of these
(D) at infinity
12. An object of height 10 cm forms a real image
5. If the object is placed 30 cm in front of a
of height 5 cm when placed at a distance of
convex mirror of focal length 15 cm, then the
6 cm from a concave mirror. Then the focal
image formed is ____.
length of the mirror is ____.
(A) real and diminished
(A) 2 cm (B) 4 cm
(B) virtual and diminished
(C) 5 cm (D) 2.5 cm
(C) virtual and enlarged
(D) real and enlarged 13. A metre stick is placed 7 cm from a convex
6. An object is placed at a distance of 6 cm from mirror of focal length 28 cm. The height of
a concave mirror. If the image is formed 8 cm the image is ____.
on the same side, then the focal length is ____. (A) 1.2 m (B) 0.8 m
(A) 5 cm (B) 4 cm (C) 0.7 m (D) 0.9 m
(C) 3.43 cm (D) 4.5 cm 14. The refractive index of a medium is defined
7. Radius of curvature of a convex mirror is as ____.
10 cm. If image is formed 10 cm on the other (A) V (B) C
side of the mirror, then ____. C V
(A) both the object and the image are virtual (C) VC (D) None of these
(B) the object is real and image is virtual

Ray Optics 175

 15. µ2 is best represented as ____. 1 2
1 (A)  Re al depth (B)  Re al depth
2 1
µ2 µ
(A) (B) 1
µ1 µ2 (C) Real depth (D) Data insufficient
µ
(C) 1 2 µ (D) None of these 21. The velocity of light in a medium, where
16. Snell’s law at the surface AB is best absolute refractive index is 2, will be ____.
represented by the expression ____. (A) 3 × 109 m/s (B) 6 × 109 m/s
(C) 1.5 × 109 m/s (D) None of these

i 1 22. Total angular dispersion is given by where

B symbols have their usual meaning ____.
A
2 (A) A(   1)1 (B) A(   1)(1  2 )
r
(C) A(   1)(1  2 ) (D) None of these

(A)  1 sin i  2 sin r 23. In case of combination of prism, total

deviation is given by ____.
(B)  1 sin r  2 sin i (A) A(   1) (B) A(1  2 )
(C) sin i sin r   1 2   
(C) A(1  2 ) (D) A(   1)  1  1 
 2 
(D) sin i   1  sin r  2 
24. If A is the angle of prism, r and r’ are the angle
17. When a ray travels from a rarer medium to of refraction, then ____.
denser medium____. (A) r = r′ (B) r = r′ = A
(A) it bends toward the normal (C) r + A = r′ (D) r + r′ = A
(B) it bends away from the normal
(C) it moves along its path 25. If A is the angle of prism, i1 and i2 are the angle
(D) nothing can be said of incidence, then the basic prism equation
is ____.
18. When a ray travels from a denser medium to (A)   i1  i2 (B)   (i1  i2 )  A
rarer medium ____.
(C)   A (D)   i1  i2  A
(A) it bends toward the normal
(B) it bends away from the normal 26. The condition for minimum deviation is
(C) it moves along its path for the symbols having their usual meaning
(D) nothing can be said ____.
19. In the following Fig. 1.67, the condition for (A) i1 = i2 (B) r1 = r2
total internal reflection is ____. (C) A = 2r (D) All of these

x 27. The condition for minimum deviation is

for the symbols having their usual meanings
y ____.
(A)   2i  A (B)   2i
   Fig. 1.67 (C)   2r (D)   2A

28. A prism splits a beam of white light into its

(A) sinC   x (B) sinC   y
constituent colours. This is so because ____.
1 (A) phase of each colour is different
(C) sinC  x  y (D) sinC 

x y (B) a mplitude of different colours is
20. The apparent depth in terms of real depth can different
be written (if µ 1 and µ2 are refractive indices (C) f requency of different colours is
of denser and rarer mediums, respectively) different
____. (D) velocity of different colours is different

176 Ray Optics

29. A prism has a refracting angle of 60°. When a ray (A) f = R (B) f = 2R
of light is incident on its face at 45°, it suffers R 3R
(C) f = (D) f =
minimum deviation. The angle of minimum 2 2
deviation is ____. 37. A convex lens is immersed in a liquid of
(A) 30° (B) 45° refractive index greater than that of the glass.
(C) 60° (D) 15° It will behave as ____.
(A) convergent lens
30. A prism of 6° is made of material of refractive
(B) divergent lens
index 5 . The deviation caused by it, is ____.
3 (C) plane glass
(A) 2° (B) 4° (D) homogeneous liquid
(C) 8° (D) None of these
38. An equi-convex lens is made of a material,
31. A glass prism with refractive index 2 has a which has a refractive index of 1.6 for blue
refracting angle of 30°. One of the refracting light and 1.5 for red light. What is the ratio of
surfaces of the prism is silvered. A beam of focal length for red light to focal length for
monochromatic light will retrace its path if blue light?
the angle of incidence is ____. 5
(A) (B) 15
(A) 90° (B) 30° 6 16
(C) 45° (D) 60° 6
(C) 16 (D)
32. A prism of refracting angle 60° has a refractive 15 5
index equal to µ. For a certain wavelength of 39. A hollow double concave lens is made of very
light, the angle of minimum deviation is 30°. thin transparent material. It can be filled with
For this wavelength, µ is equal to ____. air or either of two liquids L1 or L2, having
(A) 1.531 (B) 1.593 refractive indices n1 and n2, respectively (n2 >
(C) 1.414 (D) 1.160 n1 > 1). The lens will diverge a parallel beam of
light if it is filled with ____.
33. The angle of refraction is half the angle of
(A) Air and placed in air
incidence, when the ray passes symmetrically
(B) Air and immersed in L1
through the prism (i  e). The refractive
(C) L1 and immersed in L2
index of the prism is 2. Then the angle of (D) L2 and immersed in L1
the prism (A) is equal to ____.
(A) 90° (B) 75° 40. A convex lens of focal length 40 cm is in
(C) 60° (D) 45° contact with a concave lens of focal length
25 cm. The power of the combination is ____.
34. The focal length of a lens does not depend on
(A) –1.5 D (B) –6.5 D
____.
(C) 1.5 D (D) 6.5 D
(A) the material of the lens
(B) r efractive index of the surrounding 41. A convex lens projects full image of an object.
medium If half of the lens is covered by an opaque
(C) the radii of curvature of its surface paper, then ____.
(D) the aperture of the lens (A) half image is not seen
(B) full image of same intensity is seen
35. An object is kept at a distance of 40 cm
(C) full image of reduced intensity is seen
from a convex lens of focal length 15 cm. The
(D) no image is seen
image distance and the nature of image are
42. A convex lens forms a real image of a point
respectively ____.
object placed on its principal axis. If the
(A) 24 cm and virtual (B) 24 cm and real
upper half of the lens is painted black then
(C) 48 cm and virtual (D) None of these
____.
36. In case of equi-convex lens (   1.5 ), the (A) the image will be shifted downward
relation between focal length f and radius of (B) the image will be shifted upward
curvature R is ____.
Ray Optics 177
 (C) the image will not be shifted (C) a ngle of incidence is not equal to angle
(D) the intensity of the image will increase of reflection
(D) either (B) or (C) is correct, depending on
43. If a lens is painted half black, its image will
the angle of incidence
____.
(A) not be formed 49. If the incident ray on a plane mirror strikes
(B) be half of actual size the surface at an angle of 150° to the mirror
(C) will be of same size and of same intensity surface, what is the angle of reflection?
(D) will be of same size and half the intensity (A) 150° (B) 60°
(C) 30° (D) 40°
44. If a lens is cut parallel to the principal axis,
____. 50. Two mirrors are inclined at an angle of 200°.
(A) its focal length will be half An object is placed so that the incident rays
(B) its focal length will be twice can fall on both the mirrors. Then the number
(C) nothing will happen of images formed is ____.
(D) data insufficient (A) 1 (B) 2
(C) 3 (D) 4
45. A point object is placed in front of a silvered
plano-convex lens of refractive index n and 51. Three mirrors are placed along the sides of
radius of curvature R so that its image is an equilateral triangle with reflecting sides
formed on itself. Then the object distance is facing each other. If the object is placed at the
equal to ____. centroid, then the number of images formed
is ____.
(A) 6 (B) 12
(C) 18 (D) Infinite
O 52. A man stands at a distance of 10 m in front
of a concave mirror of focal length 4 m. He
stands above the principal axis and starts
R 2R walking away from the mirror. Then his image
(A) (B)
2n n will ____.
R (A) start enlarging and moving away from the
(C) (D) None of these
n mirror
Level II (B) start diminishing and moving away from
the mirror
46. When a ray of light strikes a plane mirror, (C) start diminishing and moving towards the
its angle of deviation is 30°. The angle of mirror
reflection is ____. (D) reach the focus when the man moves 2
(A) 100° (B) 90° m away from the mirror
(C) 120° (D) 75°
53. A ray of light travels in the fashion as shown
47. The number of images formed by two mirrors in Fig. 1.68. After passing through water, the
inclined at an angle of 40° (for an object ray graces along the water-air interface. The
placed between them) is ____. value of µ2 in terms of i is ____.
(A) 4 (B) 6
(C) 8 (D) 9
Water
48. When a ray of light is incident on a concave = 4/3
mirror, the reflected ray is such that ____.
(A) it either passes through the focus or i Glass
becomes parallel to principal axis
(B) angle of incidence = Angle of reflection Fig. 1.68

178 Ray Optics

 4 1 two mediums is c. The maximum possible
(A) sin i (B)
3 sin i deviation of the ray will be ____.
(A)   c (B)   2c
4
(C) (D) 1 
3 (C) 2c (D) c
2
54. A ray of light passes through four transparent 58. The condition for achromatic combination
mediums with refractive indices µ1, µ2, µ3, and of lens is where symbols have their usual
µ4 as shown in Fig. 1.69. The surface of all meaning.
mediums are parallel. If the emergent ray CD
w1 w2
is parallel to incident ray AB, ____. (A)  0
f1 f1
1 2 3 4 D
w1 w1
C (B)  0
f1 f2
B (C) w 1  w2  0
A
w1 w2
(D)  0
Fig. 1.69 f1 f2

(A)  1  2 (B) 2  3 59. A glass prism of refractive index 1.5 is

(C) 3  4 (D)  1  4  4
immersed in water     as shown in
 3
55. Two beams of light are incident normally on Fig. 1.71. A light ray incident on face AB is
 4 reflected to reach the face BC, if ____.
water    . If beam 1 passes through a
 3 B A

glass of height h as shown in Fig. 1.70, the time
difference in both the beams on reaching the
bottom is ____.
2 C
1

Fig. 1.71
glass
3
h
(A) sin  (B) sin  2
2
2 8
(C) sin  (D) sin 
3 9
Fig. 1.70
60. A thin prism P1 of angle 4°, made of a glass
(A) Zero (B) h (   1.54), is combined with prism P2, made
6c
6h h from a material (µ = 17.2), such that it
(C) (D) produces dispersion without deviation. The
c 6c
56. A ray of light from a denser medium strikes a angle of P2 is ____.
rarer medium at an angle of incidence i. The (A) 5° (B) 4°
reflected and refracted rays make an angle of (C) 3° (D) 2°
90° with each other. The angle of reflection 61. A given ray of light suffers minimum
and refraction are r and , respectively. The deviation in an equilateral prism P, additional
critical angle is ____. identical prism Q and R are added as shown
(A) sin (tan r ) in Fig. 1.72. The ray will now suffer ____.
−1
(B) sin−1 (cot i )

(C) tan−1 (sin r ) (D) tan−1 (sin i ) Q

P R
57. A ray of light travels from an optically denser
to a rarer medium. The critical angle for the Fig. 1.72

Ray Optics 179

 (A) greater deviation (A) 4 (B) 3
(B) no deviation (C) 2 (D) None of these
(C) same deviation as before
66. A thin, symmetric double-convex lens of
(D) total internal reflection
power P is cut into three parts A, B, and C
62. What is the relationship between the focal as shown in Fig. 1.75. The power of ____.
length of a concave mirror and the refractive A
index µ?
1
(A) f 

(B) f  
(C) f is independent of µ B C
(D) None of these
63. If the behaviour of light rays through a convex Fig. 1.75
lens is as shown in Fig. 1.73, then ____.
(A) A is P (B) A is 2P
P
(C) B is P (D) B is
4
67. Diameter of a plano-convex lens is 6 cm and
2  2 thickness at the centre is 3 mm. Its speed
of light in material of lens is 2 × 108 m / s, the
focal length of the lens is ____.
 [JEE Main 2013]
Fig. 1.73
(A) 15 cm (B) 20 cm
(A) µ = µ2 (B) µ < µ2 (C) 30 cm (D) 10 cm
(C) µ > µ2 (D) µ ≤ µ2
68. The graph between angle of deviation   and
64. A concave lens of glass, refractive index 1.5 angle of incidence (i) for a triangular prism is
has both surfaces of same radius of curvature represented by ____.
R. On immersion in a medium of refractive
 [JEE Main 2013]
index 1.75, it will behave as a ____.
(A) convergent lens of focal length 3.5R  
(B) convergent lens of focal length 3.0R
(C) divergent lens of focal length 3.5R (A) (B)
(D) divergent lens of focal length 3.0R O O
i i
65. A thin lens is cut into identical parts to make
L1 and L2. Again, L1 is made into two identical  
parts; 1 and 2, and L2 is made into two parts;
3 and 4, as shown in Fig. 1.74. If L1 and L2 are (C) (D)
slightly displaced along the x-axis, the total O O
i i
number of images formed due to an object,
O is equal to ____.
69. A thin convex lens made from crown glass
 3
    has focal length f. When it is measured
L1  2
1 2
in two different liquids having refractive
O X 4 5
indices 3 and , it has the focal lengths
3
3
L2 f1 and f2, respectively. The correct relation
4
between the focal lengths is ____.
Fig. 1.74  [JEE Main 2014]

180 Ray Optics

 (A) f2 > f and f1 becomes negative
   1  
(B) f1 and f2 both become negative (C)   sin1   sin  A  sin1    
(C) f1 = f2 < f       
(D) f1 > f and f2 becomes negative    1  
(D)   sin1   sin  A  sin1    
70. A green light is incident from the water to the       
air–water interface at the critical angle (θ).
Select the correct statement. 73. An observer looks at a distant tree of height
 [JEE Main 2014] 10 m with a telescope of magnifying power
(A) The spectrum of visible light whose of 20. To the observer, the tree appears ____.
frequency is more than that of green light  [JEE Main 2016]
will come out to the air medium. (A) 10 times taller (B) 10 times nearer
(B) The entire spectrum of visible light will (C) 20 times taller (D) 20 times nearer
come out of the water at various angles
74. In an experiment for the determination of
to the normal.
refractive index of glass of a prism by i –
(C) The entire spectrum of visible light will
δ, plot, it was found that a ray incident at
come out of the water at an angle of 90°
angle 35°, suffers a deviation of 40° and
to the normal.
that it emerges at angle of 79°. In that case,
(D) The spectrum of visible light whose
which of the following is closest to the
frequency is less than that of green light
maximum possible value of the refractive
will come out to the air medium.
index?
71. Assuming human pupil to have a radius of
 [JEE Main 2016]
0.25 cm and a comfortable viewing distance
(A) 1.5 (B) 1.6
of 25 cm, the minimum separation between
(C) 1.7 (D) 1.8
two objects that human eye can resolve at
500 nm wavelength is ____. 75. A diverging lens with magnitude of focal
length 25 cm is placed at a distance of
 [JEE Main 2015]
15 cm from a converging lens of magnitude
(A) 100 µm (B) 300 µm
of focal length 20 cm. A beam of parallel
(C) 1 µm (D) 30 µm light falls on the diverging lens. The final
72. Monochromatic light is incident on a glass prism image formed is ____.
of angle A. If the refractive index of the material  [JEE Main 2017]
of the prism is μ, a ray, incident at an angle θ, (A) real and at a distance of 40 cm from
on the face AB would get transmitted through convergent lens.
the face AC of the prism provided ____. (B) virtual and at a distance of 40 cm from
convergent lens.
 [JEE Main 2015] (C) real and at a distance of 40 cm from
the divergent lens.
A (D) real and at a distance of 6 cm from the
 convergent lens.

76. An upright object is placed at a distance

B C of 40 cm in front of a convergent lens of
focal length 20 cm. A convergent mirror of
   1  
(A)   cos1   sin  A  sin1     focal length 10 cm is placed at a distance
       of 60 cm on the other side of the lens. The
 position and size of the final image will be
  1  
(B)   cos1   sin  A  sin1     __________.
      
 [JEE Main 2019]

Ray Optics 181

 (A) 4 0 cm from the convergent lens, twice 4 3
the size of the object g  and a  g  .
a
3 2
(B) 40 cm from the convergent mirror, same
size as the object 3. Find the focal length of a plano-convex lens
(C) 20 cm from the convergent mirror, twice when ________.
the size of the object a. the plane surface is silvered and the
(D) 20 cm from the convergent mirror, same object is placed in front of a curved
size as the object surface.
77. In Fig. 1.76, the optical fibre is l = 2 m long and b. curved surface is silvered and the object
has a diameter of d = 20µ m. If a ray of light is in front of a plane surface.
is incident on one end of the fibre at angle 4. A ray of light is incident on a transparent glass
θ1 = 40°, the number of reflections it makes slab of refractive index 1.62. If the reflected
before emerging from the other end is close and refracted rays are mutually perpendicular,
to ____. (refractive index of fibre is 1.31 and what is the angle of incidence?
sin 40° = 0.64). [JEE Main 2019]
i r

d
2
r'
40°

Fig. 1.76 5. Light is incident from air on oil at an angle of

30°. After moving through oil 1, oil 2, and glass,
(A) 45000 (B) 55000
it enters water. If the refractive index of glass
(C) 66000 (D) 57000
and water are 1.5 and 1.3, respectively, find
78. The wavelength of the carrier waves in a the angle which the ray makes with normal
modern optical fibre communication network in water?
is close to ____.
30°
(A) 600 nm (B) 2400 nm Air
(C) 1500 nm (D) 900 nm Oil 1
 (JEE Main 2019]
Multiple Choice Question (One or More Oil 2
Choice Correct) Glass
79. Which of the following forms, virtual and erect Water
images for all positions of the object? 6. A fish is positioned at a depth h inside a lake.
(A) Convex lens (B) Concave lens Calculate the radius of the circular aperture
(C) Convex mirror (D) Concave mirror through which fish sees the outside world
( water   ).
Subjective Type Questions
7. A fish in an aquarium approaches the left
1. A convex lens of focal length 15 cm is placed in
wall at a rate of 3 m/s and observes a fly
front of a convex mirror. Both are coaxial, and
the lens is 5 cm from the apex of the mirror. approaching it at 8 m/s. If the refractive index
4
When an object is placed on the axis at a of water is , find the actual velocity of the
3
distance of 20 cm from the lens, it is found that fly.
the image coincides with the object. Calculate
the radius of curvature of the mirror. fly

2. A lens has a power of +5 diopters in air. What x

x
will be its power if completely immersed in
water?

182 Ray Optics

 8. A prism of refracting angle 30° is coated
with a thin film of transparent material of I1
refractive index 2.2 on face AC of the prism.
Such that angle of incidence is 60°. Find the d

angle of emergence  glass  3 .  O
A

30°

I2

B C 0.3 M v

9. An image y is formed of a point object x by Fig. 1.77

a lens whose optic axis is AB, as shown in
13. A quarter cylinder of radius R and refractive
the following figure. Draw a ray diagram to
index 1.5 is placed on a table. A point object
locate its focus. If the image y of the object
P is kept at a distance mR from it. Find the
x is formed by a concave mirror (having the
value of m from which a ray from P will
same optic axis AB) instead of a lens, draw
emerge parallel to the table, as shown in
another ray diagram to locate the mirror and
Fig. 1.78.
focus. Write down the steps of construction
of the ray diagram.

10. The distance between two point sources of P

light is 2.4 cm. Find out where you would mR R
place a converging lens of focal length
Fig. 1.78
9 cm, so that the images of both sources are
formed at the same point. 14. The convex lenses of focal length 20 cm
each are laced coaxially with separation of
3
11. A thin biconvex lens of refractive index is ∈0 cm between them. Find the image of a
2
placed on a horizontal mirror, as shown in distant object formed by the combination.
the following figure. The space between the 15. A convex lens of focal length f0 = 24 cm in
lens and the mirror is then filled with water air, is surrounded by four different mediums,
4 as shown in Fig. 1.79. A point object ‘O’ is
of refractive index . It is found that when a
3 placed along the principal axis at a distance
point object is placed 15 cm above the lens u = 30 cm from the lens. Find the number
on its principal axis, the object coincides with and position of the images formed.
its own image. On repeating with another
liquid, the object and image again coincide at
a distance of 25 cm from the lens. Calculate 1 = 1.0 3 = 1.2 Principal
the refractive index of the liquid. O 2 = 1.2 4 = 1.6
axis

12. A point object is placed at a distance of 0.3

m from a convex lens (focal length 0.2 m) cut
Fig. 1.79
into two halves, each of which is displaced
by 0.0005 m, as shown in Fig. 1.77. Find the 16. Fig. 1.80 shows a silvered lens, µA = 1.6 and
position of the image. If more than one image µB = 1.2, R1 = 80 cm, R1 = 40 cm, and R3 =
is formed, find the number and distance 20 cm. An object is placed at a distance of
between them. 12 cm from this lens. Find the image
position.

Ray Optics 183

 between them. What is the value of d, if a
parallel beam of light incident on A leaves B
R1 A B as a parallel beam?

A B 19. A ray of light refracted through a sphere,

R2 whose material has a refractive index
µ , in such a way that it passes through the
R3
extremities of two radii, which make an angle
Fig. 1.80 with each other. If α is the deviation of the
ray caused by its passage through the sphere,
17. An object is placed at a distance of 10 cm to
1
the left on the axis of a convex lens L1 of focal then prove that cos
2
      cos 2
length 20 cm. A second convex lens L2 of focal
length 10 cm is placed coaxially to the right 20. A ray of light passes through a transparent
of the lens L1 at a distance of 5 cm from it. sphere of refractive index µ and radius R. If b
Find the position of the final image and its is the distance between the incident ray and a
magnification. Trace the path of the rays. parallel diameter of the sphere, show that the
angle of deviation δ is given by the expression
18. A convex lens A of focal length 20 cm and
 b  2 2
    
1/ 2 1/ 2
a concave lens B of focal length 5 cm are sin  2 
 R  b2  R2  b2  .
2 R 
kept along the same axis with a distance d

184 Ray Optics

 Previous Years’ Questions

Multiple Choice Questions (Single Choice m

1
Correct)
(A) x
1. If we need a magnification of 375 from a f 2f
compound microscope of tube length 150
mm and an objective of focal length 5 mm, m
the focal length of the eye-piece, should be 1
close to: (B) x
f 2f
[Mains, Jan 2020]
(A) 22 mm (B) 12 mm
m
(C) 2 mm (D) 33 mm 1
2. A thin lens made of glass (refractive index (C) x
f 2f
= 1.5) of focal length f = 16 cm is immersed
in a liquid of refractive index 1.42. If its focal
length in liquid is f1, then the ratio f1 / f is m
closest to the integer: 1
(D) x
[Mains, Jan 2020] f 2f
(A) 17 (B) 1
(C) 9 (D) 5 6. The aperture diameter of a telescope is 5 m.
3. The magnifying power of a telescope with The separation between the moon and the
tube length 60 cm is 5. What is the focal earth is 4 × 105 km. With light of wavelength
length of its eye piece? of 5500 Å, the minimum separation between
objects on the surface of moon, so that they
[Mains, Jan 2020]
are just resolved, is close to:
(A) 10 cm (B) 20 cm
[Mains, Jan 2020]
(C) 40 cm (D) 30 cm
(A) 600 m (B) 60 m
4. The critical angle of a medium for a specific
(C) 20 m (D) 200 m
wavelength, if the medium has a relative
4 7. A vessel of depth 2h is half filled with a liquid
permittivity 3 and relative permeability for
3 of refractive index 2 2 and the upper half
this wavelength, will be:
with another liquid of refractive index 2 .
[Mains, Jan 2020]
The liuqids are immiscible. The apparent
(A) 30° (B) 15°
depth of the inner surface of the bottom of
(C) 60° (D) 45° vessel will be:
5. An object is gradually moving away from [Mains, Jan 2020]
the focal point of a concave mirror along
h h
the axis of the mirror. The graphical (A) (B)
2 3 2
representation of the magnitude of linear
3 h
magnification (m) versus distance of the (C) h 2 (D)
4 2( 2 + 1)
object from the mirror (x) is correctly given
by: (Graphs are drawn schematically and 8. A point like object is placed at a distance of
are not to scale) 1 m in front of a convex lens of focal length
[Mains, Jan 2020] 0.5 m. A plane mirror is placed at a distance of

Ray Optics 185

 2 m behind the lens. The position and nature (A) 1.0 × 108 ms–1
of the final image formed by the system is: (B) 0.5 × 108 ms–1
[Mains, Sep 2020]
(C) 1.5 × 108 ms–1
(A) 1 m from the mirror, virtual
(B) 2.6 m from the mirror, virtual (D) 3.0 × 108 ms–1
(C) 1 m from the mirror, real
13. The power of a lens (biconvex) is 1.25 m–1 in
(D) 2.6 m from the mirror, real
a particular medium. Refractive index of the
9. A double convex lens has power P and same lens is 1.5 and radii of curvature are 20 cm
radii of curvature R for both the surfaces. The and 40 cm respectively. The refractive index
radius of curvature of a surface of a plano- of surrounding medium:
convex lens made of the same material with [Mains, Jul 2022]
power 1.5 P is:
9
[Mains, Sep 2020] (A) 1:0 (B) 
7
R 3R 3 4
(A) (B) (C)  (D) 
3 2 2 3
R
(C) (D) 2R 14. The focal length f is related to the radius of
2
curvature r of the spherical convex mirror
10. The refracting angle of a prism is A and
by:
refractive index of the material of the prism
[Mains, Feb 2021]
is cot. Then the angle of minimum deviation
will be: (A) f = r

[Mains, Jun 2022] 1

(B) f r
2
(A) 180 –2A (B) 90 – A
1
(C) 180 + 2A (D) 180 – 3A (C) F   r
2
11. The speed of light in media ‘A’ and ‘B’ are (D) f = −r
2.0 × 1010 cm/s and 1.5 × 1010 cm/s respectively.
A ray of light enters from the medium B to A 15. Given below are two statements: One is
at an incident angle ‘θ’. If the ray suffers total labeled as Assertion A and the other is
internal reflection, then: labeled as Reason R.
[Mains, Jun 2022] Assertion A: For a simple microscope, the
angular size of the object equals the angular
3 2
(A)   sin1   (B)   sin1   size of the image.
4 3 Reason R: Magnification is achieved as the
3 3 small object can be kept much closer to the
(C)   sin1   (D)   sin1  
4 4 eye than 25 cm and hence it subtends a large
angle.
12. As shown in the figure, after passing through
In the light of the above statements, choose
the medium 1. The speed of light v2 in medium
the most appropriate answer from the
2 will be:
options given below:
(Given c = 3 × 108 ms–1)
[Mains, Feb 2021]
[Mains, Jul 2022]
(A) Both A and R are true but R is NOT the
Air Medium 1 Medium 2
correct explanationof A
µr = 1 µr = 1
(B) Both A and R are true and R is the correct
r = 4 r = 9
explanation of A
(C) A is true but R is false
C v1 v2
(D) A is false but R is true

186 Ray Optics

16. The angle of deviation through a prism is deviation is also A, then in terms of refractive
minimum when: index value of A is:
[Mains, Jul 2021]
 
 (A) sin1   (B) 2 cos1  
2 2
  1 
(C) (D) cos1  
sin1  
 2  2

(A) Incident ray and emergent ray are
19. A ray of light entering from air into a denser
symmetric to the prism.
4
(B) The refracted ray inside the prism medium of refractive index , as shown in
3
becomes parallel to its base.
figure. The light ray suffers total internal
(C) Angle of incidence is equal to that of the
reflection at the adjacent surface as shown.
angle of emergence.
The maximum, value of angle θ should be
(D) When angle of emergence is doubled the
equal to:
angle of incidence.
Choose the correct answer from the options
given below: 
[Mains, Mar 2021]
(A) Only statement (D) is true
'
(B) Statements (A), (B) and (C) are true
(C) Statements (B) and (C) are true 4
= –
(D) Only statements (A) and (B) are true 3
''
17. Three rays of light, namely red (R), green (G)
and blue (B) are incident on the face PQ of a
right angled prism PQR as shown in the figure. [Mains, Jul 2021]

P 5 7
(A) sin−1 (B) sin−1
3 3

B 7 5
(C) sin−1 (D) sin−1
4 4
G
20. An object is placed beyond the centre of
R curvature C of the given concave mirror. If
the distance of the object is d1 from C and
R the distance of the image formed is d2 from
Q
C, the radius of curvature of this mirror is:
The refractive indices of the material of the [Mains, Aug 2021]
prism for red, green and blue wavelength are 2d1d2 2d1d2
(A) (B)
1.27, 1.42 and 1.49 respectively. The colour d1 − d2 d1 + d2
of the ray(s) emerging out of the face PR is:
d1d2 d1d2
[Mains, Mar 2021] (C) (D)
A) Blue (B) Green d1 − d2 d1 + d2
(C) Red (D) Blue and green 21. Two plane mirrors M1 and M2 are at right
18. A prism of refractive index µ and angle of angle to each other shown. A point source
prism A is placed in the postion of minimum ‘P’ is placed at ‘a’ and ‘2a’ meter away
angle of deviation. If minimum angle of from M1 and M2 respectively. The shortest

Ray Optics 187

 distance between the images thus formed is: shown in the figure. A ray of light is incident
(Take = 5 = 2.3 ) from medium n1 to n2 at an angle θ, where
sin θ is slightly larger than 1/n1. Take refractive
a P
index of air as 1. Which of the following
statement(s) is(are) correct?

air
M1 2a
n2

M2 n1 

[Mains, Aug 2021]

[Advanced, Oct 2021]
(A) 2 10 a (B) 2.3a (A) The light ray enters air if n2 = n1.
(C) 3a (D) 4.6a (B) The light ray is finally reflected back
22. A glass tumbler having inner depth of 17.5 cm into the medium of refractive index n1 if
is kept on a table. A student starts pouring n2 < n1.
water (   4 / 3 ) into it while looking at the (C) The light ray is finally reflected back
surface of water from the above. When he into the medium of refractive index n1 if
feels that the tumbler is half filled, he n2 > n1.
stops pouring water. Up to what height, the (D) The light ray is reflected back into the
tumbler is actuallly filled? medium of refractive index n1 if n2 = 1.
[Mains, Sep 2021] 25. For a prism of prism angle θ = 60°, the
(A) 11.7 cm (B) 7.5 cm refractive indices of the left half and the right
(C) 8.75 cm (D) 10 cm half are, respectively, n1 and n2 (n2 ≥ n1) as
shown in the figure. The angle of incidence
23. An extended object is placed at point O, 10 cm
i is chosen such that the incident light rays
in front of a convex lens L1 and a concave lens
will have minimum deviation if n1 = n2 = n =
L2 is placed 10 cm behind it, as shown in the
1.5. For the case of unequal refractive indices,
figure. The radii of curvature of all the curved
n1 = n and n2 = n + ∆n (where ∆n << n), the
surfaces in both the lenses are 20 cm. The
angle of emergence e = i + ∆e. Which of the
refractive index of both the lenses is 1.5. The
following statement(s) is (are) correct?
total magnification of this lens system is:
L1 L2


i
O

10 cm 10 cm
n1 n2
[Advanced, Oct 2021]
(A) 0.4 (B) 0.8 [Advanced, Oct 2021]
(C) 1.3 (D) 1.6 (A) The value of ∆e (in radians) is greater
than that of ∆n.
(B) ∆e is proportional to ∆n.
Multiple Choice Questions (One or More
(C) ∆e lies between 2.0 and 3.0 milliradians,
Choices Correct)
if ∆n = 2.8 × 10−3.
24. A wide slab consisting of two media of (D) ∆e lies between 1.0 and 1.6 milliradians,
refractive indices n1 and n2 is placed in air as if ∆n = 2.8 × 10−3.

188 Ray Optics

Subjective Questions 32. An object ‘O’ is placed at a distance of 100 cm
in front of a concave mirror of radius of
26. A point object in air is i front of the curved
curvature 200 cm as shown in the figure. The
surface of a plano-convex lens. The radius of
object starts moving towards the mirror at a
curvature of the curved surface is 30 cm, and
speed 2 cm/s. The position of the image from
the refractive index of the lens material is 1.5,
the mirror after 10 s will be at ________ cm.
then the focal length of the lens (in cm) is
________.
[Mains, Jan 2020]
27. A light ray enters a solid glass sphere of
refractive index   3 at an angle of O
incidence 60°. The ray is both reflected and
refracted at the farther surface of the sphere.
The angle (in degrees) between the reflected [Mains, Jul 2022]
and refracted rays at this surface is ________.
33. A deviation of 2° is produced in the yellow
[Mains, Sep 2020]
ray when prism of crown and flint glass are
28. A compound microscope consists of an achromatically combined. Taking dispersive
objective lens of focal length 1 cm and an eye powers of crown and flint glass as 0.02 and
piece of focal length 5 cm with a separation 0.03 respectively and refractive index for
of 10 cm. The distance between an object yellow light for these glasses are 1.5 and 1.6
and the objective lens, at which the strain on respectively. The refracting angles for crown
n glass prism will be ________ °(in degree).
the eye is minimum is cm . The value of n
40 (Round off to the nearest integer)
is ________.
[Mains, Mar 2021]
[Mains, Sep 2020]
34. The image of an object placed in air formed
29. A prism of angle A = 1° has a refractive index
by convex refracting surface is at a distance
µ = 1.5. A good estimate for the minimum
of 10 m behind the surface. The image is real
angle of deviation (in degrees) is close to
2nd
N/10. Value of N is ________ and is at of the distance of the object
3
[Mains, Sep 2020] from the surface. The wavelength of light
30. A parallel beam of light is allowed to fall on 2
inside the surface is times the wavelength
a transparent spherical globe of diameter 30 3
cm and refractive index 1.5. The distance from in air. The radius of the curved surface is
the centre of the globe at which the beam of x
m . The value of ‘x’ is ________
light can converge is ________ mm. 13
[Mains, Mar 2021]
[Mains, Jun 2022]
35. An object viewed from a near point distance
31. A convex lens of focal length 20 cm is placed of 25 cm, using a microscopic lens with
in front of a convex mirror with principle axis magnification '6', gives an unresolved image. A
coinciding each other. The distance between resolved image is observed at infinite distance
the lens and mirror is 10 cm. A point object with a total magnification double the earlier
is placed on principal axis at a distance of using an eyepiece along with the given lens
60 cm from the convex lens. The image and a tube of length 0.6 m, if the focal length
formed by combination coincides the object of the eyepiece is equal to ________.
itself. The focal length of the convex mirror is [Mains, Jul 2021]
________ cm.
36. A ray of light passing through a prism
[Mains, Jul 2022] (  3 ) suffers minimum deviation. It is

Ray Optics 189

 found that the angle of incidence is double 37. An object is placed at a distance of 12 cm
the angle of refraction within the prism. from a convex lens. A convex mirror of focal
Then, the angle of prism is ________ (in length 15 cm is palced on other side of lens at
degrees). 8 cm as shown in the figure. Image of object
[Mains, Jul 2021] coincides with the object.

Image Object ×
Image in the absence of mirror

12 cm 8 cm

When the convex mirror is removed, a real schematically in the figure. Looking at the
and inverted image is formed at a positon. wire from this corner, another student sees
The distance of the image from the object two images of the wire, located symmetrically
will be ________ cm. on each side of the line of sight as shown.
[Mains, Aug 2021] The separation (in cm) between these images
is ________.

Integer Type Questions

38. A large square container with thin transparent Line of sight
vertical walls and filled with water (refractive
index) is kept on a horizontal table. A student 12 cm
holds a thin straight wire vertically inside the
water 12 cm from one of its corners, as shown [Advanced, Sep 2020]

190 Ray Optics

 Answer Key (Practice Exercise)

Multiple Choice Questions (Single Choice Correct)

Level I
1. (C) 2. (B) 3. (B) 4. (C) 5. (B) 6. (C) 7. (A) 8. (B) 9. (C) 10. (B)
11. (A) 12. (A) 13. (B) 14. (B) 15. (A) 16. (A) 17. (A) 18. (B) 19. (D) 20. (B)
21. (C) 22. (C) 23. (D) 24. (D) 25. (B) 26. (D) 27. (A) 28. (D) 29. (A) 30. (B)
31. (A) 32. (C) 33. (A) 34. (D) 35. (B) 36. (A) 37. (B) 38. (D) 39. (D) 40. (A)
41. (C) 42. (D) 43. (D) 44. (D) 45. (C)

Level II
46. (D) 47. (C) 48. (B) 49. (B) 50. (B) 51. (D) 52. (C) 53. (B) 54. (D) 55. (D)
56. (B) 57. (B) 58. (D) 59. (D) 60. (C) 61. (B) 62. (A) 63. (B) 64. (A) 65. (C)
66. (A) 67. (C) 68. (C) 69. (D) 70. (D) 71. (D) 72. (C) 73. (C) 74. (B) 75. (A)
76. (D) 77. (D) 78. (C)

Multiple Choice Question (One or More Choice Correct)

79. (B) and (C)

Subjective Type Questions

1. 55 cm 2. 1.25

R R
3. a. 2   1   b. 4. i = 58.3°
 1  2 µ

 1  h
5. r  sin1  . 6. r
 2.6  2  1

7. 3.75 m/s 8. Angle of emergence = 0

10. 18 cm and 6 cm

11. 1.6 12. 0.003 dioptre

4
13. m= 14. 40 cm to right of second lens
3

15. –50.5 cm (virtual) 16. 24 cm left of the silvered surface

17. 1.33 18. 15 cm

 b  2 2
    
1/ 2 1/ 2
19. sin   R  b2  R2  b2 20. *
2  R2  

Ray Optics 191

 Answer Key (Previous Years’ Questions)

Multiple Choice Questions (Single Choice Correct)

1. (A) 2. (c) 3. (a) 4. (a) 5. (d) 6. (b) 7. (c) 8. (a, b both are correct)

9. (a) 10. (a) 11. (d) 12. (a) 13. (b) 14. (c) 15. (b) 16. (b) 17. (c) 18. (b)

19. (b) 20. (a) 21. (d) 22. (d) 23. (b)

Multiple Choice Questions (One or More Choices Correct)

24. (B), (C), (D) 25. (B), (C)

Subjective Questions

26. 60 cm 27. 90 28. 50 29. 5 30. 225 31. 10 32. 400 33. 12 34. 30

35. 25 36. 60 37. 50

Integer Type Questions

38. (Official answer: bonus marks to all) Probable Answer: 3 or 4 or 5 (for 3)

192 Ray Optics

 Solutions (Practice Exercise)

Multiple Choice Questions (Single Choice u = 10 cms

Correct) So, we have virtual object 10 cm inside the
mirror.
Level I
  1. (C) 8. (B)
Refer to the textbook. i  r (Always)
So, ∠r is also doubled.
  2. (B)
Angle of incidence = Angle of reflection 9. (C)
So, i  90  30  60 The initial angle of incidence must be less
r  60 than 45°, so that the angle doubled is less
than 90°.
  3. (B) 10. (B)
 Both the object and images are moving 1 1 1
toward each other at 0.5 m/s. So, the relative  
v u f
speed of images with respect to the object
= 1.0 m/s. 1 1 1
 
v 8 7
  4. (C)
Refer to the text 1 1 1 8  7 15
   
  5. (B) v 7 8 56 56

1 1 1 56
v =
  15
v u f
v 56
1 1 1 m   0.47
  u 15  8
v 30 15
11. (A)
1 1
= Object is at infinity, image will be at focus.
v 10
12. (A)
v = 10 cm
v 5 1
So, the image is virtual and diminished. m  
u 10 2
10 1
m =
= u = –6 cm
30 3
6
  6. (C) v   3 cm
2
1 1 1
  1 1 1
v u f  
1 1 1 v u f
 
8 6 f 1 1 1
 
3  4 1 3 6 f

24 f
1 1
24 
f   3.43 cm 2 f
7
  7. (A) ∴ f = –2
1 1 1 13. (B)
 
v u f 1 1 1
 
v u f
1 1 1
 
10 u 5 1 1 1
 
v 7 28
1 1 1 1
   cm
u 5 10 10 1 1 1 1 4 5
   
v 28 7 28 28

Ray Optics 193

 28 1
v
= = 5.6 cm sinc 
5 
x y

v 5.6 I
m    20. (B)
u 7 0
Refer to the textbook
I = 0.8 m
21. (C)
14. (B)
c
Definition of refractive index 
v
15. (A)
c
c 2=
v
v2 2
2   22. (C)
1
c 1
v1
  
Total dispersion       '  '
V R v R 
16. (A)   
Definition of Snell’s law  A1  V  R   1  2 
 1 
17. (A)
= A1 (µ – 1) (ω1 – ω2)
1 sin i  2 sin r
23. (D)

sin r  1 sin i Total deviation     '
2
 A1    1 1  2 
2   1
  
A    1  1  1 
i
 2 
1

2 24. (D)
r
Refer to the text

So, sinr < sini. 25. (B)

18. (B) Basic derivation of prism
1 sin i  2 sin r 26. (D)
Basic condition for minimum deviation
1
sin r  sin i 27. (A)
2
Basic condition for minimum deviation
 1  2 28. (D)
Velocity = f * λ changes in a medium.
i 1
29. (A)
  i1  i2  A
r
2 At minimum deviation i1 = i2 = i

So, sini > sinr.  m  2i  A  2  45  60  30

19. (D) 30. (B)
 x sin 90   y  sin c 5  2
     1 A    1  A  . 6  4
x  3  3
sinc 
y

194 Ray Optics

31. (A)   1.5, R1  R, R2  R
A
1  1 1
A X  ( 1.5  1)   
90° f R R
i r 90°
f=R
37. (B)
C B Factual
For retracing the path, at point X, light must
38. (D)
fall normally on surface AB.
So, by geometry ∠r = A By lens maker’s formula
1 2

sin i
   sin i   sin A      1   [When surrounding medium is
sin A f R
air and lens is equiconvex]
i = 90° (Grazing incidence)
1 2 1 2
So,   b  1   and   r  1  
32. (C) fb R
  f R
r
 A  m
sin  
fr

 b
 1

6
2  sin 45
    fb  r
 1 5
A sin 30
sin
2 39. (D)
1
  2  2 As n2 > n1 > 1, a concave lens will behave
2 as a diverging lens when the refractive index
= 1.414 of lens is greater than refractive index of
surrounding.
33. (A)
A = r1 + r2 = 2r = i [  r1 = r2 for a symmetric ray] 40. (A)
1
 m  2i  A  i  A  P  P1  P2
f
 A  m  Focal length (and power) is –ve for concave
sin  
 2  sin A A lens.
  2    2 cos
A A 2 1 1
sin sin P    4  2.5  1.5 D
2 2 0.25 0.4
A 1 41. (C)
 cos   A  90
2 2 Part lens also acts as a full lens as far as the
image is concerned. As half the original light
34. (D) is cut off, intensity reduces.
R 42. (D)
f = is independent of µ.
2
Factual
35. (B)
43. (D)
1 1 1
  Factual
v u f
44. (D)
1 1 1
  45. (C)
v 40 15
v = 24 cm
n
36. (A)
O

 he effective focal length of the combination

T
can be given as
1  1 1 
 (   1)    1 2 1
f  R1 R2     ( Both are converging optical
feff fl fm
element)
Ray Optics 195
 1    1
51. (D)
1 2
 and  . I4
fl R fm R

1 2    1 2 C
  
feff R R
I3 I1

1 2 O

feff R A B
I2
So, the combination is behaving as a concave
R
mirror of radius of curvature , the object
µ I5
should be placed at the centre of curvature
so that the image coincides with the object.
I1 is formed due to BC.
Level II I2 is formed by I1 due to AB.
46. (D) I3 is formed by I2 due to AC.
D = 180 –2I = 30 I4 is formed by I3 at BC, and so on.
150 So, infinite images are possible.
I  75
2
52. (C)
I = R = 75° Refer to the text
47. (C)
53. (B)
360 360
n 1 1 We have
 40
4
sinc = 1
48. (B) 3
Laws of reflection are valid. 4
And  g sin i  sin c
49. (B) 3
1
Hence,  g 
sin i
60°
150° 54. (D)
Apply Snell’s law at each interface.

Fig. 1.81 1 sin 1  2 sin2  3 sin3  4 sin4

From the above Fig. 1.81, i  150  90  60  1  4
 r  i  60 Hence,  1  4
50. (B) 55. (D)
O c
200° We know  
v
As v g < v w
M1
O h h
Time difference  
M2 I1 I2 v g vw

h h h
The two images formed are I1 (due to mirror  
c /  g c / w

c g

  w 
OM1) and I2 (due to mirror OM2).
So, two images are formed. h3 4 h
    =
c 2 3 6c

196 Ray Optics

56. (B) 60. (C)
Applying Snell’s law, For the condition of dispersion without
1 sin i  2 sin r ' ... (i) deviation
A1   1  1  A2  2  1
from the given condition r + r’ = 90
Or r’ = 90 - r    1
Or sin r ' = cos r ... (ii) A2   1  A1
 2  1 
From (i) and (ii), we get
 1.54  1 
1 sin i  2 cos r ... (iii) A2     4
 1.72  1 
By law of reflection, i = r
A2  3
  1 sin i  2 cos i
2 61. (B)
tan i 
1
At the time of total internal reflection,
 There will be no deviation occurring on
sinc  1
2 interfaces 2 and 3, as there is no change in
the medium. However, interface 4 is the same
c  sin1  cot i 
as it would have been if P was alone.
57. (B) 62. (A)
R
 
f = is independent of µ.
rarer 2

denser 63. (B)
64. (A)
rarer 
A convex lens behaves as a diverging lens
when the refractive index of the material of
  the lens is less than the refractive index of
denser
the surrounding medium.
When the ray passes into the rarer medium,
the deviation is      , which can have a 65. (C)
 Since, there is separation only between the
maximum value of  c.
2 upper and lower halves of the lens and not
When the total internal reflection occurs, the amongst the subparts of the upper and lower
deviation is     2 . So, the minimum value halves of the lens, the number of images will
of θ is c for getting δ maximum. be two only.
Hence,     2c. 66. (A)
58. (D)
67. (C)
Refer to the text
1 1
59. (D) 
f R
y
B A
R
–

r
90

R–y

C
r2
Total internal reflection occurs if ic = θ R
= = 15 cm
2y
1
sin  sin ic 
 R
f   30 cm
1.5 9 8 1
Relative     sin 
4 8 9
68. (C)
3
The closest correct graph is option (C).

Ray Optics 197

 69. (D) 73. (C)
1 Magnifying power
    1 k
f Angle subtended by image from telescope at eye
=
1 3  2 Angle subtended by objject on eye
   1 k, f 
f 2  k Since, magnifying power is 20.
∴ Angle subtended by image on eye by tele-
1 3 3  8 scope increases 20 times.
    1  k , f1 
f1  2 4  k Size of the object depends upon angle sub-
f1 > f and f2 becomes negative. tended on the eye.
Thus, object appears 20 times taller.
70. (D)
74. (B)
f
 δ=i+e–A
VIBGYOR
  A = 74°
c Now, for maximum µ
A–C<C
Hence, the spectrum of visible light whose
A
frequency is less than that of green light will C>
come out of the air medium. 2
71. (D)  A
sin C > sin  
1.22 x 2
  
D 25 cm  A
µ < cosec  
1.22  500nm  25 cm 2
x 
0.25  2 cm
1

 30 m sin 37
72. (C) 5

3
µ < 1.66
∴ µmax = 1.6
75. (A)
For light to just come out, r2 must be less 15 cm
than θc.
 1
c  sin1  

But r1  r2  A
r1  A  r2

 1 f1 = 25 cm f2 = 20 cm
r2  sin1  

Image by diverging lens is formed at 25 cm,
 1
c  sin  
1 to the left of the lens. This image serves as
 an object for the converging lens.
sin ∴ u = – 40 cm; f = +20 cm.
But 
sin r1 1 1 1
 
v u f
  sin1   sin r1 
1 1 1 1 1 1 1
    1 
     
   sin    sin  A  sin1     
1 v 40 20 v 20 40 40
       
   v = +40 cm

198 Ray Optics

 ∴ Image is real and at a distance of 40 cm 1  1 1
from the convergent lens. Now,
fa
  
g  1   
 R1 R2 
a

76. (D)
60 cm 1  1 1
20 cm And
fw
  g  1   
 R1 R2 
w

Dividing equation (i) by equation (ii), we get

40 cm
10 cm  a g  1 
fw
Hence, option ‘D’ is correct.  
fa  w  g  1 
77. (D)
n1 sinθ1 = n2 sin θ2. Again
g 3/2 9
1 × 0.64 = 1.31 sin θ2. g 
a
 
0.64
w
a
w 4/3 8
sin2 
1.31

fw

 3 / 2  1   1 / 2  4
tan2 
d fa 9 / 8  1  1 / 8
x
fw  fa  4  20  4  80 cm  0.8 m
d
x
tan2 1 1
P= = = 1.25 dipotre
w
fw 0.8
l
Number of collisions   57000
x 3. a. Let the focal length of lens be fl and that
78. (C) of mirror be fm.
Wavelength used in modern optical fibre
communication is close to 1500 nm.
O

Multiple Choice Question (One or More

Choice Correct) As the lens-mirror combination is converg-
ing in nature
79. (B) and (C)
Factual 1 2 1
So, f  f  f
Subjective Type Questions eq l m

Since, refraction from lens is taken twice

1.
1 1 1
    1   
fl R 
R
 fl  and fm  
1
20 cm 5 cm R fl R
So, feq  2  2   1
For the lens,  
1 1 1
  b. Curved surface is silvered
v u f
Substituting
u = –20, f = +15, v = 5 + R O

We get, R = 55 cm
2. Let fa and fw be the focal lengths of the lens in
R R
air and water, respectively. Then fl  and fm 
1 2
1 1
Pa  or  5 
fa fa 1 2 1 2    1 2 2
    
feq fl fm R R R
=fa 0=
.2 m 20 cm

Ray Optics 199

 R h
Or feq  r
2 2  1
4. Given, 90 – r + 90 – r’ = 90
7.  or the fish the apparent distance of the
F
r’ = 90 – r fly from the wall of the aquarium is µx.
Or r’ = 90 – i (i = r) (x is the actual distance)
By Snell’s law, d
1  sin i    sin r ' Then apparent velocity will be
dt
  x .
V = µV
sin i   sin(90  i ) (apparent) fly fly
 ow, the fish observes the velocity of the fly
N
  tan i to be 8 m/s.
⇒ Apparent relative velocity will be = 8 m/s
i  tan1 (  )

Vfish  Vapp  fly
= 8 m/s
i  tan1 ( 1.62)
3  Vfly  8
i  58.3
5. As we know 3
Vfly  5   3.75 m/s
 sin i  c 4
air sin iair   glass sin rglass 8. A

30°
air 30° E
sin rglass   sin iair 60°
 glass D
 ... (i)
Again B C

 glass sin iglass  water sin iwater 1  sin 60  3  sin r

 ... (ii)
From (i) and (ii), we get 1
sin =
sin 30 = 1.3 sin r 2
1 1 r = 30°
sin r   ∆ADE
2  1.3 2.6
30  (90  30)    180
 1 
r  sin  1
   90
 2.6 
Hence, angle of emergence is zero.
6. r = h tan c 9.
X 2
sin c
r=h
cos c F
A B

r
Y
h The line joining x and y intersects the optic
C C axis AB at the optical centre O. If we draw
a line xZ || AB and then join z to y, then the
intersection point of zy and AB will be the
Fish point F. Point F is focus of the lens.
h sin c Draw a line Y ' BY ⊥ AB, such that Y ' B = YB.

1  sin c 2 Now, join yʻto x and produce it to intersect
AB at O. The point O is the pole of the mirror.
1 The line joining X to Y intersects AB at the
h
 point C, the centre curvature of the mirror.
Draw a line XZ || BA and then join z to y. The
1
1 line zy intersects AB at point F, the focus of
2 the mirror.

200 Ray Optics

10. 24 – X Substituting u = –0.3 M, f = 0.2 M, we get
v = 0.6 m
From similar triangles,
S1 S2
d 0 .3  0 .6
  d  0.003 M
y 0.0005 0.3
24 cm Y'

Suppose, the images of both the sources S1

x
and S2 are formed at y. z
For S1, u = –x, v = y, f = 9 A B
1 1 1 1 1 1 0 F c
      ... (i)
v u f y x 9
For S2, u = –(24 – x), u = –y, f = 9 Y
1 1 1 1 1 1 13.
      ...(ii)
v u f y 24  x 9 A
B
From (i) and (ii), eliminating y we get P
1 1 1 1 From the principle of reversibility of light, the
  
9 x 24  x 9 rays can be taken in opposite directions. For
the refraction on the curved surface,
 x 2  24 x  108  0
2  1 2   1
 x  18 cm, 6 cm  
v u R
11. For the first case,
Substituting 1  1, 2  1.5, u  , R  R,
 3   3 4  We get
   1    
1
 2  2    2 3  v = 3R
15  R   R   ⇒ AC = 3R
  
 

 Therefore, BC = 3R – R = 2R
 R  20 cm. BC 2R 4
Now, BP    R
 1.5 3
For the second case,
 3  3  4
  1       m 
1 3
 2  2 2 
25  R   R   14. We know,

   
   1 1 1 d
  
8 F f1 f2 f1f2
  1.6
5 1 1 60 cm
12.   
I1 20 cm 20 cm (20 cm)2
Or F  20 cm
O It is a concave lens. It will be kept at a
distance
60   20 
I2 D  60 cm
0.3 M v
20
So, an equivalent concave lens will be placed
Two images I1 and I2 are formed at a distance at a distance of ∈0 cm to the right of the sec-
v from the lens. ond lens.
1 1 1 ∴ It is (0 20)  40 cm to the right of the
Now,   . second lens.
v u f

Ray Optics 201

 15. 
Two images are formed, one by the upper fB = 200 cm
half and the other by the lower half. 1 12 2 1
  
For the upper half, the position of image is feq 800 200 10
obtained by using the lens formula.
1 = 1.0 3 = 1.2 feq = 8 cm
To apply mirror formula
 0 = 1.5 1 1 1
 
O u v f
u = 30 cm u = –12 cm, v = ?, f = –8 cm
 3  1 0   1  3  0 1 1 1
    
v1 u R1 R2 u v f
1 1 1
u  30 cm, R1  24 cm, R2  24 cm  
v 8 12
1.2  1  1 v = –24 cm
  2  1.5   1.0  1.2
v 1  30  24   –ve sign indicates that the image is formed
to the left side of the silvered surface.
Or v 1   L1 L2
17.
Similarly, for the lower half
v2 = 66.2 cm
Q1 O1 O1 Q

30 cm
F1 5 cm F2
I O 0
Here, for first lens,
4 = 1.6
3 = 1.2 u1  10 cm; F1  20 cm
1 1 1
 4  2 0   2  4  0 = =
   v 1 u1 F1
v2 u R1 R2
1 1 1
u = –30 cm, R1 = +24 cm, R2 = –24 cm   
v 1 20 10
∴ v2 = –50.5 cm (Virtual)
 v 1  20 cm
16. Since the lenses are thin, we can solve by
finding out the equivalent focal length of the i.e., the image is virtual, and hence, lies on
combination and then applying the mirror the same side of the object. This will behave
formula, as the combination will behave as as an object for the second lens.
a mirror. For second lens,
Equivalent focal length of combination 1 1 1
 
1 2 2 1 v 2 u2 F2
  
feq fA fB fm Here, u2    20  5  , F2  10 cm
Where, fA, fB, and fm are the focal lengths of 1 1 1 50 2
lens A, B, and mirror M, respectively.    v2   16 cm
v 2 2.5 10 3 3
1  1 1 
   A  1    2
fA 
 1R R 2  i.e., the final image is at a distance of 16 cm
on the right of the second lens. 3
800
fA = cm The magnification of the image is given by,
6
v v 20 50 4
1  1 1  m 1. 2  .   1.33
  B  1    u1 u2 10 3  2.5 3
fB 
 R2 R3 

202 Ray Optics

18. Deviation α = 2(i – r)

ir
I 2
   
 i  90   
 2 
d 5 cm
20 cm    
 sin i  cos    ... (ii)
Fig. 1.82  2 
As the incident beam is parallel, in absence From Snell’s law,
of a concave lens, it will form an image at a  sin r  sin i
distance v from it, such that
    
1 1 1   cos  cos 
  , i.e., v  20 cm   F  2

v  20  2 
Now, since d is the distance between convex     
and concave lens, the distance of image I Or cos     cos
 2  2
from concave lens B will be = (20 – d).
Since, the image I will act as an object for 20. i
concave lens which forms its image at ∞, so r
b r i
1 1 1 i R
 
  20  d  5

20 – d = 5
Or d = 15 cm   2(i  r )
This all is shown in Fig. 1.82 above.
19. 
 sin  sin i cos r  cos i sin r ... (i)
B C
2
r
r
 b b2
Now, sin i   cos i  1
O R R2

sin i b b2
From ∆OBC, And sin r    cos r  1
 R  2 R2
2r + θ = 180
Substituting these values in equation (i), we
1 1 1
  get
  20  d  5
 b  2 2
  
 
1/ 2 1/ 2
sin  2 
 R  b2  R2  b2 
θ 2 R 
sinr = cos  ... (i)
2

Ray Optics 203