NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 9
Chapter Name : Ray Optics And Optical Instruments

Q9.1 A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of
curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a
sharp image? Describe the nature and size of the image. If the candle is moved closer to the
mirror, how would the screen have to be moved?

Answer. Size of the candle h = 2.5 cm
Image size = h'
Object distance, u = -27 cm
Radius of curvature of the concave mirror, R = -36 cm
Focal length of the concave mirror, f = = −18cm
R

2

Image distance = v
The image distance can be obtained using the mirror formula:
1 1 1
+ =
u v f

1 1 1
= −
v f u

1 1 −3+2 1
= − = = −
−18 −27 54 54

v = -54 cm
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp
image. The magni cation of the image is given as:
′
h v
m = = −
h u
v
′
∴ h = − × h
u

−54
= −( ) × 2.5 = −5cm
−27

The height Of the candle's image is 5 cm. The negative sign indicates that the image is inverted
and virtual. If the candle is moved closer to the mirror, then the screen will have to be moved away
from the mirror in order to obtain the image.

Page : 344 , Block Name : Exercise

Q9.2 A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the
location of the image and the magni cation. Describe what happens as the needle is moved
farther from the mirror.

Answer. Height of the needle, h1 = 4.5 cm
Object distance, u = -12 cm
Focal length of the convex mirror, f = 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
1 1 1
+ =
u v f

1 1 1
= −
v f u

1 1 4+5 9
= + = =
15 12 60 60

60
∴ v = = 6.7cm
9

Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the
other side of the mirror.
The image size is given by the magni cation formula:
h2 v
m = = −
h1 u
v
∴ h2 = − × h1
u
−6.7
= × 4.5 = +2.5cm
−12

Hence, magni cation of the image, m =
h2 2.5
= = 0.56
h1 4.5

The height Of the image is 2.5 cm. The positive sign indicates that the image is erect , virtual, and
diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and
the size of the image will reduce gradually.

Page : 344 , Block Name : Exercise

Q9.3 A tank is lled with water to a height of 12.5 cm. The apparent depth of a needle lying at the
bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of
water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what
distance would the microscope have to be moved to focus on the needle again?

Answer. Actual depth of the needle in water, h1 = 12.5 cm
Apparent depth Of the needle in water, h2 = 9.4 cm
Refractive index of water = μ
The value of v can be obtained as follows:
h1
μ =
h2

12.5
= ≈ 1.33
9.4

Hence, the refractive index of water is about 1.33.
Water is replaced by a liquid of refractive index, μ = 1.63 ′

The actual depth of the needle remains the same, but it’s apparent depth changes.
Let y be the new apparent depth of the needle. Hence, we can write the relation:

h1
′
μ =
y

h1
∴ y =
′
μ

12.5
= = 7.67cm
1.63

Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2.
Therefore, to focus the needle again, the microscope should be moved up.
E-Distance by which the microscope should be moved up = 9.4 — 7.67 = 1.73 cm

Page : 344 , Block Name : Exercise

Q9.4 Figures (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air
and water-air interface, respectively. Predict the angle of refraction in glass when the angle of
incidence in water is 45° with the normal to a water-glass interface [(c)].

Answer. As per the given gure, for the glass — air interface:
Angle of incidence, i = 60°
Angle of refraction, r — 35°
The relative refractive index of glass with respect to air is given by Snell's law as:
sin i
a
μg =
sin r
∘
sin 60 0.8660
= ∘
= = 1.51 … (1)
sin 35 0.5736

As per the given gure, for the air — water interface:
Angle of incidence, i = 60°
Angle of refraction, r = 47°
The relative refractive index Of water with respect to air is given by Snell's law as:
sin i
a
μw =
sin r
sin 60 0.8660
= = = 1.184 … (2)
sin 47 0.7314

Using (1) and (2), the relative refractive index of glass with respect to water can be

a
μ
w B
μg =
a
μw

1.51
= = 1.275
1.184

The following gure shows the situation involving the glass — water interface.

Angle of incidence, i = 45°
Angle of refraction = r
From Snell's law, r can be calculated as:
sin i m
= μe
sin r
∘
sin 45
= 1.275
sin r

√2
sin r = = 0.5546
1.275

−1 ∘
∴ r = sin (0.5546) = 38.68

Hence, the angle of refraction at the water glass interface is 38.68°.

Page : 344 , Block Name : Exercise

Q9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is
the area of the surface of water through which light from the bulb can emerge out? Refractive
index of water is 1.33. (Consider the bulb to be a point source.)

Answer. Actual depth Of the bulb in water, d1 = 80 cm = 0.8 m
Refractive index of water, μ = 1.33
The given situation is shown in the following gure:

i = Angle of incidence
r = Angle of refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle of

radius
sin r
μ =
sin i
∘
sin 90
1.33 =
sin i

−1 1 ∘
∴ i = sin ( ) = 48.75
1.33

Using the given gure, we have the relation:
OC R
tan i = =
OB d1

∘
QR = tan 48.75 × 0.8 = 0.91m

Area of the surface of water =- nR2 = n (0.91)2 = 2.61 m 2

Hence, the area of the surface of water through which the light from the bulb can
emerge is approximately 2.61 m . 2

Page : 344 , Block Name : Exercise

Q9.6 A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a
face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive
index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed
in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of
light.

Answer. Angle of minimum deviation, δ = 40 m
∘

Angle of the prism, A = 60°
Refractive index of water, μ = 1.33
Refractive index of the material of the prism = μ ′

The angle Of deviation is related to refractive index (μ ) as : ′

(A+δm )
sin
′ 2
μ =
A
sin
2
∘ ∘
(60 +40 )
sin ∘
2 sin 50
= ∘
= ∘
= 1.532
60 sin 30
sin
2

Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let δ be the new angle of minimum deviation for ′
m

the same prism.
The refractive index of glass with respect to water is given by the relation:
′ ′
μ (A+δm )
w
μg = =
μ A
sin
2
′ ′
(A+δm ) μ A
sin = sin
2 μ 2
′ ∘
(A+δm )
1.532 60
sin = × sin = 0.5759
2 1.33 2

′
(A+δm )
−1 ∘
= sin 0.5759 = 35.16
2

∘ ′ ∘
60 + δm = 70.32

′ ∘ ∘ ∘
∴ δm = 70.32 − 60 = 10.32

Hence, the new minimum angle Of deviation is 10.32°.

Page : 344 , Block Name : Exercise

Q9.7 Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both
faces of the same radius of curvature. What is the radius of curvature required if the focal length is
to be 20cm?

Answer. Refractive index Of glass, μ= 1.55
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens = RI
Radius of curvature of the other face of the lens = R2
Radius of curvature of the double-convex lens = R
∴ R1 = R and R2 = −R

The value of R can be calculated as:
1 1 1
= (μ − 1) [ − ]
f R1 R2

1 1 1
= (1.55 − 1) [ + ]
20 R R

1 2
= 0.55 ×
20 R

∴ R = 0.55 × 2 × 20 = 22cm

Hence, the radius of curvature of the double-convex lens is 22 cm.

Page : 344 , Block Name : Exercise

Q9.8 A beam of light converges at a point P. Now a lens is placed in the path of the convergent
beam 12cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal
length 20cm, and (b) a concave lens of focal length 16cm?

Answer. In the given situation, the object is virtual and the image formed is real.
Object distance, u = +12 cm
(a) Focal length of the convex lens, f 20 cm
Image distance = v
According to the lens formula, we have the relation:
1 1 1
− =
v u f

1 1 1
− =
v 12 20

1 1 1 3+5 8
= + = =
v 20 12 60 60

60
∴ v = = 7.5cm
8

Hence, the image is formed 7.5 cm away from the lens, toward its right.

(b) Focal length of the concave lens, f = -16 cm
Image distance = v
According to the lens formula, we have the relation:

1 1 1
− =
v u f

1 1 1 −3+4 1
= − + = =
v 16 12 48 48

∴ v = 48cm

Hence, the image is formed 48 cm away from the lens, toward its right.

Page : 344 , Block Name : Exercise

Q9.9 An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm.
Describe the image produced by the lens. What happens if the object is moved further away from
the lens?

Answer. Size of the Object, h1 = 3
Object distance, u = -14 cm
Focal length of the concave lens, f = -21 cm
Image distance = v
According to the lens formula, we have the relation:
1 1 1
− =
v u f

1 1 1 −2−3 −5
= − − = =
v 21 14 42 42

42
∴ v = − = −8.4cm
5

Hence, the image is formed on the other side Of the lens, 8.4 cm away from it. The
negative sign shows that the image is erect and virtual.
The magni cation of the image is given as:
Image height (h2 ) v
m = =
Object height (h1 ) u

−8.4
∴ h2 = × 3 = 0.6 × 3 = 1.8cm
−14

Hence, the height Of the image is 1.8 cm. If the object is moved further away from the lens, then
the virtual image will move toward the focus of the lens, but not beyond it. The size of the image
will decrease with the increase in the object distance.

Page : 344 , Block Name : Exercise

Q9.10 What is the focal length of a convex lens of focal length 30cm in contact with a concave lens
of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the
lenses.

Answer. Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = -20 cm
Focal length of the system of lenses = f
The equivalent focal length Of a system Of two lenses in contact is given as:

1 1 1
= +
f f1 f2

1 1 1 2−3 1
= − = = −
f 30 20 60 60

∴ f = −60cm

Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the
system of lenses acts as a diverging lens.

Page : 345 , Block Name : Exercise

Q9.11 A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece
of focal length 6.25 cm separated by a distance of 15cm. How far from the objective should an
object be placed in order to obtain the nal image at (a) the least distance of distinct vision
(25cm), and (b) at in nity? What is the magnifying power of the microscope in each case?

Answer. Focal length of the objective lens, f1 = 2.0 cm
Focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d =15 cm
(a) Least distance of distinct vision, d = 25cm ′

lmage distance for the eyepiece, v2 = −25cm
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
1 1 1
− =
v2 u2 f2

1 1 1
= −
u2 v2 f2

1 1 −1−4 −5
= − = =
−25 6.25 25 25

∴ u2 = −5cm

Image distance for the objective lens, v = d + u = 15 − 5 = 10cm 1 2

Object distance for the objective lens — u1
According to the lens formula, we have the relation:
1 1 1
− =
v1 u1 f1

1 1 1
= −
u1 v1 f1

1 1 1−5 −4
= − = =
10 2 10 10

∴ u1 = −2.5cm

Magnitude of the object distance, |u | = 2.5 cm 1 =

The magnifying power of a compound microscope is given by the relation:
′
v1 d
m = (1 + )
|u1 | f2

10 25
= (1 + ) = 4(1 + 4) = 20
2.5 6.25

Hence, the magnifying power of the microscope is 20.
(b) The nal image is formed at in nity.
Image distance for the eyepiece, v = ∞ 2

Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
1 1 1
− =
v2 u2 f2

1 1 1
− =
∞ u2 6.25

∴ u2 = −6.25cm

Image distance for the objective lens, v = d + u = 15 − 6.25 = 8.75cm
1 2

Object distance for the objective lens = u1
According to the lens formula, we have the relation:
1 1 1
− =
v1 u1 f1

1 1 1
= −
u1 v1 f1

1 1 2 − 8.75
= − =
8.75 2.0 17.5

17.5
∴ u1 = − = −2.59cm
6.75

|u1 | = 2.59 cm
Magnitude of the object distance,
The magnifying power of a compound microscope is given by the relation:
′
v1 d
m = ( )
|u1 | |u2 |

8.75 25
= × = 13.51
2.59 6.25

Hence, the magnifying power Of the microscope is 13.51.

Page : 345 , Block Name : Exercise

Q9.12 A person with a normal near point (25 cm) using a compound microscope with objective of
focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm
from the objective in sharp focus. What is the separation between the two lenses? Calculate the
magnifying power of the microscope.

Answer. Focal length of the objective lens, fo = 8 mm = 0.8 cm
Focal length Of the eyepiece, fe 2.5 cm
Object distance for the objective lens, uo = - 9.0 mm = -0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, ve = -d = -25 cm
Object distance for the eyepiece = u e

Using the lens formula, we can obtain the value ofu as: e

1 1 1
− =
ve ue fc

1 1 1
= −
ue ve fc

1 1 −1−10 −11
= − = =
−25 2.5 25 25

25
∴ uc = − = −2.27cm
11

We can also obtain the value of the image distance for the objective lens (v _ { e }\) using the lens
formula.
1 1 1
− =
vo uo fo

1 1 1
= +
vo fo uo

1 1 0.9−0.8 0.1
= − = =
0.8 0.9 0.72 0.72

v0 = 7.2cm

The distance between the objective lens and the eyepiece = |u | + v
e o

= 2.27 + 7.2
= 9.47 cm
The magnifying power of the microscope is calculated as:
vo d
(1 + )
|uo | fe

7.2 25
= (1 + ) = 8(1 + 10) = 88
0.9 2.5

Hence, the magnifying power of the microscope is 88.

Page : 345 , Block Name : Exercise

Q9.13 A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal
length 6.0cm. What is the magnifying power of the telescope? What is the separation between the
objective and the eyepiece?

Answer. Focal length of the objective lens, fo = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power of the telescope is given as:
f0
m =
fe

144
= = 24
6

The separation between the objective lens and the eyepiece is calculated as:
f0 + fe

= 144 + 6 = 150cm

Hence, the magnifying power of the telescope is 24 and the separation between the
objective lens and the eyepiece is 150 cm.

Page : 345 , Block Name : Exercise

Q9.14 (a) A giant refracting telescope at an observatory has an objective lens of focal length 15m.

If an eyepiece of focal length 1.0cm is used, what is the angular magni cation of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon
formed by the objective lens? The diameter of the moon is 3.48 × 10 m, and the radius of lunar
6

orbit is 3.8 × 10 m 8

Answer. Focal length of the objective lens, fo = 15 m = 15 x 102 cm
Focal length of the eyepiece, fe = 1.0 cm
(a) The angular magni cation of a telescope is given as:
fo
α =
fe

2
15 × 10
= = 1500
1.0

Hence, the angular magni cation of the given refracting telescope is 1500.
(b) Diameter of the moon, d = 3.48 × 10 m
6

Radius of the lunar Orbit, ro 3.8 × 10 m 8

Let d be the diameter of the image of the moon formed by the objective lens. The angle subtended
′

by the diameter Of the moon is equal to the angle subtended by the image.
′
d d
=
r0 f0

6 ′
3.48×10 d
8
=
3.8×10 15

′ 3.48 −2
∴ d = × 10 × 15
3.8

−2
= 13.74 × 10 m = 13.74cm

Hence, the diameter Of the moon's image formed by the objective lens is 13.74 cm

Page : 345 , Block Name : Exercise

Q9.15 Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located
between the focus and the pole
(d) an object placed between the pole and focus of a concave mirror produces a virtual and
enlarged image.

Answer. (a) For a concave mirror, the focal length (f) is negative.
f<0
When the object is placed on the left side of the mirror, the object distance (u) is
Negative.
u<0
For image distance v, we can write the lens formula as:
1 1 1
− =
v u f

1 1 1
........ (1)
= −
v f u

The object lies between f and 2f.

∴ 2f < u < f

1 1 1
> >
2f u f

1 1 1 ........ (2)
− < − < −
2f u f

1 1 1 1
− < − < 0
f 2f f u

Using equation (1), we get:
1 1
< < 0
2f v

- 1

v
is negative, i.e., v is negative
1 1
<
2f v

2f > v

−v > −2f

Therefore, the image lies beyond 2f.

(b) For a convex mirror, the focal length (f) is positive.
f>0
When the object is placed on the left side Of the mirror, the object distance (u) is Negative.
u<0
For image distance v, we have the mirror formula:
1 1 1
+ =
v u f

1 1 1
= −
v f u

But we have u < 0

1 1
∴ >
v f

ν < f

Hence, the image formed is diminished and is located between the focus (f) and the
pole.

(d) For a concave mirror, the focal length (f) is negative.
f<0
When the object is placed on the left side of the mirror, the object distance (u) is
Negative.
u<o
It is placed between the focus (f) and the pole.
∴ f > u > 0

1 1
< < 0
f u

1 1
− < 0
f u

For image distance v, we have the mirror formula:

1 1 1
+ =
v u f

1 1 1
= −
v f u

1
∴ < 0
v

v > 0

The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:
1 1
>
u v

v > u

Magni cation, m = v

u
>1
Hence, the formed image is enlarged.

Page : 345 , Block Name : Exercise

Q9.16 A small pin xed on a table top is viewed from above from a distance of 50cm. By what
distance would the pin appear to be raised if it is viewed from the same point through a 15 cm
thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend
on the location of the slab?

Answer. Actual depth of the pin, d = 15 cm
Apparent depth of the pin = d ′

Refractive index of glass, μ= 1.5
Ratio of actual depth to the apparent depth is equal to the refractive index of glass,
d
μ = ′
d

′ d
∴ d =
μ

15
= = 10cm
1.5

The distance at which the pin appears to be raised = d − d
′

= 15 - 10 = 5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.

Page : 345 , Block Name : Exercise

Q9.17 (a) Figure shows a cross-section of a ‘light pipe’ made of a glass bre of refractive index
1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the
range of the angles of the incident rays with the axis of the pipe for which total re ections inside
the pipe take place, as shown in the gure.
(b) What is the answer if there is no outer covering of the pipe?

Answer. (a) Refractive index Of the glass bre, μ = 1.68 1

Refractive index of the outer covering of the pipe, μ = 1.44 2

Angle of incidence = i
Angle Of refraction = r
Angle of incidence at the interface = i'
The refractive index (μ) of the inner core-outer core interface is given as:
μ2 1
μ = =
μ1 sin i′

μ1
′
sin i =
μ2

1.44
= = 0.8571
1.68

′ ∘
∴ i = 59

For the critical angle, total internal re ection (TIR) takes place only when , i.e., i > i , i.e., i > 59
′ ∘

Maximum angle of re ection, r = 90 − i = 90 − 59 = 31
max
∘ ′ ∘ ∘ ∘

Let, i be the maximum angle of incidence.
max

The refractive index at the air glass interface, μ = = 1.68 1

We have the relation for the maximum angles of incidence and re ection as:
sin imax
μ1 =
sin rmax

sin imax = μ1 sin rmax

= 1.68 × 0.5150

= 0.8652
−1 ∘
∴ imax = sin 0.8652 ≈ 60

Thus, all the rays incident at angles lying in the range 0 < i < 60 will suffer total internal
∘

re ection.

(b) If the outer covering of the pipe is not present, then:
Refractive index Of the outer pipe, μ Refractive index of air = 11

For the angle of incidence i = 90 we can write Snell's law at the air — pipe interface as:
∘

sin i
= μ2 = 1.68
sin r
∘
sin 90 1
sin r = =
1.68 1.68

−1
r = sin (0.5952)

∘
= 36.5

′ ∘ ∘ ∘
∴ i = 90 − 36.5 = 53.5

Since i' > r, all incident rays will suffer total internal re ection.

Page : 345 , Block Name : Exercise

Q9.18 Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they
produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we 'see' a virtual image,
we are obviously bringing it on to the 'screen'

(i.e., the retina) Of our eye. Is there a contradiction?
(c) A diver underwater, looks obliquely at a sherman standing on the bank of a lake. Would the
sherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent
depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some
use to a diamond cutter?

Answer. (a) Yes , plane and convex mirrors can produce real images as well. If the object is virtual,
i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are re ected
to a point on a screen placed in front of the mirror, then a real image will be formed.

(b) No A virtual image is formed when light rays diverge. The convex lens of the eye causes these
divergent rays to converge at the retina. In this case, the virtual image serves as an object for the
lens to produce a real image.

(c) The diver is in the water and the sherman is on land (i.e., in air). Water is a denser medium
than air. It is given that the diver is viewing the sherman. This
indicates that the light rays are travelling from a denser medium to a rarer medium.
Hence, the refracted rays will move away from the normal. As a result, the
sherman will appear to be taller.

(d) Yes; Decrease The apparent depth Of a tank Of water changes when viewed obliquely. This is
because light bends on travelling from one medium to another. The apparent depth of the tank
when viewed obliquely is less than the near-normal viewing.

(e) Yes The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical
angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to
ensure that the light entering the diamond is totally re ected from its faces. This is the reason for
the sparkling effect Of a diamond.

Page : 346 , Block Name : Exercise

Q9.19 The image of a small electric bulb xed on the wall of a room is to be obtained onthe
opposite wall 3 m away by means Of a large convex lens. What is the maximum
possible focal length of the lens required for the purpose?

Answer. Distance between the object and the image, d — 3 m
Maximum focal length Of the convex lens = f m

For real images, the maximum focal length is given as:
d
fmax =
4
3
= = 0.75m
4

Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Page : 346 , Block Name : Exercise

Q9.20 A screen is placed 90 cm from an object. The image of the object on the screen is formed by
a convex lens at two different locations separated by 20 cm. Determine
the focal length Of the lens.

Answer. Distance between the image (screen) and the object, D 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
2 2
D − d
f =
4D
2 2
(90) − (20) 770
= = = 21.39cm
4 × 90 36

Therefore, the focal length of the convex lens is 21.39 cm.

Page : 346 , Block Name : Exercise

Q9.21 (a) Determine the 'effective focal length' of the combination of the two lenses in Exercise
9.10, if they are placed 8.0 cm apart with their principal axes coincident.Does the answer depend
on which side Of the combination a beam Of parallel light isincident? Is the notion of effective
focal length of this system useful at all?

(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above.
The distance between the object and the convex lens is 40
cm. Determine the magni cation produced by the two-lens system, and the size of the image.

Answer. Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = -20 cm
Distance between the two lenses, d = 8.0 cm

(a) When the parallel beam of light is incident on the convex lens rst:
According to the lens formula, we have:
1 1 1
− =
v1 u1 f1

Where,

u1 = Object distance = ∞

v1 = Image distance

1 1 1 1
= − =
v1 30 ∞ 30

∴ v1 = 30cm

The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:

1 1 1
− =
v2 u2 f2

where,

u2 = Object distance

= (30 − d) = 30 − 8 = 22cm

v2 = Image distance

1 1 1 10−11 −1
= − = =
v2 22 20 220 220

∴ v2 = −220cm

The parallel incident beam appears to diverge from a point that is (220 − d

2
= 220 − 4) 216cm

from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens rst:
According to the lens formula, we have:
1 1 1
− =
v2 u2 f2

1 1 1
= +
v2 f2 u2

Where,

u2 = Object distance = −∞

v2 = Image distance

v2 = Image distance

v2 = Image distance

∴ v2 = −20cm

The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
1 1 1
− =
v1 u1 f1

Where,

u1 = Object distance

= −(20 + d) = −(20 + 8) = −28cm

v1 = Image distance

1 1 1 14−15 −1
= + = =
v1 30 −28 420 420

∴ v2 = −420cm

Hence, the parallel incident beam appear to diverge from a point that is (420 - 4)
416 cm from the left Of the centre Of the combination Of the two lenses.
The answer does depend on the side of the combination at which the parallel beam of light is
incident. The notion of effective focal length does not seem to be useful for
this combination.

(b) Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens,u 1 = −40cm

|u | = 40cm= 40 cm
1

According to the lens formula:

1 1 1
− =
v1 u1 f1

Where,

v1 = Image distance

1 1 1 4−3 1
= + = =
v1 30 −40 120 120

∴ v1 = 120cm
v1
m =
|u1 |

Magnification,

120
= = 3
40

Hence, the magni cation due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula:
1 1 1
− =
v2 u2 f2

Where,

u2 = Object distance

= +(120 − 8) = 112cm

v2 = Image distance

1 1 −112+20 −92
v2 = + = =
−20 112 2240 2240

−2240
∴ v2 = cm
92

Magni cation m = ∣∣ ′ v2
∣
u2 ∣

2240 1 20
= × =
92 112 92

Hence, the magni cation due to the concave lens is 20

92

The magnification produced by the combination of the two lenses is calculated a
′
m × m

20 60
= 3 × = = 0.652
92 92

The magnification of the combination is given as:
h2
= 0.652
h1

h2 = 0.652 × h1

Where,

h = Object size = 1.5cm

h2 = size of the image

∴ h2 = 0.652 × 1.5 = 0.98cm

Hence, the height of the image is 0.98cm

Page : 346 , Block Name : Exercise

Q9.22 At what angle should a ray of light be incident on the face of a prism of refracting angle 60°
so that it just suffers total internal re ection at the other face? The refractive index of the material
of the prism is 1.524.

Answer. The incident, refracted, and emergent rays associated with a glass prism ABC are shown
in the given gure.

Angle of prism, A = 60°
Refractive index of the prism, μ = 1.524
i = Incident angle
1

r = Refracted angle
1

r = Angle of incidence at the face AC
2

e = Emergent angle = 90°
According to Snell's law, for face AC, we can have:
sin e
= μ
sin r2

1
∘
sin r2 = × sin 90
μ

1
= = 0.6562
1.524
−1 ∘
∴ r2 =sin 0.6562 ≈ 41

It is clear from the figure that angle A = r1 + r2
∘
∴ r1 = A − r2 = 60 − 41 = 19

According to Snell's law, we have the relation:
sin i1
μ =
sin r1

sin i1 = μ sin r1
∘
= 1.524 × sin 19 = 0.496
∘
∴ i1 = 29.75
∘
Hence, the angle of incidence is 29.75

Page : 346 , Block Name : Exercise

Q9.23 A card sheet divided into squares each of size 1 mm is being viewed at a distance of 9 cm
2

through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
(a) What is the magni cation produced by the lens? How much is the area of each square in the
virtual image?
(b) What is the angular magni cation (magnifying power) of the lens?
(c) Is the magni cation in (a) equal to the magnifying power in (b)? Explain.

Answer. Missing

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Q9.24 (a) At what distance should the lens be held from the gure in Question 9.29 in order to
view the squares distinctly with the maximum possible magnifying power?
(b) What is the magni cation in this case?
(c) Is the magni cation equal to the magnifying power in this case? Explain

Answer. Missing

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Q9.25 What should be the distance between the object in Question 9.24 and the magnifying glass
if the virtual image of each square in the gure is to have an area of 6.25 mm . Would you be able
2

to see the squares distinctly with your eyes very close to the magni er?
[Note: Question 9.23 to 9.25 will help you clearly understand the difference between magni cation
in absolute size and the angular magni cation (or magnifying power) of an instrument.]

Answer. Missing

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Q9.26 Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the
virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide
angular magni cation?

(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens.
Does angular magni cation change if the eye is moved back?

(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the
lens. What then stops us from using a convex lens of smaller and smaller focal length and
achieving greater and greater magnifying power?

(d) Why must both the objective and the eyepiece of a compound microscope have short focal
lengths?

(e) When viewing through a compound microscope, our eyes should be positioned not on the
eyepiece but a short distance away from it for best viewing. Why? How much should be that short
distance between the eye and eyepiece?

Answer. Missing

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Q9.27 An angular magni cation (magnifying power) of 30X is desired using an objective of focal
length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound
microscope?

Answer. Missing

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Q9.28 A small telescope has an objective lens of focal length 140cm and an eyepiece of focal
length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the nal image is at in nity)?
(b) the nal image is formed at the least distance of distinct vision (25cm)?

Answer. Missing

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Q9.29 (a) For the telescope described in Question 9.28
(a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of
the tower formed by the objective lens?
(c) What is the height of the nal image of the tower if it is formed at 25cm?

Answer. Missing

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Q9.30 A Cassegrain telescope uses two mirrors as shown in Fig. Such a telescope is built with the
mirrors 20 mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror
is 140 mm, where will the nal image of an object at in nity be?

Answer. Missing

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Q9.31 Light incident normally on a plane mirror attached to a galvanometer coil retraces
backwards as shown in Fig . A current in the coil produces a de ection of 3.5o of the mirror. What
is the displacement of the re ected spot of light on a screen placed 1.5 m away?

Answer. Missing

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Q9.32 Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on
top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until
its inverted image is found at the position of the needle. The distance of the needle from the lens
is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance
is measured to be 30.0cm. What is the refractive index of the liquid?

Answer. Missing

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