Important Questions of Ray optics

1. MULTIPLE CHOICE QUESTIONS
1. When a lens is immersed in water it’s focal length:
a) Increases b) Decreases c) Remain same d) None of these
2. Which of the following phenomena explains that light does not travel in straight line?
(a) Reflection (b) refraction (d) diffraction (d) polarization
3. The magnification produced by on astronomical telescope is normal adjusted is
(a) 𝑓°+𝑓𝑒 (b) 𝑓°x𝑓𝑒 (c)
𝑓°
𝑓𝑒
(d)
𝑓𝑒
𝑓°
4. In a white light spectrum produced by dispersion through a prism the ray that is deviated least is
(a) Yellow (b) blue (c) green (d) red
5. Dioptre is the unit of
(a) Focus of a lens (b) focal length of a lens (c) power of a lens (d) magnifying power
6. When a ray of light passes from one medium to another, which of the following does not change?
(a) Velocity (b) frequency (c) refractive index (d) wave length
7. A convex lens of focal length 12 cm is placed in contact with a concave lens of 18 cm focal length of the
combination is
(a) 50 cm (b) 45 cm (c)46 cm (d) 18 cm
8. A lens of power 5 D is placed in contant with another lens of power – 3 D. the focal length of the combination is:-
(a) 25 cm (b) 50 cm (c) 75 cm (d)1 cm
9. Angle of minimum deviation of a prism depends upon
(a) angle of prism. (b) angle of incidence
(c) nature of material of prism (d) all the above
10. Mirage is observed in deserts due to :
(a) Interference of light (b) diffraction of light
(c) total internal reflection of light (d) double refraction.
11. Diameter of the objective of a telescope is d and wavelength of light used is ⋋,then the resolving power of this
telescope will be
(a)
⋋
𝑑
(b)
1.22⋋
𝑑
(c)
𝑑
1.22⋋
(d) ⋋. 𝐷
12. . A virtual image larger than the object is formed in
a) Convex mirror b) Concave mirror c) Plane mirror d) None of these
13. The image formed by a simple microscope is
a) imaginary and erect b) imaginary and magnified c) real and erect d) real and inverted
14. With increase in wavelength, the value of refractive index
a) increase b) decreases c) remains unchanged d) none of these
15. A ray of light is incident on a glass plate at an angle of 60°. If the reflected and refracted rays are mutually
perpendicular, then the refractive index of the material is
a)
3
2
b) 3 c)
1
3
d)
1
2
16. In a comnpound microscope the image founed between the objective and the eye – piece is : -
a) Virtul erect and magnified b) Real erect & magnified
c) real inverted and magnified d) virtual erect and diminished
17. If the focal length of objective lens of an astronomical telescope is 20 cm and length of it is 25 cm.
The magnification of the telescope for normal adjustment is
a) 5 b) 4 c) 1.25 d) 1
18. Which of the following phenomenon exhibits that light is transverse wave.
a) Interference of light b) Diffraction of light
c) Polarization of light d) Refraction of light
TEST SERIES 2023 – 2024
PHYSICS: - RAY OPTICS

2. 19. A thin prism of refracting angle 6 degree produces a deviation of 3 degree the refractive index off the material of
the prism is
a) 1.33 b) 1.5 c) 0.5 d) 1.25
ANSWER
1.A 2.B 3.C 4.D 5.C 6.B 7.C 8.B 9.D 10.C 11.C 12.B 13.B 14.B 15.B
16.C 17.C 18.C 19.B
1 MARKS
1. A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65, what is the nature of the
lens.
Ans. Since 𝜇 < 𝜇𝑚𝑒𝑑𝑖𝑢𝑚 .
𝑙𝑒𝑛𝑠 It behave like a converging lens.
2. Does a beam of white light give a spectrum on passing through a hollow prism?
Ans. No, because no dispersion takes place in a hollow prism.
3. Two lenses of focal lengths 5 cm and 50 cm are to be used for making a telescope. Which lens will be used for
objective?
Ans. Lens of larger focal length i.e., 50 cm will be used as objective.
4. What is photometry
Ans. Photometry is a branch of physics that deals with the measurement of light emitted by or reflected from
objects.
5. Which of blue and red light will be deviated more by a prism and why?
Ans. We have, 𝛿 = 𝜇 − 1 𝐴
Where 𝜇 is the refractive index of the prism material and A is the angle of the prism.
It should be remembered that refractive index increase with decreasing wavelength.
i.e., 𝜇 > 𝜇𝑅
𝐵
𝛿 > 𝛿𝑅
𝐵
So, blue ray will deviate more.
6. For which colour of visible light the refractive index of a transparent material is maximum and for which
minimum?
Ans. Maximum for violet and minimum for red.
7. Does critical angle depend upon the wavelength of light used?
Ans. Yes. Since critical angle depends on refractive index which in turn depends on wavelength.
8. What should be the position of an object relatives to a biconvex lens so that it behaves like a magnifying lens?
Ans. For biconvex lens to behave as a magnifying lens, the object must be placed between the focal point and the
optical centre of the lens.
9. Give an expression for the magnifying power of a telescope.
Ans. Magnifying power of a telescope, M =
𝑓 0
𝑓𝑒
Here, 𝑓 0 and 𝑓𝑒 are the focal lengths of the objective and the eyepiece respectively.
10. Does the dispersive power of a prism depend upon the angle of the prism?
Ans. No, it is independent of the angle of prism.
11. Distant objects which appear quite small with naked eye, appear larger through a telescope, how?
Ans. The telescope forms the image of the distant object near the eye. Therefore, the image subtends a large visual
angle at the eye and the object seems bigger.
12. What is the magnification produced by a single convex lens used as a simple microscope in normal use?
Ans. The magnification produced by a single convex lens, as a simple microscope in the normal use is given by
M = 1 +
𝐷
𝑓
13. How do optical fibers transmit light without significant absorption? Mention one practical application of optical
fibers.

3. Ans. Optical fibers transmit light without significant absorption due to the total internal reflection. Optical fibers are
used in communication.
14. Mention condition for total internal reflection to occur.
Ans. The conditions for total internal reflection are:
i) The ray must travel form denser to rarer medium
ii) The angle of incidence should be greater than the critical angle.
15. State two basic differences between primary and secondary rainbows.
Ans. i) In a primary rainbow, light rays have suffered two total internal reflections.
ii) In primary rainbow, outer edge is red and inner edge is violet while in secondary rainbow, outer edge is violet and
inner edge is red
16. Define power of a lens. Name the unit in which it is expressed.
Ans. The power (P) of a lens is its ability to deviate the rays towards its principle axis and is defined as the reciprocal
of its focal length (expressed in metres), i.e.,
𝑃 =
1
𝑓(𝑖𝑛 𝑚𝑒𝑡𝑟𝑒 )
The unit of power is m-1
, which is also called the dioptre (D)
3 MARKS
1. Write three properties o a plane mirror.
Ans. Three properties of plane mirror are:
i) The image formed is laterally inverted
ii) The image formed is as far behind the mirror as the object is in front of it.
iii) The size of the image equals the size of the object.
iv) Image formed is always virtual and erect.
2. What are paraxial rays?
Ans. Paraxial rays are those, that subtend a small angle with normal to the surface at the point where they strike.
The rays are paraxial:
i) If they are closer to the principal axis, or
ii) If the mirror has a small aperture.
3. State the three conditions for the formation of pure spectrum
Ans. The three conditions for the formation of pure spectrum are:
i) The slit should be narrow
ii) The rays falling on the prism should be parallel to each other.
iii) The prism should be placed in minimum deviation position with respect to the mean ray and the refracting edge
of the prism should be parallel to the slit.
4. Find the focal length and nature of a lens whose optical power is – 5 D
Ans. Nature of lens is concave lens.
5. i) Define resolving power of a simple astronomical telescope.
ii) State one advantage of a reflecting telescope over refracting telescope.
Ans. Resolving power: it is the ability of an optical instrument to produce separate images of objects, which are very
close to each other.
i) One advantage of a reflecting telescope over refracting telescope is that the image formed by reflecting telescope
is larger and brighter than the image formed by a refracting telescope. Reflecting telescope is cheaper.
6. Why are the danger signals are red in colour?
Ans. According to Rayleigh’s scattering law, the scattering of light is inversely proportional to the fourth power of
the wavelength of light. So, the scattering of red light is much less than other light and the signals of red light can be
seen upto a longer distance. As a result whether it is fog or smoke, red light passes comparatively easily through
them. Therefore, the danger signals are red in colour.
7. Why is a reflecting type telescope preferred in astronomy?

4. Ans. Reflecting type telescopes are preferred because images formed by reflection are usually brighter than images
formed by lenses under similar conditions. There is no chromatic aberration in mirrors. It is not possible to make
large aperture lenses free of all aberrations. A large aperture is required to increase the light gathering power, to be
able to view very faint stars. Spherical aberration, in case of large concave mirrors can be eliminated by using
parabolic mirros.
8. Very distant stars which are not visible to the eye, are visible in telescope. Why?
Ans. The aperture of the objective of the telescope is much larger than the aperture of eye, therefore, it collects
sufficient light from the star and forms a bright image, which is visible to the eye. Thus, feeble stars are visible in the
telescope because of the ‘large’ aperture of the objective.
9. How can you produce dispersion without deviation?
Ans. Dispersion without deviation: A single prism is known to produce both dispersion and deviation However, we
can produce dispersion without deviation by using two prisms one of crown glass and the other of flint glass so that
the deviations produced by them for a particular wavelength of light are equal and opposite.
10. Why does the sky appear dark instead of blue to an astronaut?
Ans. The sky appears blue, when viewed from Earth, because of the scattering of the light. When sunlight strikes
molecules in our atmosphere, the light is redirected in many directions. The blue light is scattered more than the red
light causing the sky to be blue. In the space, there is no atmosphere. So, light will not scatter in space. This is the
cause of dark sky when viewed from space.
11. Distinguish between magnification and angular magnification produced by a lens?
Ans. The magnification produced by a lens is the ratio of the size of the image to the size of the object. It is also equal
to the ratio of the ratio of the distance of the image from the lens to the distance of the object from the lens. Thus,
magnification,𝑚 = 𝑣/𝑢
On the other hand, angular magnification produced by a lens is defined as the ratio of the angle subtended by image
at the eye to the angle subtended by object seen directly, When both lie at the least distance of distinct vision. It is
also called magnifying power of the lens. The angular magnification produced = 𝐷/𝑢; where, D is least distance of
distinct vision. The magnification becomes equal to the angular magnification, when 𝑣 = D.
12. What is dispersion and angular dispersion? Derive an expression for the angular dispersion produced by a
prism of small angle.
Ans. Dispersion: The phenomenon of splitting up of white light into its constituent colours is called ‘dispersion of
light’.
Angular dispersion: The angular dispersion is defined as the difference between
the deviation on the angles of deviation suffered by two colours while passing
Through a prism. For red and violet colours the angular dispersion is:
𝜃 = 𝛿 −𝛿 .
𝑅
𝑉
If 𝜇 𝑎𝑛𝑑
𝑉 𝜇𝑅 are the refractive indices of the material of the prism for violet and red colours,
𝛿 −𝛿 = 𝐴 (
𝑅
𝑉 𝜇 – 1) − 𝐴 ( 𝜇 −1)
𝑅
𝑉
𝜃 = (𝜇 – 𝜇 )
𝑅
𝑉
𝐴.
13. For a ray of monochromatic light passing through a glass prism, draw a graph to show the variation of the
angle of deviation (D) with the angle of incidence i). When does the ray suffer minimum deviation?
Ans. The graph showing the variation of the angle of deviation (D) with the angle of incidence i) is as follows:
The ray of light suffers minimum deviation, when

5. Angle of incidence = Angle of emergence
14. You are given three lenses having powers P and apertures A as follows:
P1 = 6 D, A1 = 3 cm
P2 = 3 D, A2 = 15 cm
P3 = 12 D, A3 = 1.5 cm
Which two of these will you select to construct, i) A telescope and ii) A microscope? State the basis for your
answer in each case.
Ans. Given: Three lenses having powers P and apertures A as follows :
Lens 1 P1 = 6 D A1 = 3 cm
Lens 2 P2 = 3 D A2 = 15 cm
Lens 3 P3 = 12 D A3 = 1.5 cm
i) For constructing a telescope the objective lens should have large aperture. Hence, lens 2 will be preferred as an
objective, Moreover, the eyepiece should have small length. So lens 3 will be preferred as an eye piece. [∴ 𝑃 ∝ 1/𝑓]
ii) For constructing a microscope, both the objective and the objective and the eyepiece should have short focal
lerigths and the focal length of the objective should be shorter than the focal length of eye piece. Therefore, lens 3
will be used as the objective and lens 1 will be used as an eyepiece.
15. Draw ray diagrams to illustrate the formation of:
a) A primary rainbow by refraction and total internal reflection of sun light in water drops.
ii) Image at infinity in a refracting telescope
Ans.
16. What is optical fiber.
Ans. Optical fiber is one of the practical examples to show total internal reflection. As shown in figure, the fiber
consists of two different transparent mediums. Medium of higher refractive index is at the middle and is surrounded
by another medium of relatively low refractive index. The fiber is covered with an opaque covering. When the light
enters form one opening it suffers multiple total internal reflections from the surface separating the two medium as
each time I (angle of incidence) is more than critical angle iC and ultimately light ray reaches to the other end.
17. What are the laws of reflection?
Ans. i) The incident ray, the reflected ray and the normal at the point of incidence, all lie in the same plane.
ii) The angle of incidence equals the angle of reflection.
18. What are the laws of refraction?
Ans. i) The incident ray, the normal and the refracted ray at the point of incidence all lie in the same plane.
ii) The ratio of the since of the angle of incidence to the since of the angle of refraction is a constant and is called the
refractive index of the medium. This is known as Snell’s law.
sin 𝑖
sin 𝑟
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

6. 19. The critical angle for a given transparent medium and air is iC . A ray of light travelling in air is incident on this
transparent medium at an angle of incidence equal to the polarising angle ip. What is the relation between the
two angles ic and ip?
Ans. We know that, 𝑛 = tan 𝑖𝑝
tan 𝑖 =
1
sin 𝑖𝑐
𝑝 ∴ 𝑛 =
1
sin 𝑖𝑐
𝑖 = tan (𝑐𝑜𝑠𝑒𝑐 𝑖 )
𝑐
−1
𝑝
5 MARKS
1. Define the focal length and radius of curvature of a spherical mirror. Deduce the relation between them.
Ans. Focal length: The distance between the pole and the principal focus of the mirror is called the ‘focal length’ of
the mirror.
Radius of curvature: The radius of sphere, of which the mirror is a part is called the ‘radius of curvature’ of the
mirror.
Relation between the focal length and the radius of curvature : Consider a ray of light OA, parallel to the principal
axis, incident on a spherical mirror at point A. The normal to the surface at point A is CA and CP = CA = R, is the
radius of curvature. They ray OA, after reflection from mirror will pass through F and obeys law of reflection i.e., i = r.
From the figure,
∠𝐴𝐶𝑃 = 𝜃 = 𝐼
In ∆ 𝐶𝐴𝐹,
𝜃 = 𝑟
∴ 𝐴𝐹 = 𝐹𝐶
If the aperture of the mirror is small, A lies close to P,
∴ 𝐴𝐹 = 𝑃𝐹 = 𝑅 𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟
and 𝐹𝐶 = 𝐹𝑃
Now, 𝑃𝐶 = 𝑃𝐹 + 𝐹𝐶
𝑃𝐶 = 𝑃𝐹 + 𝑃𝐹
𝑃𝐶 = 2𝑃𝐹
Or 𝑅 = 2𝑓
𝑓 =
𝑅
2
This is the required relation between 𝑓 𝑎𝑛𝑑 𝑅.
2. Deduce the relation 𝝁 =
𝑹𝒆𝒂𝒍 𝒅𝒆𝒑𝒕𝒉
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝒅𝒆𝒑𝒕𝒉
Ans. Consider an object O is placed in a liquid. A ray of light OA is incident along the normal to the surface of the
liquid. Another ray OB, on entering air bends away from the normal. The two refracting ray appear to come from
point I, therefore, object though lying at depth OA, appears to be at depth A Accordingly, AO is called ‘real depth’
and AI is called ‘apparent depth’ of the object O.
The refractive index of the medium air ′𝑎′ with respect to liquid ‘I’ is given by
𝐼𝜇 =
sin 𝑖
sin 𝑟
𝑎 𝑎𝑛𝑑 𝑎 𝜇 =
𝑠𝑖𝑟 𝑟
sin 𝑖
𝐼
But from ∆𝐼𝐴𝐵 sin𝑟 =
𝐴𝐵
𝐼𝐵
And from ∆𝑂𝐴𝐵 sin𝑖 =
𝐴𝐵
𝑂𝐵

7. 𝑎 𝜇 =
𝐴𝐵/ 𝐼𝐵
𝐴𝐵/𝑂𝐵
𝐼
=
𝑂𝐵
𝐼𝐵
𝑎 𝜇 =
𝐵𝑂
𝐼𝐵
𝐼
But BO ~ 𝐴𝑂 𝑎𝑛𝑑 𝐵𝐼 ~ 𝐴𝐼, 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝐴 𝑖𝑠 𝑣𝑒𝑟𝑦 𝑐𝑙𝑜𝑠𝑒 𝑡𝑜 𝐵
∴ 𝑎 𝜇 =
𝐴𝑂
𝐴𝐼
𝐼
∴ 𝑎𝜇 =
𝑅𝑒𝑎𝑙 𝑑𝑒𝑝𝑡 𝑕
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑑𝑒𝑝𝑡 𝑕
𝐼
3. Prove the relation 𝒂𝝁 =
𝟏
𝒂𝝁𝒃
𝒃
Ans. Consider a ray of light AO incident on the transparent surface and refracted along the path OB. Let the ray OB is
incident on the plane mirror and retraces back in the same medium when light travels from medium
′𝑎′
𝑡𝑜 𝑚𝑒𝑑𝑖𝑢𝑚′𝑏′
𝑎 𝜇 =
sin 𝑖
sin 𝑟
𝐼 (𝐹𝑟𝑜𝑚 𝑆𝑛𝑒𝑙𝑙′
𝑠 𝑙𝑎𝑤 … 𝑖)
Now, when the light travels from medium ‘b’ to medium ′𝑎′ after reflection from mirror MM’:
𝑏𝜇 =
sin 𝑟
sin 𝑖
𝑎 … 𝑖𝑖)
Multiplying equation i) and ii), we get
𝑎 𝜇𝑏 × 𝑏𝜇𝑎 =
sin 𝑖
sin 𝑟
×
sin 𝑟
sin 𝑖
𝑎 𝜇𝑏 × 𝑏𝜇𝑎 = 1
𝑎 𝜇𝑏 =
1
𝑏 𝜇 𝑎
This is also called as principle of reversibility of light.
4. Give reasons :
i) Why do the sun and the sky turn red during sunrise and sunset?
ii) Why is the sky blue?
i) According to Rayleigh’s law, the intensity of scattered light varies inversely as the fourth power of its wavelength.
The sunlight consists of seven colours. Of these, red, has the maximum wavelength. During sunrise and sunset, the
rays have to travel a larger part of the atmosphere because they are very close to the horizon. Therefore, light other
than red is mostly scattered away. Most they are light, which is the least scattered, enters our eyes. Hence, the sun
and the sky appear red.
ii) The blue colour of the sky is due to Rayleight scattering. As light moves through the atmosphere , most of the
longer wavelengths pass straight through. Little of the red, orange and yellow light is affected by the air. However,
much of the shorter wavelength light is absorbed by the gas molecules. The absorbed blue light is then radiated in
different directions. It gets scattered all around the sky in whatever direction you look. Since we see the blue light
from everywhere overhea, the sky looks blue.
Define angular magnification of a telescope. Draw a neat diagram with proper arrows to show the image formation
in an astronomical refracting telescope in normal adjustment for relaxed eye. Obtain the expression for angular
magnification of the telescope in this adjustment.

8. The angular magnification or magnifying power of telescope is defined as the ratio of the angle subtended by the
final image at the eye to the angle subtended by the eye when the object is in its actual position, i.e.,
𝑀 =
𝐴𝑛𝑔𝑙𝑒 𝑠𝑢𝑏𝑡𝑒𝑛𝑑𝑒𝑑 𝑏𝑦 𝑡𝑕𝑒 𝑓𝑖𝑛𝑎𝑙 𝑖𝑚𝑎𝑔𝑒 𝑡𝑕𝑒 𝑒𝑦𝑒
𝐴𝑛𝑔𝑙𝑒 𝑠𝑢𝑏𝑡𝑒𝑛𝑒𝑑 𝑏𝑦 𝑡𝑕𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑎𝑡 𝑡𝑕𝑒 𝑒𝑦𝑒 𝑤𝑕𝑒𝑛 𝑡𝑕𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑖𝑠 𝑖𝑛 𝑖𝑡𝑠 𝑎𝑐𝑡𝑢𝑎𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛
Because the position of eye is very close to the eyepiece E, hence the angle ′𝛽′ subtended by the final image AB of
the eyepiece may be taken as the angle subtended at the eye. In the same way, when the as the angle subtended at
the eye, then M = 𝛽/𝛼.
Since angles 𝛼 and 𝛽 are very small, so 𝛽 = tan 𝛼 = tan 𝛼.
Now, tan 𝛽
𝐴𝐵
𝐴𝐸
and tan 𝛼 =
𝐴𝐵
𝑂𝐴
∴ 𝑀 =
𝛽
𝑎
=
tan 𝛽
tan 𝛼
=
𝐴𝐵
𝐴𝐸
𝐴𝐵
𝐴𝑂
=
𝐴𝑂
𝐴𝐸
If the focal length of the object lens (or objective) be 𝑓 𝑎𝑛𝑑 𝐴𝐸 = −𝑢 𝑡𝑕𝑒𝑛
𝑒
°
𝑀 = −
𝑓°
𝑢𝑒
This is the general formula for the magnifying power of a telescope
Note : When the final image is formed at infinity then 𝑢𝑒 = 𝑓𝑐
𝑀 = −𝑓 𝐼𝑓𝑐
°
5. Derive the thin formula which relates object distance, image distance and the focal length of a convex lens.
Ans. Derivation of Lens formula: In the ray diagram, A’B’ is the image of the object AB formed by a thin convex lens
of focal length 𝑓.
Here, ∆ ABP similar to ∆ 𝐴, 𝐵, 𝑃 𝐴𝐴𝐴 𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦
So,
𝑃𝑄
𝐴 ′𝐵′
=
𝐴𝑃
𝐴′ 𝑃
∆𝑃𝑄𝐹 is similar to ∆ 𝐴′
𝐵′
𝐹 (𝐴𝐴𝐴 𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦)
So,
𝑃𝑄
𝐴′𝐵′
=
𝑃𝐹
𝐴 ′𝐹
Since, 𝑃𝑄 = 𝐴𝐵,
∴
𝐴𝐵
𝐴 ′𝐵′
=
𝑃𝐹
𝐴 ′𝐹
From equations i) and ii)
𝐴𝑃
𝐴′ 𝑃
=
𝑃𝐹
𝐴′𝐹
As per the sign convention,
𝐴𝑃 = − 𝑢, 𝑃𝐹 = 𝑓, 𝐴′
𝑃 = 𝑣, 𝐴′
𝐹 = 𝑣 − 𝑓
−𝑢
𝑣
=
𝑓
𝑣−𝑓
∴ −𝑢𝑣 + 𝑢𝑓 = 𝑣𝑓
On dividing the above equation by 𝑢𝑣𝑓, 𝑤𝑒 𝑔𝑒𝑡
1
−𝑓
+
1
𝑣
=
1
𝑢
Or
1
𝑣
+
1
𝑢
=
1
𝑓

9. 6. Derive the formula connecting the object distance 𝒖, image distance 𝒗 and radius of curvature R of a spherical
refracting surface.
Ans. Let a spherical refracting, surface XY separate a rarer medium of refractive index 𝜇1
From a denser medium of refractive index 𝜇2.
Suppose the surface is convex towards the rarer medium side.
OA is the incident ray for which AI is the refracted ray which meets the principles axis OPI at I. It is the image of the
point object. The angle of incidence is I and the angle of refraction is r. Let the angles made by the incident ray, the
refracted ray and normal at A with principal axis be 𝑎, 𝛽 𝛾 respectively as shown in figure. The angles are assumed to
be very smaller.
From A, draw AL ⊥ 𝑂𝐼. 𝐿𝑒𝑡 ∠𝐴𝑂𝐿 = 𝛼, ∠𝐴𝐼𝐿 = 𝛽 𝑎𝑛𝑑 ∠𝐴𝐶𝐿 = 𝛾.
In∠𝐼𝐶𝐴, as the external angle is equal to sum of internal opposite angles,
∴ 𝑟 + 𝛽 = 𝛾 …i)
Or 𝑟 = 𝛾 − 𝛽
Similarly, in ∆𝑂𝐴𝐶, 𝑖 = 𝛼 + 𝛾 …ii)
According to Snell’s law,
𝜇 2
𝜇 1
=
sin 𝑖
sin 𝑟
=
𝑖
𝑟
(∵ 𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑟𝑒 𝑠𝑚𝑎𝑙𝑙)
𝜇1 𝑖 = 𝜇2 𝑟
Using equations i) and ii), we get
𝜇1 𝛼 + 𝛾 𝑚 = 𝜇 ( 𝛾 − 𝛽
2 )
As angles 𝛼, 𝛽 𝑎𝑛𝑑 𝛾 are small, they can be taken as equivalent to their tangents.
∴ 𝜇 (tan𝛼 + tan 𝛾)
1 = 𝜇 (tan 𝛾 − tan 𝛽
2 … iii)
𝐼𝑛 ∆𝐴𝑂𝐿, tan 𝛼 =
𝐴𝐿
𝐿𝑂
𝐼𝑛 ∆𝐴𝐼𝐿, tan 𝛽 =
𝐴𝐿
𝐿𝐼
𝐼𝑛 ∆𝐴𝐶𝐿, tan 𝛾 =
𝐴𝐿
𝐿𝐶
∴ 𝜇1
𝐴𝐿
𝐿𝑂
+
𝐴𝐿
𝐿𝐶
= 𝜇2
𝐴𝐿
𝐿𝐶
+
𝐴𝐿
𝐿𝐼
…iv)
As aperture of the surface is small, L is close to P. Therefore,
𝐿𝑂 ≃ 𝑃𝑂
𝐿𝐼 ≃ 𝑃𝐼
𝐿𝐶 ≃ 𝑃𝐶
∴ 𝜇1
1
𝑃𝑂
+
1
𝑃𝐶
= 𝜇2
1
𝑃𝐶
+
1
𝑃𝐼
Or
𝜇 1
−𝑢
+
𝜇 2
𝑣
=
𝜇 2 −𝜇 1
𝑃𝐶
This is the required relation.
7. For any prism, show that refractive index of its material is given by:
𝒏 𝒐𝒓 𝝁
𝐬𝐢𝐧
𝑨+𝜹𝒎
𝟐
𝐬𝐢𝐧
𝑨
𝟐
Where the terms have their usual meaning.
Ans. In the figure, a ray of light PQ is incident at an angle i on the face AB of prism ABC. This ray is refracted along QR
at an angle r. This refracted ray is incident on the face AC at an angle r’ and emerges along RS at an angle i.
In ∆ 𝑄𝐷𝑅,
𝛿 = 𝑖 − 𝑟 + (𝑖′
− 𝑟′
) (By exterior angle property)
= 𝑖 + 𝑖′
− (𝑟 + 𝑟′
) …i)

10. 𝐼𝑛 𝑄𝑢𝑎𝑑. 𝐴𝑄𝐸𝑅, 𝐴 + 𝐸 = 180°
𝐼𝑛 ∆ 𝑄𝐸𝑅, 𝑟 + 𝑟′
+ 𝐸 = 180° …ii)
∴ 𝑟 + 𝑟′
= 𝐴 … iii)
Putting value of r + r’ in equation (i), we get [from eq. ii) and iii)]
𝛿 = 𝑖 + 𝑖′
− 𝐴 … iv)
In the position of minimum deviation condition,
𝑖 = 𝑖′
, 𝑟 = 𝑟′
, 𝛿 = 𝛿𝑚𝑣
So, 𝑟 + 𝑟′
= 𝐴
2𝑟 = 𝐴
Or 𝑟 = 𝐴/2 …v)
𝛿 = 2𝑖 − 𝐴
𝑚
𝑖 =
𝐴+𝛿𝑚
2
…vi)
Putting value of i and r from (v), (vi), in snell’s law,
𝑛 =
sin 𝑖
sin 𝑟
𝑛 =
sin
(𝐴+𝛿 )
𝑚
2
sin 𝐴/2
8. Prove that in case of a prism, I + e = A + 𝜹, where the symbols have their usual meanings.
Ans. ∆𝐴𝐵𝐶 is a principal section of the prism. A ray of light KL is incident on the face AB of the prism and the
refracted ray LM is incident at ∠𝑟2 on the face AC of the prism. The angle of deviation is ∠𝛿.
𝐼𝑛 ∆𝑃𝐿𝑀, 𝛿 = 1 + 2 …i)
𝑖 = 1 + 𝑟1 …ii)
𝑒 = 2 + 𝑟2 … iii)
On adding equation (ii) and (iii), we get
Or 𝑖 + 𝑒 = 1 + 2 + 𝑟 + 𝑟
1 2
𝑖 + 𝑒 = 𝛿 + 𝑟 + 𝑟2
1 *From eq. i)+… (iv)
Again 𝐴 + ∠𝐿𝑂𝑀 = ∠180° *opposite angles of a cyclic quadrilateral+ …(v)
and 𝑟 + 𝑟
1 + ∠𝐿𝑂𝑀 = 180°
2 …vi)
From equations (v) and vi)
𝐴 + ∠ 𝐿𝑂𝑀 = 𝑟 + 𝑟 + ∠𝐿𝑂𝑀
2
1
Or 𝐴 = 𝑟 + 𝑟 2
1 …(vii)
Putting this value in (iv), we get
𝑖 + 𝑒 = 𝛿 + 𝐴
This is the required relation.
9. Derive an expression for the angle of deviation of a small prism in terms of the refractive index and the angle of
the prism. Give a proper diagram.
Ans. Derive an expression for the angle of deviation of a small prism in terms of the refractive index and the angle of
the prism. Give a proper diagram.
Ans. The figure shows the passage of a light ray PQRS through a prism. The deviation of the emergent ray RS from
the incident ray PQ is𝛿.
The quadrilateral AQNR has right angles at the vertices Q and R.
Therefore, 𝐴 + ∠ 𝑄𝑁𝑅 = 180°

11. From the triangle QNR,
𝑟 + 𝑟′
+ ∠𝑄𝑁𝑅 = 180°
Comparing these equations, we get
𝐴 = 𝑟 + 𝑟′
From the triangle QMR, 𝛿 = ∠𝑀𝑄𝑅 + ∠𝑀𝑅𝑄
= 𝑖 − 𝑟 + 𝑖′
− 𝑟′
= 𝑖 + 𝑖′
− (𝑟 + 𝑟′
)
For the small angles,
sin 𝑖 ≃ 𝑖, sin 𝑟 ≃ 𝑟
sin 𝑖′ ≃ 𝑖′, sin𝑟 ≃ 𝑟′
⟹ 𝜇 ≃
𝑖
𝑟
≃
sin 𝑖′
sin 𝑟′
⟹ 𝑖 = 𝜇𝑟 𝑎𝑛𝑑 𝑖′
= 𝜇𝑟′
Thus, the deviation will be 𝛿 = 𝑖 − 𝑟 + 𝑖′
+ 𝑟′
= 𝜇𝑟 − 𝑟 + 𝜇𝑟′
− 𝑟′
= 𝜇 − 1 𝑟 + 𝑟′
But, 𝑟 + 𝑟′
= 𝐴
∴ 𝛿 = 𝜇 − 1 𝐴
Which is the required relation?
10. Explain the working of a simple microscope and show that its angular magnification is given by
𝑴 = 𝟏 +
𝑫
𝒇
, 𝒘𝒉𝒆𝒓𝒆 𝑫 𝒊𝒔 𝒕𝒉𝒆 𝒍𝒆𝒂𝒔𝒕 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒐𝒇 𝒅𝒊𝒔𝒕𝒊𝒏𝒄𝒕 𝒗𝒊𝒔𝒊𝒐𝒏.
Ans. A simple microscope is an optical instrument which forms large image of a close and minute object.
This image subtends a large visual angle at the eye so that the object looks large.
In the simplest form, a simple microscope, or magnifying glass is just a thin, short – focus convex lens placed
between the lens and its focus and the eye is placed just behind the lens.
Then, image on the same side as the object. The position of the object
between the lens and its focus is so adjusted that the image is formed
at the least distance of distinct vision (D) from the eye. The image is
then seen most distinctly.
In Fig. i) AB is a small a object placed between a lens L and its first
Focus ‘F’. Its magnified virtual image A’B’ is formed at distance D from
The lens. Since the eye is just behind the lens, the distance of the image
A ‘B’ at the eye and ′𝑎′ be the angle subtended by the object OA at the
eye when placed directly at a distance D from the eye [fig. ii) ]. Then, the magnifying power of the simple
microscope is given by
Angle subtended by the image at the eye when placed at least distance of district vision (𝛼)
Since the object is small, the angles 𝛼 𝑎𝑛𝑑 𝛽 are also small. Then, we may write
tan 𝛽 = 𝛽 𝑎𝑛𝑑 tan 𝑎 ≈ 𝑎
From the triangle IBC,

12. tan 𝛽 =
𝐼𝐵
𝐼𝐶
From the triangle IA’C,
tan 𝑎 =
𝐼𝐴′
𝐼𝐶
M =
𝐼𝐵
𝐼𝐶
×
𝐼𝐶
𝐼𝐴′
𝑀 =
𝐼𝐶
𝐼𝐴′
𝑀
𝐼𝐵
𝑂𝐴
But
𝐼𝐵
𝑂𝐴
= 𝐿𝑖𝑛𝑒𝑎𝑟 𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛
Hence, when the simple microscope is adjusted such that the image is formed at the near point, the angular
magnification is equal to the linear magnification. Applying new Cartesian sign convention, 𝑢
Therefore, the lens equations is,
1
𝑣
−
1
𝑢
=
𝑖𝑓
𝑓
𝑣
𝑣
−
𝑣
𝑢
=
𝑣
𝑓
1 − 𝑀 = −
𝐷
𝑓
𝑀 = 1 +
𝐷
𝑓
11. Draw a labelled diagram to illustrate the action of a compound microscope in normal use. Define
angular magnification and it to find the magnification when the final image is formed at least distance of distinct
vision.
Ans. The labelled diagram, illustrating the action of a compound microscope is given below:
Object AB is placed just beyond the focus of the objective Fa its image A’B’ is formed within the focal plane of the
eye lens, i.e., at distance less than Fe, The final virtual, magnified and upright image A “B” is formed. The real image
A’B’ acts as an object for the eyepiece. Thus a clear upright and magnified image is seen by the eye.
Angular magnification: The angular magnification (also called magnifying power) of a compound microscope is the
ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye, when
both are placed at the least distance of distinct vision.
𝑀 =
𝛽
𝑎
Since, the angle are small, 𝛽 = tan 𝛽 𝑎𝑛𝑑 𝑎 ≃ tan 𝑎
Therefore, 𝑀 =
tan 𝛽
tan 𝑎
=
𝐴 ′ 𝐵′ /𝐸𝐴′
𝐴𝐵/𝐷
=
𝐴 ′ 𝐵′
𝐴𝐵
×
𝐷
𝐸𝐴′
If the distances of the object AB and the image A ‘B’ from the objective O is 𝑢 𝑎𝑛𝑑 𝑣0
0 respectively, then from the
magnification formula, we have
𝐴 ′ 𝐵′
𝐴𝐵
=
+𝑣0
−𝑢0
Similarly, if the distance of A’B’ from the eyepiece be 𝑢 𝑡𝑕𝑒𝑛 𝐸𝐴′
= −𝑢 . 𝐴𝑙𝑠𝑜 𝐷 𝑖𝑠 𝑡𝑜 𝑏𝑒 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 − 𝐷.
𝑒
𝑒
Then from the above formula, we have

13. 𝑀 = −
𝑣0
𝑢0
−𝐷
− 𝑢𝑒
= −
𝑣0
𝑢0
𝐷
𝑢𝑒
Now, if the distance of distance of the final image A ‘’B’’ from the eyepiece be D, then using
1
𝑣
−
1
𝑢
=
1
𝑓
, 𝑓𝑜𝑟 𝑡𝑕𝑒
eyepiece, we will have
𝑣 = −𝐷, 𝑢 = −𝑢 𝑎𝑛𝑑 𝑓 = +𝑓𝑒
𝑒
⇒ −
1
𝐷
+
1
𝑢𝑒
=
1
𝑓𝑒
⇒ −
1
𝑢𝑒
=
1
𝐷
=
1
𝑓𝑒
⇒
𝐷
𝑢𝑒
= 1 +
𝐷
𝑓 𝑒
⇒ 𝑀 = −
𝑣0
𝑢0
1 +
𝐷
𝑓𝑒
12. Derive the Lens – maker’s formula for a convex lens.
Ans. Let O is an object on the principal axis of the lens. If the lens material were continuous and there were no
boundary/second surface XP2 Y, the refracted ray AB would go straight meeting the first refracted ray rat I1.
Therefore, I1 would have been a real image of O formed after refraction at XP1 Y. If CI1 = P1 I1 =
𝑣 𝑎𝑛𝑑 𝐶𝐶 = 𝑃 𝐶 = 𝑅 𝑤𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒
1,
1
1
1
1
−
𝜇 1
𝑢
+
𝜇 1
𝑣
=
𝜇 − 𝜇 2
2
𝑅1
In fact the lens material is not continuous.
Therefore, the refracted ray AB suffers furthers refractions at B and emerges along BI, meeting the principal axis at I.
Thus, I is the final real image of O, formed after refractive at the lens.
For refraction at the second surface XP2TY, we can regard I1 as a virtual object whose real image is formed at I.
Therefore, object, 𝑢 = 𝐶𝐼 = 𝑃 𝐼 = 𝑣 ,
1
1
2
1
Let CI ≃ 𝑃 𝐼 = 𝑣1,
1
2
and CC2 ≃ 𝑃 𝐶 = −𝑅2
2
2
As refraction is now taking place from denser to rarer medium, interchanging 𝜇 𝑎𝑛𝑑
1 𝜇2 𝑖𝑛 𝑖), 𝑤𝑒 𝑕𝑎𝑣𝑒
−𝜇 2
𝑣1
+
𝜇 1
𝑣
=
𝜇 1−𝜇 2
− 𝑅2
Or
−𝜇 2
𝑣1
+
𝜇 1
𝑣
=
𝜇 2−𝜇 1
𝑅2
…ii)
Adding equations i) and ii), we have
𝜇 1
−𝜔
+
𝜇 1
𝑣
= ( 𝜇 −𝜇
1
𝑅1
+
1
𝑅2
1)
2
𝜇1
1
𝑣
−
1
𝑢
= ( 𝜇 −𝜇
1
𝑅1
+
1
𝑅2
1)
2
1
𝑣
−
1
𝑢
=
𝜇 2
𝜇 1
− 1
1
𝑅1
+
1
𝑅2
Take
𝜇 2
𝜇 1
= 𝜇 (refractive index of material of the lens).
1
𝑣
−
1
𝑢
= 𝜇 − 1
1
𝑅1
+
1
𝑅2
∴
When the object on the left of the lens is at ∞, image is formed at the principal focus
Therefore, when 𝑢 = − ∞, 𝑣 = 𝑓 ( = 𝑓𝑜𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑙𝑒𝑛𝑠)

14. 1
𝑓
−
1
−∞
= 𝜇 − 1
1
𝑅1
+
1
𝑅2
This is the Lens – maker’s formula.
13. Draw a neat labelled ray diagram to show how the image of a distant object is formed at least distance of
distinct vision by an astronomical refracting telescope. Obtain an expression for its magnifying power.
Ans. Astronomical telescope is an instrument used to see distant objects clearly. It is mainly used to see distant
objects clearly. It is mainly used to see heavenly bodies and since he heavenly bodies are like a sphere, the inverted
image does not affect the observation.
Construction : It consists of a long cylindrical metallic tube having an achromatic convex lens of large focal length. At
its one end having a large aperture called ‘objective lens’ and at the other end of the tube is a small tube, which can
slide in or out in the bigger tube by a rack and pinion arrangement . This smaller tube, at its outer end, contains and
achromatic convex lens of small focal length and small aperture. This lens is called an ‘eyepiece’. Cross wires are
mounted in the smaller tube at the focus of the eyepiece.
Adjustment and working : The objective forms the real and inverted image of a distant object. The position of the eyepiece is so
adjusted that the final image is formed at the least distance of distinct vision. In case, position of the eyepiece is so adjusted
that the final image is formed at infinity, the telescope is said to be in normal adjustment.
Angular magnification or magnifying power : It is defined as the ratio between the angle subtended by the final image at the eye
and the angle subtended by the object at the eye when the object is in its actual position, i.e.,
𝑀 =
𝐴𝑛𝑔𝑙𝑒 𝑠𝑢𝑏𝑡𝑒𝑛𝑑𝑒𝑑 𝑏𝑦 𝑡𝑕𝑒 𝑓𝑖𝑛𝑎𝑙 𝑖𝑚𝑎𝑔𝑒 𝑎𝑡 𝑡𝑕𝑒 𝑒𝑦𝑒
𝐴𝑛𝑔𝑙𝑒 𝑠𝑢𝑏𝑡𝑒𝑛𝑑𝑒𝑑 𝑏𝑦 𝑡𝑕𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑎𝑡 𝑡𝑕𝑒 𝑒𝑦𝑒 𝑤𝑕𝑒𝑛 𝑡𝑕𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑖𝑠 𝑖𝑛 𝑖𝑡𝑠 𝑎𝑐𝑡𝑢𝑎𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛
Hence, M =
𝛽
𝛼
But angles 𝛽 𝑎𝑛𝑑 𝛼 are very small, we can write
𝛽 = tan 𝛽, 𝑎𝑛𝑑 𝛼 = tan 𝛼,
From the figure, tan 𝛽 =
𝐴 ′ 𝐵′
𝐴′ 𝐸
and tan 𝑎 =
𝐴 ′ 𝐵′
𝑂 𝐴′
M =
𝐴 ′ 𝐵′ / 𝐴′ 𝐸′
𝐴 ′ 𝐵′ / 𝐴′ 𝑂
=
𝑂𝐴′
𝐴′ 𝐸
If 𝑓 𝑖𝑠 𝑡𝑕𝑒
0 focal length of the objective and 𝜇𝑒 is the distance of A ‘B’ from the eyepiece E, then with proper sign, OA’ = +
𝑓 𝑎𝑛𝑑 𝐴′
𝐸 = − 𝜇 ; 𝑠𝑜 𝑤𝑒 𝑕𝑎𝑣𝑒 𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛,
𝑒
0
𝑀 = −
𝑓0
𝑢 𝑒
If the distance of the final image A’’B’’ from the eyepiece be D, then applying the Lens formula,
1
𝑣
−
1
𝑢
=
1
𝑓
For the eyepiece, we shall have
𝑣 = −𝐷, 𝑢 = −𝑢 𝑎𝑛𝑑 𝑓 = + 𝑓𝑒
𝑒
Where, 𝑓𝑒 is the focal length of the eyepiece.
−
1
𝐷
=
1
−𝑢𝑒
=
1
𝑓𝑒
Or
1
𝑢𝑒
=
1
𝑓𝑒
+
1
𝐷
=
1
𝑓𝑒
1 +
𝑓𝑒
𝐷
Thus, the magnifying power of the astronomical telescope is,
𝑀 = −
𝑓0
𝑓𝑒
1 +
𝑓𝑒
𝐷