NCERT CLASS 12 PHYSICS

CHAPTER 9 – RAY OPTICS AND OPTICAL INSTRUMENTS

PART 1: IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS

1. What is ray optics?

Ray optics is the branch of optics that describes the propagation of light in terms of straight-line rays.

It is also known as geometrical optics.

It is used to study reflection, refraction and image formation by mirrors, lenses and optical instruments.

2. What is reflection of light?

Reflection of light is the phenomenon in which light returns to the same medium after striking a surface.

Examples include the formation of images in plane mirrors and spherical mirrors.

3. State the laws of reflection.

There are two laws of reflection:

The angle of incidence is equal to the angle of reflection.
∠i = ∠r
The incident ray, reflected ray and normal at the point of incidence all lie in the same plane.

4. What is a spherical mirror?

A spherical mirror is a mirror whose reflecting surface forms a part of a hollow sphere.

There are two types:

Concave mirror
Convex mirror

5. What is a concave mirror?

A concave mirror is a spherical mirror whose reflecting surface is curved inward.

It is also called a converging mirror because it can converge parallel rays of light at its principal focus.

6. What is a convex mirror?

A convex mirror is a spherical mirror whose reflecting surface is curved outward.

It is also called a diverging mirror because parallel rays of light diverge after reflection.

7. Define the pole of a spherical mirror.

The pole is the geometric centre of the reflecting surface of a spherical mirror.

It is represented by P.

8. What is the centre of curvature?

The centre of curvature is the centre of the sphere of which the spherical mirror forms a part.

It is represented by C.

9. What is the radius of curvature?

The radius of curvature is the radius of the sphere of which the spherical mirror forms a part.

It is represented by R.

The distance between the pole and centre of curvature is the radius of curvature.

10. What is the principal axis of a spherical mirror?

The straight line passing through the pole and centre of curvature of a spherical mirror is called its principal axis.

11. What is the principal focus of a concave mirror?

The principal focus of a concave mirror is the point on its principal axis where rays parallel to the principal axis converge after reflection.

It is represented by F.

12. What is the principal focus of a convex mirror?

The principal focus of a convex mirror is the point on its principal axis from which rays parallel to the principal axis appear to diverge after reflection.

It is a virtual focus.

13. What is the relation between focal length and radius of curvature of a spherical mirror?

For a spherical mirror:

f = R/2

Therefore:

R = 2f

This relation is valid for paraxial rays and mirrors with a small aperture.

14. What is refraction of light?

Refraction is the phenomenon in which light changes its direction when it travels obliquely from one transparent medium to another due to a change in its speed.

15. State the laws of refraction.

There are two laws of refraction:

The incident ray, refracted ray and normal at the point of incidence lie in the same plane.
For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
sin i/sin r = constant

This is known as Snell's law.

16. What is refractive index?

Refractive index is a measure of how much the speed of light changes when it enters a medium from vacuum.

The absolute refractive index is:

n = c/v

Where:

c = Speed of light in vacuum
v = Speed of light in the medium

17. What is optical density?

Optical density describes the relative ability of a medium to slow down light.

A medium with a higher refractive index is optically denser than a medium with a lower refractive index.

Optical density is different from mass density.

18. What is total internal reflection?

Total internal reflection is the phenomenon in which a light ray travelling from an optically denser medium to an optically rarer medium is completely reflected back into the denser medium when the angle of incidence exceeds the critical angle.

19. What are the conditions for total internal reflection?

The following two conditions must be satisfied:

Light must travel from an optically denser medium to an optically rarer medium.
The angle of incidence must be greater than the critical angle of the two media.

20. What is the critical angle?

The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium becomes 90°.

For a denser medium of refractive index n₁ and a rarer medium of refractive index n₂:

sin C = n₂/n₁

For a medium surrounded by air:

sin C = 1/n

21. What is an optical fibre?

An optical fibre is a thin, flexible transparent fibre made of glass or plastic that transmits light through repeated total internal reflection.

It is used in communication, medical instruments and optical devices.

22. What is apparent depth?

Apparent depth is the depth at which an object appears to be located when viewed through a refracting surface.

For an object in a denser medium viewed normally from air:

Refractive index = Real depth/Apparent depth

23. What is lateral displacement?

Lateral displacement is the perpendicular distance between the original direction of an incident ray and the emergent ray when light passes through a parallel-sided glass slab.

The emergent ray is parallel to the incident ray but is laterally displaced.

24. What is a lens?

A lens is a transparent refracting medium bounded by two surfaces, at least one of which is spherical.

Lenses are mainly of two types:

Convex lens
Concave lens

25. What is a convex lens?

A convex lens is thicker at the centre and thinner at the edges.

It is also called a converging lens because it converges parallel rays of light.

26. What is a concave lens?

A concave lens is thinner at the centre and thicker at the edges.

It is also called a diverging lens because it diverges parallel rays of light.

27. What is the optical centre of a lens?

The optical centre is a point on the principal axis of a lens through which a ray of light passes without undergoing any appreciable deviation.

It is represented by O.

28. What is the focal length of a lens?

The focal length of a lens is the distance between its optical centre and its principal focus.

It is represented by f.

29. What is the lens maker's formula?

The lens maker's formula gives the relationship between the focal length of a thin lens, its refractive index and the radii of curvature of its two surfaces.

For a lens in air:

1/f = (n − 1)(1/R₁ − 1/R₂)

Where:

n = Refractive index of lens material
R₁ = Radius of curvature of first surface
R₂ = Radius of curvature of second surface

30. What is the power of a lens?

The power of a lens is the reciprocal of its focal length expressed in metres.

P = 1/f

Its SI unit is dioptre (D).

A convex lens has positive power, while a concave lens has negative power.

31. What is dispersion of light?

Dispersion is the phenomenon in which white light splits into its constituent colours when it passes through a prism.

The colours are:

Violet, Indigo, Blue, Green, Yellow, Orange and Red.

32. What is the angle of deviation in a prism?

The angle of deviation is the angle between the direction of the incident ray produced forward and the emergent ray.

It is represented by δ.

33. What is the angle of minimum deviation?

The angle of minimum deviation is the smallest angle through which a prism deviates a ray of light.

At minimum deviation:

The angle of incidence equals the angle of emergence.
The angles of refraction inside the prism are equal.
The light path inside the prism is symmetrical.

34. What is a simple microscope?

A simple microscope is an optical instrument consisting of a single convex lens of short focal length.

It is used to obtain a magnified virtual and erect image of a small object.

35. What is a compound microscope?

A compound microscope is an optical instrument consisting of two convex lenses:

Objective lens
Eyepiece lens

The objective forms a real, inverted and magnified intermediate image. The eyepiece acts as a simple microscope and produces the final magnified virtual image.

36. What is an astronomical telescope?

An astronomical telescope is an optical instrument used to observe distant objects such as stars and planets.

It consists of:

An objective lens of large focal length and large aperture.
An eyepiece lens of small focal length.

37. What is normal adjustment of an astronomical telescope?

Normal adjustment is the condition in which the final image formed by the telescope is at infinity.

In this condition, the intermediate image formed by the objective lies at the principal focus of the eyepiece.

The separation between the objective and eyepiece is:

L = fₒ + fₑ

38. What is a reflecting telescope?

A reflecting telescope is an astronomical telescope that uses a concave mirror as its objective instead of a convex lens.

It collects light from distant objects and forms their images using reflection.

39. What is magnification?

Magnification is the ratio of the height of the image to the height of the object.

For a mirror:

m = hᵢ/hₒ = −v/u

For a thin lens:

m = hᵢ/hₒ = v/u

40. What is magnifying power?

Magnifying power is the ratio of the angle subtended by the image at the eye to the angle subtended by the object when viewed directly at the least distance of distinct vision.

It is also called angular magnification.

PART 2: NCERT EXERCISE QUESTIONS AND ANSWERS

QUESTION 9.1

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?

Solution:

Given:

Object height, hₒ = 2.5 cm

Object distance, u = −27 cm

Radius of curvature, R = −36 cm

Focal length:

f = R/2 = −36/2 = −18 cm

Using mirror formula:

1/v + 1/u = 1/f

1/v = 1/f − 1/u

1/v = −1/18 + 1/27

1/v = (−3 + 2)/54

1/v = −1/54

v = −54 cm

Therefore, the screen should be placed 54 cm in front of the mirror.

Magnification:

m = −v/u

m = −(−54)/(−27)

m = −2

Image height:

hᵢ = m × hₒ

hᵢ = −2 × 2.5

hᵢ = −5 cm

Final Answer:

Screen distance = 54 cm in front of the mirror.
Image size = 5 cm.
Image is real, inverted and magnified.

When the candle moves closer to the focus, the image moves farther away. Therefore, the screen must also be moved away from the mirror. If the candle moves inside the focal length, the image becomes virtual and cannot be obtained on a screen.

QUESTION 9.2

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 magnification. Describe what happens as the needle is moved farther from the mirror.

Solution:

Given:

Object height, hₒ = 4.5 cm

Object distance, u = −12 cm

Focal length, f = +15 cm

Using mirror formula:

1/v + 1/u = 1/f

1/v = 1/f − 1/u

1/v = 1/15 + 1/12

1/v = (4 + 5)/60

1/v = 9/60

v = 60/9

v = 6.67 cm

The positive sign indicates that the image is behind the mirror.

Magnification:

m = −v/u

m = −6.67/(−12)

m = +0.556

Image height:

hᵢ = 0.556 × 4.5

hᵢ ≈ 2.5 cm

Final Answer:

Image distance = 6.67 cm behind the mirror.
Magnification = +0.556.
Image size ≈ 2.5 cm.
Image is virtual, erect and diminished.

When the needle is moved farther away, the image moves towards the focus and becomes smaller.

QUESTION 9.3

A tank is filled 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?

Solution:

Given:

Real depth = 12.5 cm

Apparent depth = 9.4 cm

Formula:

Refractive index = Real depth/Apparent depth

n = 12.5/9.4

n ≈ 1.33

For the new liquid:

Refractive index = 1.63

Apparent depth = Real depth/Refractive index

Apparent depth = 12.5/1.63

Apparent depth ≈ 7.67 cm

Difference in apparent depths:

Distance = 9.4 − 7.67

Distance = 1.73 cm

Final Answer:

Refractive index of water = 1.33.
The microscope must be moved upwards by approximately 1.73 cm.

QUESTION 9.4

Figures 9.27(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 [Fig. 9.27(c)].

Solution:

From the given figures:

For glass:

Angle of incidence = 60°

Angle of refraction = 35°

Refractive index of glass:

n₉ = sin 60°/sin 35°

n₉ ≈ 0.866/0.574

n₉ ≈ 1.51

For water:

Angle of incidence = 60°

Angle of refraction = 41°

Refractive index of water:

nᵥ = sin 60°/sin 41°

nᵥ ≈ 0.866/0.656

nᵥ ≈ 1.32

At the water-glass interface:

nᵥ sin i = n₉ sin r

1.32 × sin 45° = 1.51 × sin r

sin r = (1.32 × 0.707)/1.51

sin r ≈ 0.618

r ≈ 38°

Final Answer:

Angle of refraction in glass ≈ 38°.

QUESTION 9.5

A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. 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.)

Solution:

Given:

Depth of water, h = 80 cm = 0.80 m

Refractive index, n = 1.33

Light can emerge from the water surface only when its angle of incidence is less than or equal to the critical angle.

For water-air interface:

sin C = 1/n

sin C = 1/1.33

sin C ≈ 0.752

C ≈ 48.8°

The emergent rays form a circular area on the water surface.

Radius of the circle:

r = h tan C

r = 0.80 × tan 48.8°

r ≈ 0.80 × 1.14

r ≈ 0.91 m

Area:

A = πr²

A = 3.14 × (0.91)²

A ≈ 2.6 m²

Final Answer:

Area of the water surface through which light can emerge ≈ 2.6 m².

QUESTION 9.6

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.

Solution:

Given:

Angle of prism, A = 60°

Minimum deviation, δₘ = 40°

Refractive index of water = 1.33

Formula:

n = sin[(A + δₘ)/2]/sin(A/2)

Substituting:

n = sin[(60° + 40°)/2]/sin 30°

n = sin 50°/sin 30°

n = 0.766/0.5

n ≈ 1.53

Now, when the prism is placed in water, its relative refractive index is:

n' = 1.53/1.33

n' ≈ 1.15

Using the prism formula:

1.15 = sin[(60° + δ')/2]/sin 30°

1.15 × 0.5 = sin[(60° + δ')/2]

0.575 = sin[(60° + δ')/2]

(60° + δ')/2 ≈ 35.1°

60° + δ' ≈ 70.2°

δ' ≈ 10.2°

Final Answer:

Refractive index of glass ≈ 1.53.
New angle of minimum deviation in water ≈ 10°.

QUESTION 9.7

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 20 cm?

Solution:

Given:

Refractive index, n = 1.55

Focal length, f = 20 cm

For an equiconvex lens:

R₁ = R

R₂ = −R

Using lens maker's formula:

1/f = (n − 1)(1/R₁ − 1/R₂)

1/f = (1.55 − 1)(1/R + 1/R)

1/f = 0.55 × 2/R

1/20 = 1.1/R

R = 1.1 × 20

R = 22 cm

Final Answer:

Radius of curvature of each surface = 22 cm.

QUESTION 9.8

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

Solution:

The original beam would converge at P, 12 cm beyond the lens.

Therefore, the object distance is:

u = +12 cm

(a) Convex lens

Given:

f = +20 cm

Using lens formula:

1/v − 1/u = 1/f

1/v = 1/f + 1/u

1/v = 1/20 + 1/12

1/v = (3 + 5)/60

1/v = 8/60

v = 60/8

v = 7.5 cm

(b) Concave lens

Given:

f = −16 cm

1/v = 1/f + 1/u

1/v = −1/16 + 1/12

1/v = (−3 + 4)/48

1/v = 1/48

v = 48 cm

Final Answer:

(a) The beam converges 7.5 cm from the lens.

(b) The beam converges 48 cm from the lens.

QUESTION 9.9

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

Solution:

Given:

Object height, hₒ = 3 cm

Object distance, u = −14 cm

Focal length, f = −21 cm

Using lens formula:

1/v − 1/u = 1/f

1/v = 1/f + 1/u

1/v = −1/21 − 1/14

1/v = (−2 − 3)/42

1/v = −5/42

v = −42/5

v = −8.4 cm

Magnification:

m = v/u

m = −8.4/−14

m = 0.6

Image height:

hᵢ = 0.6 × 3

hᵢ = 1.8 cm

Final Answer:

Image distance = 8.4 cm on the object's side.
Image size = 1.8 cm.
Image is virtual, erect and diminished.

When the object is moved farther away, the image moves towards the focus and becomes smaller.

QUESTION 9.10

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

Solution:

Given:

Focal length of convex lens:

f₁ = +30 cm

Focal length of concave lens:

f₂ = −20 cm

For two thin lenses in contact:

1/F = 1/f₁ + 1/f₂

1/F = 1/30 − 1/20

1/F = (2 − 3)/60

1/F = −1/60

F = −60 cm

Final Answer:

Effective focal length = −60 cm.

Since the focal length is negative, the combination behaves as a diverging lens.

QUESTION 9.11

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 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Solution:

Given:

Objective focal length, fₒ = 2 cm

Eyepiece focal length, fₑ = 6.25 cm

Separation between lenses = 15 cm

Least distance of distinct vision, D = 25 cm

(a) Final image at 25 cm

For the eyepiece:

vₑ = −25 cm

fₑ = +6.25 cm

Using lens formula:

1/uₑ = 1/vₑ − 1/fₑ

1/uₑ = −1/25 − 1/6.25

1/uₑ = −0.04 − 0.16

1/uₑ = −0.20

uₑ = −5 cm

The intermediate image formed by the objective is 5 cm from the eyepiece.

Therefore:

vₒ = 15 − 5 = 10 cm

For the objective:

1/uₒ = 1/vₒ − 1/fₒ

1/uₒ = 1/10 − 1/2

1/uₒ = −0.4

uₒ = −2.5 cm

Magnifying power:

M = (vₒ/|uₒ|)(1 + D/fₑ)

M = (10/2.5)(1 + 25/6.25)

M = 4 × 5

M = 20

(b) Final image at infinity

For the final image at infinity, the intermediate image must lie at the focus of the eyepiece.

Therefore:

uₑ = −6.25 cm

Distance of intermediate image from objective:

vₒ = 15 − 6.25

vₒ = 8.75 cm

Using objective lens formula:

1/uₒ = 1/vₒ − 1/fₒ

1/uₒ = 1/8.75 − 1/2

1/uₒ ≈ −0.386

uₒ ≈ −2.59 cm

Magnifying power:

M = (vₒ/|uₒ|)(D/fₑ)

M = (8.75/2.59)(25/6.25)

M ≈ 13.5

Final Answer:

(a) Object distance = 2.5 cm; magnifying power = 20.

(b) Object distance ≈ 2.59 cm; magnifying power ≈ 13.5.

QUESTION 9.12

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 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Solution:

Given:

fₒ = 8 mm = 0.8 cm

fₑ = 2.5 cm

uₒ = −9 mm = −0.9 cm

D = 25 cm

For the objective:

1/vₒ − 1/uₒ = 1/fₒ

1/vₒ = 1/0.8 − 1/0.9

1/vₒ = 1.25 − 1.111

1/vₒ ≈ 0.139

vₒ ≈ 7.2 cm

For the eyepiece, final image is at the near point:

vₑ = −25 cm

1/uₑ = 1/vₑ − 1/fₑ

1/uₑ = −1/25 − 1/2.5

1/uₑ = −0.04 − 0.4

1/uₑ = −0.44

uₑ ≈ −2.27 cm

Separation between lenses:

L = vₒ + |uₑ|

L = 7.2 + 2.27

L ≈ 9.5 cm

Magnifying power:

M = (vₒ/|uₒ|)(1 + D/fₑ)

M = (7.2/0.9)(1 + 25/2.5)

M = 8 × 11

M = 88

Final Answer:

Separation between lenses ≈ 9.5 cm.

Magnifying power = 88.

QUESTION 9.13

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

Solution:

Given:

fₒ = 144 cm

fₑ = 6 cm

For normal adjustment:

Magnifying power:

M = fₒ/fₑ

M = 144/6

M = 24

Separation between lenses:

L = fₒ + fₑ

L = 144 + 6

L = 150 cm

Final Answer:

Magnifying power = 24.

Separation between objective and eyepiece = 150 cm.

QUESTION 9.14

(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification 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 × 10⁶ m, and the radius of lunar orbit is 3.8 × 10⁸ m.

Solution:

Given:

fₒ = 15 m = 1500 cm

fₑ = 1 cm

Diameter of Moon = 3.48 × 10⁶ m

Distance of Moon = 3.8 × 10⁸ m

(a) Angular magnification

M = fₒ/fₑ

M = 1500/1

M = 1500

(b) Diameter of image

Angular diameter of Moon:

α = Diameter of Moon/Distance of Moon

α = (3.48 × 10⁶)/(3.8 × 10⁸)

α ≈ 9.16 × 10⁻³ rad

Image diameter:

d = fₒα

d = 15 × 9.16 × 10⁻³

d ≈ 0.137 m

d = 13.7 cm

Final Answer:

(a) Angular magnification = 1500.

(b) Diameter of image of Moon ≈ 13.7 cm.

QUESTION 9.15

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.

Solution:

Mirror formula:

1/v + 1/u = 1/f

(a) Concave mirror: Object between F and C

Let the magnitude of focal length be F.

For a concave mirror:

f = −F

Let:

u = −x

where:

F < x < 2F

Using mirror formula:

1/v = 1/f − 1/u

1/v = −1/F + 1/x

1/v = (F − x)/Fx

Therefore:

v = Fx/(F − x)

Since x is greater than F, the denominator is negative.

Therefore, v is negative and the image is real.

Also, because F < x < 2F, the magnitude of v is greater than 2F.

Hence, the image forms beyond C.

(b) Convex mirror always produces a virtual image.

For a convex mirror:

f = +F

For a real object:

u = −x

Using mirror formula:

1/v = 1/F + 1/x

The right-hand side is positive.

Therefore, v is positive.

A positive image distance for a mirror indicates that the image is behind the mirror and is virtual.

Hence, a convex mirror always produces a virtual image.

(c) Convex mirror produces a diminished image between P and F.

From the previous result:

1/v = 1/F + 1/x

Therefore, v is positive and smaller than F.

Thus, the image lies between the pole and focus.

Magnification:

m = −v/u

Since v is smaller than x:

m = v/x < 1

Therefore, the image is diminished.

(d) Concave mirror: Object between P and F

For a concave mirror:

f = −F

u = −x, where x < F.

Using mirror formula:

1/v = −1/F + 1/x

Since x is smaller than F, the right-hand side is positive.

Therefore, v is positive, which means the image is virtual.

The magnification is greater than one.

Hence, the image is enlarged.

Final Answer:

All four results are established using the mirror formula.

QUESTION 9.16

A small pin fixed on a table top is viewed from above from a distance of 50 cm. 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?

Solution:

Given:

Thickness of slab, t = 15 cm

Refractive index, n = 1.5

Formula for apparent upward shift:

Shift = t(1 − 1/n)

Substituting:

Shift = 15(1 − 1/1.5)

Shift = 15(1 − 0.667)

Shift = 15 × 0.333

Shift = 5 cm

Final Answer:

The pin appears to be raised by 5 cm.

The shift does not depend on the position of the slab between the pin and the eye, for normal viewing.

QUESTION 9.17

(a) Figure 9.28 shows a cross-section of a 'light pipe' made of a glass fibre 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 reflections inside the pipe take place, as shown in the figure?

(b) What is the answer if there is no outer covering of the pipe?

Solution:

Given:

Refractive index of core:

n₁ = 1.68

Refractive index of cladding:

n₂ = 1.44

(a) With outer covering

Critical angle:

sin C = n₂/n₁

sin C = 1.44/1.68

sin C = 0.857

C ≈ 59°

For total internal reflection:

90° − r ≥ 59°

r ≤ 31°

At the entrance face, using Snell's law:

sin i = 1.68 sin 31°

sin i ≈ 1.68 × 0.515

sin i ≈ 0.865

i ≈ 60°

Therefore, the incident angle must be approximately between 0° and 60° with the axis.

(b) Without outer covering

The fibre is surrounded by air.

Critical angle:

sin C = 1/1.68

sin C ≈ 0.595

C ≈ 36.5°

The corresponding maximum internal angle with the axis is approximately:

r ≤ 90° − 36.5°

r ≤ 53.5°

At the entrance:

sin i = 1.68 sin 53.5°

The right-hand side is greater than one, so every physically possible incident angle at the entrance satisfies the total internal reflection condition inside the fibre.

Final Answer:

(a) Incident angles approximately from 0° to 60° with the axis.

(b) All incident angles from 0° to 90° can undergo total internal reflection inside the fibre.

QUESTION 9.18

The image of a small electric bulb fixed on the wall of a room is to be obtained on the 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?

Solution:

Given:

Distance between object and screen:

D = 3 m

For a convex lens to form a real image on a screen:

D ≥ 4f

The maximum focal length occurs when:

D = 4f

Therefore:

f = D/4

f = 3/4

f = 0.75 m

Final Answer:

Maximum focal length = 0.75 m.

QUESTION 9.19

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.

Solution:

Given:

Distance between object and screen:

D = 90 cm

Distance between two lens positions:

d = 20 cm

Using the lens displacement method:

f = (D² − d²)/4D

Substituting:

f = (90² − 20²)/(4 × 90)

f = (8100 − 400)/360

f = 7700/360

f ≈ 21.4 cm

Final Answer:

Focal length of the convex lens ≈ 21.4 cm.

QUESTION 9.20

(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 is incident? 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 magnification produced by the two-lens system, and the size of the image.

Solution:

Given:

Convex lens focal length = +30 cm

Concave lens focal length = −20 cm

Separation = 8 cm

(a) Parallel light incident first on convex lens

The convex lens would converge parallel rays at 30 cm from itself.

The concave lens is 8 cm away.

Therefore, the converging point is 22 cm beyond the concave lens.

This acts as a virtual object for the concave lens:

u₂ = +22 cm

For the concave lens:

f₂ = −20 cm

Using lens formula:

1/v₂ = 1/f₂ + 1/u₂

1/v₂ = −1/20 + 1/22

1/v₂ = (−22 + 20)/440

1/v₂ = −1/220

v₂ = −220 cm

The final beam diverges as if it came from a point 220 cm before the concave lens.

Parallel light incident first on concave lens

The concave lens causes the parallel beam to diverge as if it came from its focus, 20 cm before the lens.

Since the convex lens is 8 cm away, this point is 28 cm before the convex lens.

For the convex lens:

u₁ = −28 cm

f₁ = +30 cm

1/v₁ = 1/f₁ + 1/u₁

1/v₁ = 1/30 − 1/28

1/v₁ = (28 − 30)/840

1/v₁ = −1/420

v₁ = −420 cm

The final beam diverges as if it came from a point 420 cm before the convex lens.

Therefore, the result depends on the direction from which parallel light enters the system.

A single effective focal length is not useful for this separated, non-symmetric combination.

(b) Calculate magnification

Object distance from convex lens:

u₁ = −40 cm

For the convex lens:

1/v₁ = 1/30 − 1/40

1/v₁ = 1/120

v₁ = 120 cm

Magnification by first lens:

m₁ = v₁/u₁

m₁ = 120/−40

m₁ = −3

The first image is 120 cm beyond the convex lens.

Distance from concave lens:

120 − 8 = 112 cm

This acts as a virtual object for the concave lens:

u₂ = +112 cm

For the concave lens:

1/v₂ = −1/20 + 1/112

1/v₂ = (−112 + 20)/2240

1/v₂ = −92/2240

v₂ ≈ −24.35 cm

Magnification by second lens:

m₂ = v₂/u₂

m₂ = −24.35/112

m₂ ≈ −0.217

Total magnification:

m = m₁ × m₂

m = (−3)(−0.217)

m ≈ +0.65

Image height:

hᵢ = m × hₒ

hᵢ = 0.65 × 1.5

hᵢ ≈ 0.98 cm

Final Answer:

(a) The apparent focal distance depends on the direction of incident light: 216 cm from the system centre in one direction and 416 cm in the other. A single effective focal length is not useful here.

(b) Magnification ≈ +0.65.

Image size ≈ 0.98 cm. The final image is erect relative to the original object.

QUESTION 9.21

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 reflection at the other face? The refractive index of the material of the prism is 1.524.

Solution:

Given:

Angle of prism:

A = 60°

Refractive index:

n = 1.524

For just total internal reflection, the angle of incidence at the second face must equal the critical angle.

sin C = 1/n

sin C = 1/1.524

sin C ≈ 0.656

C ≈ 41°

For a prism:

r₁ + r₂ = A

r₂ = 41°

r₁ = 60° − 41°

r₁ = 19°

Using Snell's law at the first face:

sin i = n sin r₁

sin i = 1.524 × sin 19°

sin i ≈ 1.524 × 0.326

sin i ≈ 0.496

i ≈ 29.8°

Final Answer:

Angle of incidence ≈ 29.8°.

QUESTION 9.22

A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

Solution:

Given:

Object distance:

u = −9 cm

Focal length:

f = +9 cm

Near point:

D = 25 cm

Object area = 1 mm²

(a) Linear magnification and image area

Since:

u = −f

The object is at the principal focus of the lens.

The image is formed at infinity.

Therefore, the linear magnification tends towards infinity, and the ideal geometrical image area also tends towards infinity.

(b) Magnifying power

For an image at infinity:

M = D/f

M = 25/9

M ≈ 2.78

(c) Are magnification and magnifying power equal?

No.

Linear magnification compares image size with object size.

Magnifying power compares the angle subtended by the image with the angle subtended by the object at the eye.

They represent different physical quantities.

Final Answer:

(a) Image is at infinity; ideal linear magnification and image area tend towards infinity.

(b) Magnifying power ≈ 2.78.

QUESTION 9.23

(a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power?

(b) What is the magnification in this case?

Solution:

Given:

f = 9 cm

Near point:

D = 25 cm

For maximum magnifying power, the final image should be at the near point.

Therefore:

v = −25 cm

Using lens formula:

1/v − 1/u = 1/f

1/u = 1/v − 1/f

1/u = −1/25 − 1/9

1/u = (−9 − 25)/225

1/u = −34/225

u = −225/34

u ≈ −6.62 cm

Therefore, the lens should be held approximately 6.6 cm from the card.

Magnification:

m = v/u

m = −25/−6.62

m ≈ 3.78

Magnifying power:

M = 1 + D/f

M = 1 + 25/9

M ≈ 3.78

Final Answer:

(a) Distance of lens from card ≈ 6.6 cm.

(b) Magnification ≈ 3.8.

QUESTION 9.24

What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm²? Would you be able to see the squares distinctly with your eyes very close to the magnifier?

Solution:

Given:

Object area = 1 mm²

Image area = 6.25 mm²

Focal length:

f = 9 cm

The ratio of image area to object area is:

6.25/1 = 6.25

Linear magnification:

m = √6.25

m = 2.5

For a virtual erect image:

m = v/u

Therefore:

v = 2.5u

Using lens formula:

1/v − 1/u = 1/f

1/(2.5u) − 1/u = 1/9

(0.4 − 1)/u = 1/9

−0.6/u = 1/9

u = −5.4 cm

Therefore:

v = 2.5 × (−5.4)

v = −13.5 cm

The virtual image is only 13.5 cm from the lens.

Since the near point of a normal eye is 25 cm, the image is too close to be seen distinctly with the eye very close to the magnifier.

Final Answer:

Object distance from lens = 5.4 cm.

No, the squares cannot be seen distinctly in this arrangement because the virtual image is only 13.5 cm away, which is less than the near point of 25 cm.

QUESTION 9.25

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 magnification?

(b) In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification 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 that short distance between the eye and eyepiece be?

Solution:

(a) How does a magnifying glass provide angular magnification?

Without a magnifying glass, the eye cannot see an object clearly when it is placed closer than the least distance of distinct vision, normally 25 cm.

A magnifying glass allows the object to be placed closer to the eye while still producing a clear virtual image.

Therefore, the angle subtended by the object at the eye is increased compared with direct viewing at the near point.

This produces angular magnification.

(b) Does angular magnification change when the eye is moved away?

Yes.

When the eye is moved farther from the magnifying glass, the angle subtended by the image at the eye generally decreases.

Therefore, angular magnification decreases slightly.

(c) Why cannot we use lenses of extremely small focal length?

Lenses of very small focal length are difficult to manufacture accurately.

They also suffer from spherical and chromatic aberrations.

The object must be placed extremely close to the lens, making observation difficult.

Therefore, practical limitations prevent unlimited increase in magnifying power.

(d) Why must both lenses of a compound microscope have short focal lengths?

The approximate magnifying power of a compound microscope is:

M = (L/fₒ)(D/fₑ)

Where:

L = Tube length
fₒ = Focal length of objective
fₑ = Focal length of eyepiece
D = Least distance of distinct vision

A larger magnifying power requires smaller focal lengths of both lenses.

(e) Why should the eye be placed slightly away from the eyepiece?

The eye should be placed near the eye-ring, which is the image of the objective aperture formed by the eyepiece.

This position allows the eye to receive more of the light emerging from the microscope and provides a wider field of view.

The eye should be placed close to the eye-ring, a short distance beyond the eyepiece. The exact distance depends on the microscope's design and lens separation.

QUESTION 9.26

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

Solution:

Given:

Required magnifying power:

M = 30

Objective focal length:

fₒ = 1.25 cm

Eyepiece focal length:

fₑ = 5 cm

Near point:

D = 25 cm

For maximum magnifying power, take the final image at the near point.

Magnification by eyepiece:

Mₑ = 1 + D/fₑ

Mₑ = 1 + 25/5

Mₑ = 6

Required objective magnification:

Mₒ = 30/6

Mₒ = 5

Let the object distance from objective be x.

Then:

uₒ = −x

vₒ = 5x

Using lens formula:

1/vₒ − 1/uₒ = 1/fₒ

1/5x + 1/x = 1/1.25

6/5x = 0.8

x = 6/(5 × 0.8)

x = 1.5 cm

Therefore:

uₒ = −1.5 cm

vₒ = 7.5 cm

For the eyepiece:

vₑ = −25 cm

1/uₑ = 1/vₑ − 1/fₑ

1/uₑ = −1/25 − 1/5

1/uₑ = −6/25

uₑ ≈ −4.17 cm

Separation between lenses:

L = vₒ + |uₑ|

L = 7.5 + 4.17

L ≈ 11.67 cm

Final Answer:

Object should be placed 1.5 cm in front of the objective.
Separation between objective and eyepiece ≈ 11.7 cm.
The final image is formed at the near point and the magnifying power is 30.

QUESTION 9.27

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when:

(a) The telescope is in normal adjustment (i.e., when the final image is at infinity).

(b) The final image is formed at the least distance of distinct vision (25 cm).

Solution:

Given:

fₒ = 140 cm

fₑ = 5 cm

D = 25 cm

(a) Final image at infinity

Magnifying power:

M = fₒ/fₑ

M = 140/5

M = 28

(b) Final image at near point

Magnifying power:

M = (fₒ/fₑ)(1 + fₑ/D)

M = (140/5)(1 + 5/25)

M = 28 × 1.2

M = 33.6

Final Answer:

(a) Magnifying power = 28.

(b) Magnifying power = 33.6.

QUESTION 9.28

(a) For the telescope described in Exercise 9.27 (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?

Solution:

Given:

fₒ = 140 cm

fₑ = 5 cm

Height of tower = 100 m

Distance of tower = 3000 m

(a) Separation between lenses

For normal adjustment:

L = fₒ + fₑ

L = 140 + 5

L = 145 cm

(b) Height of image formed by objective

For a distant object:

Angular size:

α = h/d

α = 100/3000

α = 1/30 rad

Image height:

hᵢ = fₒ × α

hᵢ = 140 × 1/30

hᵢ ≈ 4.67 cm

(c) Height of final image at 25 cm

Magnifying power of eyepiece:

Mₑ = 1 + D/fₑ

Mₑ = 1 + 25/5

Mₑ = 6

Final image height:

h = 6 × 4.67

h ≈ 28 cm

Final Answer:

(a) Separation = 145 cm.

(b) Intermediate image height ≈ 4.67 cm.

QUESTION 9.29

A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Solution:

Given:

Radius of curvature of large mirror:

R₁ = 220 mm

Radius of curvature of small mirror:

R₂ = 140 mm

Distance between mirrors = 20 mm

Focal length of large concave mirror:

f₁ = R₁/2

f₁ = 220/2

f₁ = 110 mm

Focal length of small convex mirror:

f₂ = R₂/2

f₂ = 140/2

f₂ = 70 mm

An object at infinity forms an image at the focus of the large mirror, 110 mm from it.

Since the small mirror is 20 mm away, this point is:

110 − 20 = 90 mm

behind the small mirror.

Thus, for the small convex mirror:

u = +90 mm

f = +70 mm

Using mirror formula:

1/v + 1/u = 1/f

1/v = 1/f − 1/u

1/v = 1/70 − 1/90

1/v = (90 − 70)/6300

1/v = 20/6300

v = 315 mm

Final Answer:

The final image is formed 315 mm from the small mirror, on the side of the large mirror. It is approximately 295 mm behind the large mirror.

QUESTION 9.30

Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.29. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Solution:

Given:

Deflection of mirror:

θ = 3.5°

Distance of screen:

L = 1.5 m

When a plane mirror rotates through an angle θ, the reflected ray rotates through an angle 2θ.

Therefore:

Angle of rotation of reflected ray:

2θ = 2 × 3.5°

= 7°

Displacement of light spot:

d = L tan 2θ

d = 1.5 × tan 7°

d = 1.5 × 0.123

d ≈ 0.1845 m

Converting into centimetres:

d ≈ 18.45 cm

Final Answer:

Displacement of the light spot ≈ 18.4 cm.

QUESTION 9.31

Figure 9.30 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 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Solution:

Given:

Refractive index of glass:

n = 1.50

Focal length without liquid:

f₁ = 30 cm

Focal length with liquid:

F = 45 cm

For an equiconvex lens:

R₁ = R

R₂ = −R

Using lens maker's formula:

1/f₁ = (n − 1)(1/R₁ − 1/R₂)

1/30 = (1.50 − 1)(1/R + 1/R)

1/30 = 0.5 × 2/R

1/30 = 1/R

Therefore:

R = 30 cm

When the liquid is present, it forms a plano-concave lens in contact with the glass lens.

For the liquid lens:

1/f₂ = (μ − 1)(1/R₁ − 1/R₂)

Taking the curved surface radius as −30 cm and the plane surface radius as infinity:

1/f₂ = (μ − 1)(−1/30 − 0)

1/f₂ = −(μ − 1)/30

For two thin lenses in contact:

1/F = 1/f₁ + 1/f₂

1/45 = 1/30 − (μ − 1)/30

Multiplying by 30:

30/45 = 1 − (μ − 1)

2/3 = 2 − μ

μ = 2 − 2/3

μ = 1.33

Final Answer:

Refractive index of the liquid ≈ 1.33.

PART 3: IMPORTANT DERIVATIONS

1. Derive the mirror formula.

The mirror formula is:

1/v + 1/u = 1/f

Where:

u = Object distance
v = Image distance
f = Focal length

For a spherical mirror, consider a paraxial ray striking the mirror at a small height h above the principal axis.

Using the geometry of the reflected ray and the small-angle approximation:

tan α ≈ α

For small angles, the relevant triangles give:

α ≈ h/u

β ≈ h/v

γ ≈ h/f

From the law of reflection and the geometry of the diagram:

α + β = 2γ

Therefore:

h/u + h/v = 2h/f

Dividing by h:

1/u + 1/v = 2/f

For a spherical mirror:

R = 2f

Thus the standard mirror equation is:

1/v + 1/u = 1/f

2. Derive the relation between focal length and radius of curvature of a spherical mirror.

For a spherical mirror, consider a paraxial ray parallel to the principal axis striking the mirror at a small height.

Let:

P = Pole
C = Centre of curvature
F = Principal focus
R = Radius of curvature
f = Focal length

For paraxial rays, the principal focus lies approximately midway between the pole and centre of curvature.

Therefore:

PF = PC/2

Hence:

f = R/2

3. Derive the refraction formula at a spherical surface.

For refraction at a spherical surface separating two media of refractive indices n₁ and n₂:

n₂/v − n₁/u = (n₂ − n₁)/R

Where:

u = Object distance
v = Image distance
R = Radius of curvature of the spherical surface

This formula is obtained by applying Snell's law to a paraxial ray and using the small-angle approximation.

4. Derive the lens maker's formula.

For a thin lens of refractive index n placed in air:

For the first spherical surface:

n/v₁ − 1/u = (n − 1)/R₁

For the second spherical surface:

1/v − n/u₂ = (1 − n)/R₂

The image formed by the first surface acts as the object for the second surface.

For a thin lens:

u₂ ≈ −v₁

Combining the two equations:

1/v − 1/u = (n − 1)(1/R₁ − 1/R₂)

Using the thin lens formula:

1/v − 1/u = 1/f

Therefore:

1/f = (n − 1)(1/R₁ − 1/R₂)

This is the lens maker's formula.

5. Derive the formula for the equivalent focal length of two thin lenses in contact.

For the first lens:

1/v₁ − 1/u = 1/f₁

For the second lens:

1/v − 1/u₂ = 1/f₂

Since the lenses are in contact:

u₂ ≈ −v₁

Adding the equations:

1/v − 1/u = 1/f₁ + 1/f₂

For the equivalent lens:

1/v − 1/u = 1/F

Therefore:

1/F = 1/f₁ + 1/f₂

For two lenses in contact, their powers also add:

P = P₁ + P₂

6. Derive the formula for the refractive index of a prism at minimum deviation.

For a prism:

A = r₁ + r₂

The angle of deviation is:

δ = i + e − A

At minimum deviation:

i = e

r₁ = r₂

Therefore:

r₁ = r₂ = A/2

Also:

i = (A + δₘ)/2

Using Snell's law:

n = sin i/sin r

Substituting:

n = sin[(A + δₘ)/2]/sin(A/2)

This is the formula for the refractive index of a prism at minimum deviation.

7. Derive the magnifying power of a simple microscope.

Magnifying power is:

M = β/α

Where:

β = Angle subtended by the image at the eye
α = Angle subtended by the object when viewed directly at the near point

For an image at infinity:

The object is placed at the focal point of the lens.

Using the small-angle approximation:

α ≈ h/D

β ≈ h/f

Therefore:

M = β/α

M = (h/f)/(h/D)

M = D/f

For the final image at the near point:

M = 1 + D/f

8. Derive the magnifying power of a compound microscope.

A compound microscope consists of an objective and an eyepiece.

Total magnifying power:

M = mₒ × Mₑ

Where:

mₒ = Linear magnification of objective
Mₑ = Angular magnification of eyepiece

For the objective:

mₒ ≈ L/fₒ

For the eyepiece, when the final image is at infinity:

Mₑ = D/fₑ

Therefore:

M = (L/fₒ)(D/fₑ)

For the final image at the near point:

M = (L/fₒ)(1 + D/fₑ)

9. Derive the magnifying power of an astronomical telescope.

For normal adjustment:

The objective forms an image at its focal plane.

The eyepiece is adjusted so that this image lies at its first focal plane.

Let:

fₒ = Focal length of objective
fₑ = Focal length of eyepiece

For small angles:

α ≈ h/fₒ

β ≈ h/fₑ

Magnifying power:

M = β/α

M = (h/fₑ)/(h/fₒ)

M = fₒ/fₑ

The magnitude of angular magnification is fₒ/fₑ. The image is inverted, so the signed angular magnification is negative under the usual convention.

PART 4: IMPORTANT NUMERICAL QUESTIONS

1. A concave mirror has a focal length of 15 cm. Find its radius of curvature.

Given:

f = −15 cm

Formula:

R = 2f

R = 2 × (−15)

R = −30 cm

2. An object is placed 30 cm in front of a concave mirror of focal length 20 cm. Find the image distance.

Given:

u = −30 cm

f = −20 cm

Mirror formula:

1/v + 1/u = 1/f

1/v = −1/20 + 1/30

1/v = −1/60

v = −60 cm

3. A convex lens has a focal length of 25 cm. Calculate its power.

Given:

f = 25 cm = 0.25 m

Formula:

P = 1/f

P = 1/0.25

P = +4 D

4. A concave lens has a focal length of 50 cm. Calculate its power.

Given:

f = −50 cm = −0.50 m

P = 1/f

P = 1/(−0.50)

P = −2 D

5. The refractive index of glass is 1.5. Calculate its critical angle when surrounded by air.

Given:

n = 1.5

Formula:

sin C = 1/n

sin C = 1/1.5

sin C = 0.667

C ≈ 41.8°

6. A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 40 cm. Calculate the equivalent focal length.

Given:

f₁ = +20 cm

f₂ = −40 cm

Formula:

1/F = 1/f₁ + 1/f₂

1/F = 1/20 − 1/40

1/F = 1/40

F = +40 cm

7. A prism has an angle of 60° and a minimum deviation of 30°. Calculate its refractive index.

Given:

A = 60°

δₘ = 30°

Formula:

n = sin[(A + δₘ)/2]/sin(A/2)

n = sin 45°/sin 30°

n = 0.707/0.5

n ≈ 1.414

PART 5: IMPORTANT FORMULA SHEET

Spherical Mirrors

Mirror formula:
1/v + 1/u = 1/f
Radius of curvature:
R = 2f
Magnification:
m = −v/u = hᵢ/hₒ

Refraction

Snell's law:
n₁ sin i = n₂ sin r
Absolute refractive index:
n = c/v
Relative refractive index:
n₂₁ = n₂/n₁
Apparent depth:
n = Real depth/Apparent depth
Critical angle:
sin C = n₂/n₁
For a medium in air:
sin C = 1/n

Lenses

Lens formula:
1/v − 1/u = 1/f
Lens maker's formula:
1/f = (n − 1)(1/R₁ − 1/R₂)
Magnification:
m = v/u
Power of lens:
P = 1/f (f in metres)
Equivalent focal length of lenses in contact:
1/F = 1/f₁ + 1/f₂
Equivalent power:
P = P₁ + P₂

Prism

Angle of prism:
A = r₁ + r₂
Angle of deviation:
δ = i + e − A
Refractive index at minimum deviation:
n = sin[(A + δₘ)/2]/sin(A/2)

Optical Instruments

Simple microscope, final image at infinity:
M = D/f
Simple microscope, final image at near point:
M = 1 + D/f
Compound microscope, final image at infinity:
M = (L/fₒ)(D/fₑ)
Compound microscope, final image at near point:
M = (L/fₒ)(1 + D/fₑ)
Astronomical telescope, normal adjustment:
M = fₒ/fₑ
Telescope length in normal adjustment:
L = fₒ + fₑ

PART 6: IMPORTANT SI UNITS

Physical Quantity	SI Unit
Focal length	Metre (m)
Radius of curvature	Metre (m)
Object distance	Metre (m)
Image distance	Metre (m)
Refractive index	No unit
Power of lens	Dioptre (D)
Angle of incidence	Degree or radian
Angle of refraction	Degree or radian
Angle of deviation	Degree or radian
Magnification	No unit
Magnifying power	No unit
Lateral displacement	Metre (m)
Wavelength	Metre (m)

PART 7: QUICK REVISION POINTS

Ray optics studies the propagation of light using rays.
The angle of incidence is equal to the angle of reflection.
A concave mirror is a converging mirror.
A convex mirror is a diverging mirror.
The focal length of a spherical mirror is half its radius of curvature.
Refraction occurs due to a change in the speed of light between media.
Snell's law describes the relationship between angles of incidence and refraction.
Refractive index is the ratio of the speed of light in vacuum to its speed in the medium.
Total internal reflection occurs when light travels from an optically denser medium to a rarer medium and the angle of incidence exceeds the critical angle.
Optical fibres work on the principle of total internal reflection.
A convex lens is a converging lens.
A concave lens is a diverging lens.
The power of a lens is the reciprocal of its focal length in metres.
The SI unit of power of a lens is dioptre.
A prism can disperse white light into its constituent colours.
At minimum deviation, the path of light through a prism is symmetrical.
A simple microscope uses a single convex lens.
A compound microscope uses an objective and an eyepiece.
An astronomical telescope is used to observe distant objects.
In normal adjustment of a telescope, the final image is formed at infinity.
The objective of an astronomical telescope has a large focal length.
The eyepiece of an astronomical telescope has a small focal length.
A reflecting telescope uses a concave mirror as its objective.
Magnification is the ratio of image height to object height.
Magnifying power is the ratio of the angular size of the image to the angular size of the object.