Selina Solutions Concise Physics Class 10 Chapter 4 Refraction of Light at Plane Surfaces

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Exercise 4(A) page no: 80

Question: 1

What do you understand by the refraction of light?

Solution:

The change in the direction of the path of light, when it passes from one transparent medium to another transparent medium, is called refraction of light.

Question: 2

Draw diagrams to show the refraction of light from (i) air to glass, (ii) glass to air. In each diagram, label the incident ray, refracted ray, the angle of incidence (i) and the angle of refraction (r).

Solution:

(i) Refraction of light from air to glass

(ii) Refraction of light from glass to air

Question: 3

A ray of light is incident normally on a plane glass slab. What will be (i) the angle of refraction and (ii) the angle of deviation for the ray?

Solution:

The ray of light incident normally on a plane glass slab passes undeviated, that means such rays suffers no bending at the surface because the angle of incidence is 0⁰. Hence, the angle of incidence ∠i = 0⁰, then the angle of refraction ∠r = 0⁰. Also, the angle of deviation of the ray will be 0⁰.

Question: 4

An obliquely incident light ray bends at the surface due to change in speed, when passing from one medium to other. The ray does not bend when it is incident normally. Will the ray have different speed in the other medium?

Solution:

Yes, the ray has different speed in the other medium because when the medium changes, the speed of light changes.

Question: 5

What is the cause of refraction of light when it passes from one medium to another?

Solution:

Change in direction of the path of light in other medium or the refraction of light occurs because light travels with different speeds in different media. When a ray of light passes from one medium to another medium, its direction or path (except ∠i = 0⁰ changes because of the change in speed of light.

Question: 6

A light ray suffers reflection and refraction at the boundary in passing from air to water. Draw a neat labelled diagram to show it.

Solution:

Air is a rarer medium while water is denser medium than air with a refractive index of 1: 33. Therefore when a ray of light travels from air to water (i.e rarer to denser) it bends towards the normal.

Question: 7

A ray of light passes from medium 1 to medium 2. Which of the following quantities of the refracted ray will differ from that of incident ray: speed, intensity, frequency and wavelength?

Solution:

Speed, intensity and wavelength are the quantities in which refracted ray will differ from that of the incident ray.

Question: 8

State the Snell’s laws of refraction of light.

Solution:

The Snell’s laws of refraction of light are

(i) The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.

(ii) The ratio of the sine of the angle of incidence i to the sine of the angle of refraction r is constant for the pair of given media.

Sin i / sin r = constant ₁μ₂

where constant ₁μ₂ is called the refractive index of the second medium with respect to the first medium

Question: 9

Define the term refractive index of a medium. Can it be less than 1?

Solution:

The refractive index of the second medium with respect to the first medium is defined as the ratio of the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium.

The refractive index of a transparent medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum.

Question: 10

(a) Compare the speeds of light of wavelength 4000 
 (i.e. violet light) and 8000 
 (i.e. red light) in vacuum.

(b) How is the refractive index of a medium related to the speed of light in it and in vacuum or air?

Solution:

(a) Speed of light depends on the medium, not on a wavelength so, speed will be the same for both.

(b) Denser medium has a higher refractive index and hence the speed of light in such medium is lower in comparison to the speed of light in a medium which has a lower refractive index.

Question: 11

A light ray passes from water to (i) air, and (ii) glass. In each case, state how does the speed of light change?

Solution:

Refractive index of water, μ_w = 1.33

Refractive index of air, μ_a = 1.0003

Refractive index of glass, μ_g = 1.5

This shows that μ_{a <} μ_{w <} μ_g

The speed of light decreases when it enters from a rarer medium to denser medium and increases when it enters from a denser medium to rarer medium

Hence, the speed of light increases when a light ray passes from water to air and the speed of light decreases when the light ray passes from water to glass.

Question: 12

A light ray in passing from water to a medium (a) speeds up (b) slows down. In each case, (i) give one example of the medium, (ii) State whether the refractive index of a medium is equal to, less than or greater than the refractive index of water.

Solution:

(a) A light ray in passing from water to air speeds up

(b) A light ray in passing from water to a glass medium slows down

(c) It means that the refractive index of the medium is less than that of water when the speed increases.

(d) Similarly, the refractive index of the medium is greater than that of water when the speed decreases.

Question: 13

What do you understand by the statement ‘the refractive index of the glass is 1.5 for white light’?

Solution:

This statement indicates that white light travels in air 1.5 times faster than in glass.

Question: 14

A monochromatic ray of light passes from air to glass. The wavelength of light in air is λ, the speed of light in air is c and in glass is v. If the refractive index of glass is 1.5, write down (a) the relationship between c and v, (b) the wavelength of light in the glass.

Solution:

(a) The relation between the speed of light in air c and in glass v is given by

c / v = μ

or c = 1.5 v

(b) The wavelength of light in glass (λ¹)

μ = λ / λ¹

= λ / 1.5

Question: 15

In an experiment of finding the refractive index of glass, if blue light is replaced by the red light, how will the refractive index of the glass change? Give reason in support of your answer.

Solution:

In glass, the speed of red light is more than that of blue light. The refractive index is c / v. Therefore the refractive index of glass for red light will decrease as compared to blue light.

Question: 16

(a)For which colour of white light, is the refractive index of a transparent medium (i) the least (ii) the most?

(b) Which colour of light travels fastest in any medium except air?

Solution:

(a)

(i) The refractive index of a transparent medium is least for red colour

(ii) The refractive index of a transparent medium is most for violet colour

(b) The wavelength of red colour is the highest. Hence, red colour travels fastest in any medium except air.

Question: 17

Name two factors on which the refractive index of a medium depends? State how does it depends on the factor state by you.

Solution:

The factors on which the refractive index of a medium depends are as follows

(i) Nature of the medium (e.g. μ_g = 1.5, μ_w = 1.33): Less the speed of light in the medium as compared to that in air, more is the refractive index of the medium

(ii) Physical condition such as temperature: With an increase in temperature, the speed of light in medium increases, so the refractive index of the medium decreases.

Question: 18

How does the refractive index of a medium depend on the wavelength of light used?

Solution:

Refractive index of a medium decreases with an increase in wavelength of light.

Refractive index of a medium for violet light which has the least wavelength is greater than that for red light which has the greatest wavelength.

Question: 19

How does the refractive index of a medium depend on its temperature?

Solution:

Refractive index decreases with the increase in temperature of the medium.

The speed of light in that medium increases with increase in temperature. Hence, the refractive index = [velocity of light in vacuum / velocity of light in medium] decreases

Question: 20

Light of a single colour is passed through a liquid having a piece of glass suspended in it. On changing the temperature of the liquid, at a particular temperature, the glass piece is not seen.

(a) When is the glass piece not seen?

(b) Why is the light of a single colour used?

Solution:

(a) The glass piece is not visible when the refractive index of the liquid becomes equal to the refractive index of glass.

(b) The refractive index of a medium (glass or liquid) is different for the light of different colours. Hence, the light of a single colour is used.

Question: 21

In the figure below, a ray of light A incident from air suffers partial reflection and refraction at the boundary of water.

(a) Complete the diagram showing (i) the reflected ray B and (ii) the refracted ray C.

(b) How are the angles of incidence i and refraction r related?

Solution:

By Snell’s law, the angle of incidence and refraction are related to each other.

sin i / sin r = μ_w

Question: 22

The diagram alongside shows the refraction of a ray of light from air to liquid.

(a) Write the values of (i) angle of incidence, and (ii) angle of refraction.

(b) Use Snell’s law to find the refractive index of liquid with respect to air.

Solution:

(a)

(i) The angle of incidence is the angle which the incident ray makes with the normal.

Hence, ∠i = 90⁰ – 30⁰ = 60⁰

(ii) Angle of refraction is the angle which the refracted ray makes with the normal

Hence, ∠r = 90⁰ – 45⁰ = 45⁰

(b) According to Snell’s law

_air μ_liquid = sin i / sin r

= sin 60⁰ / sin 45⁰

_air μ_liquid = (√3 / 2) / (1 / √2)

= (√3/2)

= 1.22

Question: 23

The refractive index of water with respect to air is _aμ_w and of glass with respect to air is _aμ_g. Express the refractive index of glass with respect to water.

Solution:

The refractive index of glass with respect to water is given by

_w μ_g = _aμ_g / _aμ_w

Question: 24

What is lateral displacement? Draw a ray diagram showing the lateral displacement of a ray of light when it passes through a parallel-sided glass slab.

Solution:

The lateral displacement is the perpendicular distance between the path of emergent ray and the direction of the incident ray

Here, XY is the perpendicular distance or the lateral displacement in the above figure

Question: 25

A ray of light strikes the surface at a rectangular glass slab such that the angle of incidence is (i) 0^o, (ii) 45^o. In each case, draw a diagram to show the path taken by the ray as it passes through the glass slab and emerges from it.

Solution:

(i) When the angle of incidence is 0⁰

(ii) When the angle of incidence is 45⁰

Question: 26

In the adjacent diagram, AO is a ray of light incident on a rectangular glass slab.

(a) Complete the path of the ray until it emerges out of the slab.

(b) In the ray diagram, mark the angle of incidence (i) and the angle of refraction (r) at the first interface. How is the refractive index of glass related to angles i and r?

(c) Mark angle of emergence by the letter e. How are the angles i and e related?

(d) Which two rays are parallel to each other? Name them.

(e) Indicate in the diagram the lateral displacement between the emergent ray and the incident ray. State one factor that affects the lateral displacement.

Solution:

(a) The complete path of incident ray in glass block is given below

(b) In part A, the angle of incidence (i) and the angle of refraction (r) are marked.

Refractive index of glass is related to the angles i and r as

sin i / sin r = μ

(c) The angle of emergence (e) is marked in part (a)

The two angles i and e are related by the relation

∠i = ∠e

(d) The two rays which are parallel to each other are incident ray and emergent ray.

(e) XY is the lateral displacement between the incident ray and the emergent ray in the above figure.

More the thickness of the medium, more is the lateral displacement

Question: 27

A ray of green light enters a liquid from air, as shown in the figure. The angle 1 is 45^o and angle 2 is 30^o.

(a) Find the refractive index of liquid.
(b) Show in the diagram the path of the ray after it strikes the mirror and re-enters in air. Mark in the diagram the angles wherever necessary.
(c) Redraw the diagram if plane mirror becomes normal to the refracted ray inside the liquid. State the principle used.

Solution:

(a) Refractive index of the liquid is given by

_a μ_l = sin i / sin r

_a μ_l = sin 45⁰ / sin 30⁰

= 1 / √2 / 1 / 2

= √2

= 1.414

(b)

(c)

The principle of reversibility is used here.

Question: 28

When an illuminated object is held in front of a thick plane glass mirror, several images are seen, out of which the second image is the brightest. Give reason.

Solution:

When the ray of the light falls on the surface of the mirror from lighted candle, a small part of light (nearly 4%) is reflected forming a faint virtual image, while a large part of light (nearly 96%) is refracted inside the glass. Now this ray is strongly reflected back by the silvered surface inside the glass. This ray is then partially refracted in air and this refracted ray forms another virtual image. This image is the brightest image because it is due to the light suffering a strong first reflection at the silvered surface.

Question: 29

Fill in the blanks to complete the following sentences:

(a) When light travels from a rarer to a denser medium, its speed ………………….

(b) When light travels from a denser to a rarer medium, its speed ………………….

(c) The refractive index of glass with respect to air is 3/2. The refractive index of air with respect to glass will be ………………….

Solution:

(a) When light travels from a rarer to a denser medium, its speed decreases

(b) When light travels from a denser to a rarer medium, its speed increases

(c) The refractive index of glass with respect to air is 3 / 2. The refractive index of air with respect to glass will be 2 / 3

MULTIPLE TYPE CHOICE

Question: 1

When a ray of light from air enters a denser medium, it:

a. Bends away from the normal

b. Bends towards the normal

c. Goes undeviated

d. Is reflected back

Solution:

When a ray of light from air enters a denser medium, it bends towards the normal.

Question: 2

A light ray does not bend at the boundary in passing from one medium to the other medium if the angle of incidence is:

a. 0° 

b. 45° 

c. 60° 

d. 90°

Solution:

A light ray does not bend at the boundary in passing from one medium to the other medium if the angle of incidence is 0⁰

Question: 3

The highest refractive index is of:

a. Glass

b. Water

c. Diamond

d. Ruby

Solution:

Diamond has the highest refractive index

NUMERICAL

Question: 1

The speed of light in air is 3 x 10⁸ m s^-1. Calculate the speed of light in glass. The refractive index of glass is 1.5.

Solution:

Given

Speed of light in air, C = 3 × 10⁸ m / s

Refractive index of glass, μ = 1.5

Speed of light in glass, v =?

Now,

c / v = μ

Hence, v = c / μ

v = 3 × 10⁸ m / s / 1.5

v = 2 × 10⁸ m / s

Question: 2

The speed of light in diamond is 125,000 km s^-1. What is the refractive index? (speed of light in air = 3 x 10⁸ m s^-1).

Solution:

Given,

Speed of light in diamond = 125,000 km s^-1 i.e 125 × 10⁶ m / s

Speed of light in air, c = 3 × 10⁸ m / s

Refractive index of diamond, μ_D = ?

We know that,

c / v = μ

μ = c / v

μ = (3 × 10⁸) m / s / (125 × 10⁶) m / s

μ = 2.4

Question: 3

The refractive index of water with respect to air is 4/3. What is the refractive index of air with respect to water?

Solution:

Given,

The refractive index of water with respect to air is 4 / 3

_air μ_water = 4 / 3

Hence, the refractive index of air with respect to water is

_water μ_air = 1 / _air μ_water

_water μ_air = 1 / (4 / 3)

_water μ_air = 3 / 4

_water μ_air = 0.75

Question: 4

A ray of light of wavelength 5400
  suffers refraction from air to glass. Taking _aμ_g = 3/2, find the wavelength of light in glass.

Given,

Wavelength of light in air = 5400

Refractive index of glass with respect to air is given by

_aμ_g = 3 / 2

Also, _a μ_g = (wavelength of light in air) / (wavelength of light in glass)

3 / 2 = (wavelength of light in air) / (wavelength of light in glass)

3 / 2 = 5400
/ wavelength of light in glass

Wavelength of light in glass = 2 / 3 × 5400

Wavelength of light in glass = 3600

Exercise 4(B) page no: 87

Question: 1

What is a prism?

With the help of a diagram of a prism, indicate its refracting surfaces, refracting angle and base.

Solution:

A prism is defined as a transparent medium bounded by five plane surfaces with a triangular cross section

Question: 2

The diagrams (a) and (b) in Fig. below show the refractions of a ray of light of single colour through a prism and a parallel-sided glass and prism, respectively.

(a) In each diagram, label the incident, refracted, emergent rays and the angle of deviation.

(b) In what way the direction of emergent ray in the two cases differ with respect to the incident ray? Explain your answer.

Solution:

(a)

(b)

For the prism, the emergent ray is not parallel to the incident ray while for the glass the emergent ray is parallel to the incident ray. This is because, refraction takes place at two inclined surfaces in a prism while in a glass refraction takes place at two parallel surfaces.

Question: 3

Define the term angle of deviation.

Solution:

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

Question: 4

Complete the following sentence:

Angle of deviation is the angle which the ________ ray makes with the direction of ________ ray.

Solution:

Angle of deviation is the angle which the emergent ray makes with the direction of incident ray

Question: 5

What do you understand by the deviation produced by a prism? Why is it caused? State three factors on which angle of deviation depends.

Solution:

In a prism, the ray of light suffers refraction at two inclined faces. The prism produces a deviation at the first surface and another deviation at the second surface. Hence, in the path of light, a prism produces a deviation.

The value of the angle of deviation depends on the following four factors

(i) The angle of incidence (i)

(ii) The material of prism (i.e on refractive index μ)

(iii) The angle of prism (A) and

(iv) The colour or wavelength (λ) of light used.

Question: 6

(a) How does the angle of deviation produced by a prism change with increase in the angle of incidence. Draw a curve showing the variation in the angle of deviation with the angle of incidence at a prism surface.

(b) Using the curve in part (a) above, how do you infer that for a given prism, the angle of minimum deviation δ_min is unique for the given light.

Solution:

Variation of angle of deviation ? with angle of incidence (i)

For a given prism and given colour of light, angle of minimum deviation (?_min) is unique since only one horizontal line can be drawn parallel to i – axis at the lowest point of i – ? curve i.e only for one value of angle of incidence i, the refracted ray inside the prism is parallel to its base.

Question: 7

State whether the following statement is ‘true’ or ‘false’.

The deviation produced by a prism is independent of the angle of incidence and is same for all the colours of light.

Solution:

False. As the angle of incidence increases, the angle of deviation first decreases and then increases. A given prism deviates the violet light most and the red light least.

Question: 8

How does the deviation produced by a prism depend on

(i) the refraction index of its material, and 

(ii) the wavelength of incident light

Solution:

(i) The prism with a higher refractive index produces a greater deviation than a prism with a lower refractive index for a given angle of incidence.

(ii) The refractive index of a given transparent medium is different for the light of different colours. It decreases with the increase in the wavelength of light. Thus the refractive index of the material of a prism for visible light is maximum for the violet colour and minimum for the red colour. Hence, a given prism deviates the violet the most and the red light least.

Question: 9

How does the angle of minimum deviation produced by a prism change with increase in (i) the wavelength of incident light and (ii) the refracting angle of the prism?

Solution:

(i) As we increase the wavelength of incident light, the angle of deviation decreases.

(ii) The angle of deviation increases with the increase in the refracting angle of the prism

Question: 10

Write a relation for the angle of deviation (?) for a ray of light passing through an equilateral prism in terms of angle of incident (i), angle of emergence (e), angle of prism (A).

Solution:

The relation between the angle of incident (i), angle of emergence (e), angle of prism (A) and angle of deviation (?) for a ray of light passing through an equilateral prism is

? = (i + e) – A

Question: 11

A ray of light incident at an angle of incidence i₁ passes through an equilateral glass prism such that the refracted ray inside the prism is parallel to its base and emerges at an angle of emergence i₂.

(a) How is the angle of emergence ‘i₂‘ related to the angle of incidence ‘i₁‘.

(b) What can you say about the angle of deviation in such a situation?

Solution:

(i) The relation between angle of emergence i₂ to the angle of incidence i₁ is

i₂ = i₁

(ii) The angle of deviation is minimum

Question: 12

Draw a ray diagram to show the refraction of a monochromatic ray through a prism when it suffers minimum deviation. How is the angle of emergence related to the angle of incidence in this position.

Solution:

In the equilateral prism, when the prism is in minimum deviation, the angle of incidence i₁ is equal to the angle of emergence i₂

i₁ = i₂ = i

Question: 13

A light ray of yellow colour is incident on an equilateral glass prism at an angle of incidence equal to 48^o and suffers minimum deviation by an angle of 36^o. (i) What will be the angle of emergence? (ii) If the angle of incidence is changed to (a) 30^o, (b) 60^o, state whether the angle of deviation will be equal to less than or more than 36^o.

Solution:

(i) In an equilateral glass prism the ray suffers minimum deviation. So

i₁ = i₂

i₂ = 48⁰

(ii) (a) If the angle of incidence is changed to 30⁰, the angle of deviation will be more than 36⁰

(b) If the angle of incidence is changed to 60⁰, the angle of deviation will be more than 36⁰

Question: 14

Name the colour of white light which is deviated (i) the most, (ii) the least, on passing through a prism.

Solution:

On passing through a prism, violet colour will deviate the most and red colour will deviate the least.

Question: 15

Which of the two prisms, A made of crown glass and B made of flint glass, deviate a ray of light more?

Solution:

The B made of flint glass has higher refractive index. So, B made of flint glass deviate a ray of light more.

Question: 16

How does the angle of deviation depend on refracting angle of the prism?

Solution:

With the increase in the angle of prism (A), the angle of deviation (?) increases.

Question: 17

An object is viewed through a glass prism with its vertex pointing upwards. Draw a ray diagram to show the formation of its image seen by the observer.

Solution:

Let two rays OA and OL from a source of light O are incident on the prism. They are refracted along AB and LM from the first face of the prism respectively. These two rays again refract from the second face of the prism emerge out along BC and MN such that they appear to come from a point I.

Hence, the observer sees the object O raised to the position I

Question: 18

A ray of light is normally incident on one face of an equilateral glass prism. Answer the following

(a)What is the angle of incidence on the first face of the prism?

(b)What is the angle of refraction from the first face of the prism?

(c)What will be the angle of incidence at the second face of the prism?

(d)Will the light ray suffer minimum deviation by the prism?

Solution:

(a) When the incident ray normal to prism then the angle of incidence is 0⁰.

(b) The angle of refraction from the first face of the prism r₁ = 0⁰

(c) The prism is equilateral so A = 60⁰ and r₁ = 0⁰. Hence at the second face of the prism, the angle of incidence will be 60⁰

(d) No, the light ray will not suffer minimum deviation.

Question: 19

The diagram below shows two identical prisms A and B placed with their faces parallel to each other. A ray of light of single colour PQ is incident at the face of the prism A. Complete the diagram to show the path of the ray till it emerges out of the prism B

Solution:

MULTIPLE CHOICE TYPE

Question: 1

In refraction of light through a prism, the light ray:

a. Suffers refraction only at one face of the prism

b. Emerges out from the prism in a direction parallel to the incident ray

c. Bends at both the surfaces of prism towards its base

d. Bends at both the surfaces of prism opposite to its base.

Solution:

In refraction of light through a prism, the light ray bends at both the surfaces of prism towards its base.

Question: 2

A ray of light suffers refraction through an equilateral prism. The deviation produced by the prism does not depend on the:

(a) angle of incidence

(b) colour of light

(c) material of prism

(d) size of prism

Solution:

The ray of light suffers refraction through an equilateral prism. The deviation produced by the prism does not depend on the size of prism.

NUMERICAL

Question: 1

A ray of light incident at an angle 48^o on a prism of refracting angle 60^o suffers minimum deviation. Calculate the angle of minimum deviation.

Solution:

Given,

Angle of incidence, i = 48⁰

Refracting angle, A = 60⁰

Angle of minimum deviation, ?_min =?

We know that

?_min = 2i – A

?_min = 2(48) – 60

?_min = 96 – 60

?_min = 36⁰

Question: 2

What should be the angle of incidence for a ray of light which suffers a minimum deviation of 36^o through an equilateral prism?

Solution:

Given,

Angle of prism, A = 60⁰

Angle of minimum deviation, ?_min = 36⁰

Angle of incidence, i = ?

We know that

?_min = 2i – A

36⁰ = 2i – 60⁰

i = 48⁰

exercise 4(c) page no: 91

Question: 1

How is the refractive index of a medium related to the real and apparent depths of an object in that medium?

Solution:

The relation of refractive index μ with real and apparent depths is

μ = Real Depth / Apparent Depth

Question: 2

Prove that

Refractive index = Real depth / Apparent depth

Solution:

Consider a ray of light OA is incident on the surface PQ normally. It passes straight along AA^’. Consider another ray from O, incident at angle i along OB. This ray gets refracted and passes along BC. The ray BC appears to be coming from point I which is the virtual image of O, obtained on producing A^’A and BC backwards. Hence, AI represents the apparent depth, which is less than the real depth.

Since, AO and BN^’ are parallel and OB is transversal line, so

∠AOB = ∠OBN¹ = i

Similarly, IA^’ and BN are parallel and IC is the transversal line, so

∠BIA^’ = ∠CBN = r

In right-angle triangle BAO,

sin i = BA / OB and

In right-angle triangle IAB,

sin r = BA / IB

For refraction from medium to air, by Snell’s law

_m μ_a = sin i / sin r = (BA / OB) / (BA / IB) = IB / OB

Hence, refractive index of medium with respect to air is,

_a μ_m = 1 / _m μ_a = OB / IB

The object is viewed from a point vertically above the object O, since point B is very close to the point A.

∴ IB = OA

Hence _a μ_m = OA / IA = Real depth / Apparent depth

Question: 3

A tank of water is viewed normally from above.

(a) State how the depth of tank appears to change.

(b) Draw a labelled ray diagram to explain your answer.

Solution:

(a) Due to the refraction of light from a denser medium to a rarer medium, the depth of the tank appears to be lesser than its real depth.

(b)

Question: 4

Water in a pond appears to be only three-quarters of its actual depth. (a) What property of light is responsible for this observation? Illustrate your answer with the help of a ray diagram. (b) How is the refractive index of water calculated from its real and apparent depths?

Solutions:

(a) Refraction of light is responsible for this observation

Due to refraction of light from denser medium to rarer medium, it is bent away from the normal.

(b)

Let an object say B is at the bottom of a pond. Consider a ray of light BC from the object that moves from water to air. The ray moves away from the normal N along the path CD, after refraction from the water surface. The produce of CD appears from the point B^’. The virtual image of the object B appears at B^’.

Refractive index of water = Real depth / Apparent depth

Question: 5

Draw a ray diagram to show the appearance of a stick partially immersed in water. Explain your answer.

Solution:

The above figure shows that a stick appears bent or raised which is partially immersed in water in a glass container. This is due to the rays appears to come from P^’ which is the virtual image of the tip P of the stick. This is due to the refraction of light from denser medium to rarer medium at the surface separating two media.

Question: 6

A fish is looking at a 1.0 m high plant at the edge of the pond. Will the plant appear shorter or taller than its actual height? Draw a ray diagram to support your answer.

Solution:

The plant appears to be taller than its actual height

Let the fish is looking from the point O. Since air is a rarer medium in comparison of water, the ray will bend away from the normal MN when the ray OP emerges out from the water to air. But when we extend the ray OP it will meet at Q. Hence due to this, the plant AB will look taller than its actual height.

Question: 7

A student puts his pencil into an empty trough and observes the pencil from the position as indicated in the Fig.

 (i) What change will be observed in the appearance of the pencil when water is poured into the trough?

(ii) Name the phenomenon which accounts for the above-started observation.

(iii) Complete the diagram showing how the student’s eye sees the pencil through water.

Solution:

(a) When water is poured into the trough, part of the pencil which is immersed in water will look short and raised up.

(b) Refraction of light is responsible for the above observation

(c) The required diagram is shown below:

Question: 8

An object placed in one medium when seen from the other medium, appears to be vertically shifted. Name the factors on which the magnitude of shift depends and state how does it depend on them.

Solution:

The shift by which the object appears to be raised, depends on:

(i) The refractive index of the medium.

(ii) The thickness of the denser medium and

(iii) The colour or wavelength of incident light

With the increase in refractive index of the medium the shift increases. It also increases with the increase in thickness of the denser medium. But the shift decreases with the increase in the wavelength of light used.

MULTIPLE CHOICE TYPE

Question: 1

A small air bubble in a glass block when seen from above appears to be raised because of:

a. Refraction of light

b. Reflection of light

c. Reflection and refraction of light

d. None of the above

Solution:

A small air bubble in a glass block when seen from above appears to be raised because of refraction of light.

Question: 2

An object in a denser medium when viewed from a rarer medium appears to be raised. The shift is maximum for:

a. Red light

b. Violet light

c. Yellow light

d. Greenlight

Solution:

The shift is maximum for(b) violet light.

NUMERICALS

Question: 1

A water pond appears to be 2.7 m deep. If the refractive index of water is 4/3, find the actual depth of the pond.

Solution:

Given,

Apparent depth = 2.7 m

Refractive index of water μ_w = 4 / 3

Real depth = Apparent depth × μ_w

Real depth = 2.7 × 4 / 3

Real depth = 3.6 m

Question: 2

A coin is placed at the bottom of a beaker containing water (refractive index = 4/3) to a depth of 12 cm. By what height the coin appears to be raised when seen from vertically above?

Solution:

Given,

Refractive index of the water, μ_w = 4 / 3

Real depth at which coin is placed = 12 cm

Shift in the image = ?

Shift = Real depth × (1 – 1 / μ)

Shift = 12 × (1 – 3 / 4)

Shift = 12 / 4

Shift = 3 cm or R = 3 cm

Question: 3

A postage stamp kept below a rectangular glass block or refractive index 1.5 when viewed from vertically above it, appears to be raised by 7.0 mm. Calculate the thickness of the glass block.

Solution:

Given,

Refractive index of the glass block, μ_g = 1.5

Shift in the image = 7 mm or 0.7 cm

Thickness of glass block or real depth = ?

Shift = Real depth × (1 – 1 / μ)

0.7 = R × (1 – 1 / 1.5)

R = (0.7 × 1.5) / 0.5

R = 2.1 cm

Exercise 4(D) page no: 100

Question: 1

Explain the term critical angle with the aid of a labelled diagram.

Solution:

Critical angle is defined as the angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90⁰

Question: 2

How is the critical angle related to the refractive index of a medium?

Solution:

The relation by which critical angle is related to the refractive index of a medium is

μ = 1 / sin C = cosec C

Question: 3

State the approximate value of the critical angle for

(a) glass-air surface  

(b) water-air surface.

Solution:

(a) The critical angle for glass – air surface is

Refractive index _aμ_g = 3 / 2

∴ sin i_c = 1 / _aμ_g = 2 / 3

i_c = 42⁰

(b) The critical angle for water – air surface is

Refractive index × _aμ_w = 4 / 3

∴ sin i_c = 1 / _aμ_g = 3 / 4

I_c = 49⁰

Question: 4

What is the meant by the statement ‘the critical angle for diamond is 24°?

Solution:

The statement ‘the critical angle for diamond is 24⁰’ implies that at an incident angle of 24⁰, the angle of refraction in the air will be 90⁰ within the diamond. If incident angle is more than the angle of refraction then the ray will suffer total internal reflection without any refraction.

Question: 5

A light ray is incident from a denser medium on the boundary separating it from a rarer medium at an angle of incidence equal to the critical angle. What is the angle of refraction for the ray?

Solution:

The angle of refraction becomes 90⁰ when a ray is incident from a denser medium to a rarer medium at an angle of incidence equal to the critical angle.

Question: 6

Name two factors which affect the critical angle for a given pair of media. State how do the factors affect it.

Solution:

The two factors which affect the critical angle are

(a) The colour or wavelength of light, and

(b) The temperature

Effect on colour of light: The critical angle for a pair of media is least for the violet light and most for the red light. Thus the critical angle increases with the increase in wavelength of light.

Effect on temperature: On increasing the temperature of medium, its refractive index decreases. So, the critical angle increases with increase in temperature.

Question: 7

The critical angle for glass-air is 45° for the light of yellow colour. State whether it will be less than, equal to, or more than 45° for (i) red light, (ii) blue light?

Solution:

As the wavelength of light decreases (or increases) refractive index becomes more (or less) and critical angle becomes less (or more)

(i) For red light the critical angle will be more than 45⁰

(ii) For blue light the critical angle will be less than 45⁰

Question: 8

(a)What is total internal reflection?

(b)State two conditions necessary for total internal reflection to occur.

(c)Draw diagrams to illustrate critical angle and total internal reflection.

Solution:

(a) Total internal reflection: When a ray of light travelling in a denser medium, is incident at the surface of a rarer medium at the angle of incidence greater than the critical angle for the pair of media, the ray is totally reflected back into the denser medium. This phenomenon is known as total internal reflection.

(b) For total internal reflection, the two necessary conditions are

(i) The light must travel from a denser medium to a rarer medium.

(ii) The angle of incidence must be greater than the critical angle for the pair of media.

(c) Below diagram shows the total internal reflection when incidence angle is more than the critical angle

Question: 9

Fill in the blanks to complete the following sentences:

(a)Total internal reflection occurs when a ray of light passes from a ………………..medium to a …………………. medium.

(b)Critical angle is the angle of ……………..in the denser medium for which the angle of …………………in rarer medium is ………………

Solution:

(a) Total internal reflection occurs when a ray of light passes from a denser medium to a rarer medium.

(b) Critical angle is the angle of incidence in denser medium for which the angle of refraction in rarer medium is 90⁰

Question: 10

State whether the following statement is true or false:

If the angle of incidence is greater than the critical angle, light is not refracted at all, when it falls on the surface from a denser medium to a rarer medium.

Solution:

True

Question: 11

The refractive index of air with respect to glass is expressed as _gμ_a = sin i / sin r

(a)Write down a similar expression for _aμ_g in terms of angle i and r.

(b)If angle r = 90⁰, what is the corresponding angle i called?

(c)What is the physical significance of the angle i and part (b)?

Solution:

(a) _aμ_g = sin r / sin i

(b) The corresponding angle of incidence i will be equal to critical angle, if refractive angel, r = 90⁰

(c) Total internal reflection occurs, if the angle of incidence exceeds the value of i obtained in part (b)

Question: 12

Figure below show two rays A and B travelling from water to air. If the critical angle for water- air surface is 48°, complete the ray diagram showing the refracted rays for each. State conditions when the ray will suffer total internal reflection.

Solution:

Two necessary conditions for the total internal reflection are

(i) Light must travel from a denser medium to a rarer medium.

(ii) The angle of incidence must be greater than the critical angle.

In this case, the angle of incidence ∠i > 48⁰

Question: 13

Fig. shows a point source P inside a water container. Three rays A, B and C starting from the source P are shown up to the water surface.

(a) Show in the diagram, the path of these rays after striking the water surface. The critical angle for water-air surface is 48°.

(b) Name the phenomenon which the rays A, B and C exhibit.

Solution:

(a)

(b) Rays A and B shows the phenomenon of ‘refraction of light’

(c) Ray C shows the phenomenon of ‘total internal reflection’

Question: 14

In the figure, PQ and PR are the two light rays emerging from an object P. The ray PQ is refracted as QS.

(a) State the special name given to the angle of incidence ∠PQN of the ray PQ.

(b) What is the angle of refraction for the refracted ray QS?

(c)Name the phenomenon that occurs if the angle of incidence ∠PQN is increased.

(d) The ray PR suffers partial reflection and refraction on the water-air surface. Give reason.

(e) Draw in the diagram the refracted ray for the incident ray PR and hence show the position of image of the object P by the letter P’ when seen vertically from above.

Solution:

(a) Critical angle

(b) The angle of refraction is 90⁰ for the refracted ray QS

(c) The phenomenon that occurs if the angle of incidence ∠PQN is increased is total internal reflection

(d) For the ray PR, the angle of incidence is less than ∠PQN i.e the critical angle. Hence, as per the laws of reflection at the interface of two media, ray PR suffers partial reflection and refraction.

(e)

Question: 15

The refractive index of glass is 1.5. From a point P inside a glass block, draw rays PA, PB and PC incident on that glass-air surface at an angle of 30^o, 42^o and 60^o respectively.

(a)In the diagram show the approximate direction of these rays as they emerge out of the block.

(b)What is the angle of refraction for the ray PB?

Solution:

(a) Given,

Refractive index of glass, μ = 1.5

Sin i_c = 1 / μ

= 1 / 1.5

= 0.667

i_c = 41.8 ≈  42⁰

(b) Since, the angle of incidence inside the glass block is 42⁰

sin i / sin r = _aμ_g

sin r = _aμ_g × sin i

sin r = _aμ_g × sin 42⁰

Take sin 42⁰ = 2 / 3 and _aμ_g = 3 / 2

sin r = (3 / 2) × (2 / 3)

sin r = 1

r = 90⁰

This shows that the ray PB is incident at the critical angle

Question: 16

A ray of light enters a glass ABCD as shown in Fig. and strikes at the Centre O of the circular part AC of the slab. The critical angle of glass is 42°. Complete the path of the ray till it emerges out from the slab. Mark the angles in the diagram wherever necessary.

Solution:

The ray is incident on the glass at its critical angle. So, the angle of refraction will be 90⁰

Question: 17

What is a total reflecting prism? State three actions that it can produce. Draw a diagram to show one action of the total reflecting prism.

Solution:

Total reflecting prism is defined as a prism having an angle of 90⁰ between its two refracting surfaces and the other two angles each equal to 45⁰, is called a total reflecting prism. Here the light is incident normally on any of its faces, suffers total internal reflection inside the prism.

Because of this, a total reflecting prism is used for the following three purposes.

(i) To deviate a ray of light through 90⁰

(ii) To deviate a ray of light through 180⁰, and

(iii) To erect the inverted image without producing deviation in its path.

An erecting prism is used to erect the inverted image without producing deviation in its path.

Question: 18

Show with the help of a diagram how a total reflecting prism can be used to turn a ray of light through 90°. Name one instrument in which such a prism is used.

Solution:

In the above figure, a beam of light is incident normally at the face AB. So, it passes undeviated into the prism and strikes at the face AC making an angle of incidence equal to 45⁰. Here the incidence is greater than the critical angle. So, the beam of light suffers total internal reflection and reflect at an angle of 45⁰. The reflected beam then strikes the face BC inside the prism where it is incident normally and thus passes undeviated. Hence, the incident beam gets deviated through 90⁰.

This type of prism is used in periscope.

Question: 19

A ray of light OP passes through a right-angled prism as shown in the adjacent diagram.

(a)State the angles of incidence at the faces AC and BC.

(b)Name the phenomenon which the ray suffers at the face AC.

Solution:

(a) The angle of incidence at the face AC = 45⁰ and the angle of incidence at the face BC = 0⁰

(b) The ray suffers total internal reflection at the face AC.

Question: 20

In Fig., a ray of light PQ is incident normally on the hypotenuse of an isosceles right angle prism ABC.

(a) Complete the path of the ray PQ until it emerges from the prism. Mark in the diagram the angle wherever necessary.

(b) What is the angle of deviation of the ray PQ?

(c) Name a device in which this action is used.

Solution:

(a)

(b) The angle of deviation of the ray PQ is 180⁰

(c) Prism binocular is the device in which it is used

Question: 21

In Fig., a ray of light PQ is incident normally on the face AB of an equilateral glass prism. Complete the ray diagram showing its emergence into air after passing through the prism.

Take critical angle for glass = 42°

(a)Write the angles of incidence at the faces AB and AC of the prism.

(b)Name the phenomenon which the ray of light suffers at the face AB, AC and BC of the prism.

Solution:

(a) At the face AB, i = 0⁰ and at the face AC, i = 60⁰

(b) At the face AB – refraction,

At the face AC – total internal reflection

At the face BC – refraction

Question: 22

Draw a neat labelled ray diagram to show the total internal reflection of a ray of light normally incident on one face of a 30°, 90°, 60° prism.

Solution:

Question: 23

Two isosceles right-angled glass prisms P and Q are placed near each other as shown in Fig. Complete the path of the light ray entering the first prism till it emerges out of the second prism Q.

Solution:

`

Question: 24

What device other than a plane mirror, can be used to turn a ray of light through 180°? Draw a diagram in support of your answer. Name an instrument in which this device is used.

Solution:

A total reflecting prism is used to turn a ray of light through 180⁰. Below diagram make it further clear.

Binocular is an instrument in which this action of the prism is used

Question: 25

Mention one difference between the reflection of light from a plane mirror and total internal reflection of light from a prism.

Solution:

The entire incident light is reflected back into the denser medium in the total internal reflection of light from a prism whereas, in ordinary reflection from a plane mirror, only a part of the light is reflected while rest is refracted and absorbed and thus the reflection is partial.

Question: 26

State one advantage of using a total reflecting prism as a reflector in place of a plane mirror.

Solution:

A total reflecting prism gives the image much brighter and the brightness remains unchanged even after the long use, while the image is less bright and the brightness gradually decreases which is obtained by a plane mirror.

MULTIPLE CHOICE TYPE

Question: 1

The critical angle for the glass-air interface is :

a. 24°

b. 48°

c. 42°

d. 45°

Solution:

The critical angle for the glass-air interface is 42⁰

Question: 2

A total reflecting right-angled isosceles prism can be used to deviate a ray of light through

a. 30°

b. 60°

c. 75°

d. 90°

Solution:

A total reflecting right angled isosceles prism can be used to deviate a ray of light through 90⁰

Question: 3

A total reflecting equilateral prism can be used to deviate a ray of light through:

a. 30°

b. 60°

c. 75°

d. 90°

Solution:

A total reflecting equilateral prism can be used to deviate a ray of light through 60⁰

Selina Solutions Concise Physics Class 10 Chapter 4 Refraction of Light at Plane Surfaces

Selina Solutions Concise Physics Class 10 Chapter 4 Refraction of Light at Plane Surfaces consists of basic concepts pertaining to the light. Refraction of Light, Laws of Refraction, Speed of Light in different Media, the principle of reversibility of the path of Light are some topics covered in this chapter. Students understand the chapter in-depth by practising Selina Solutions, as the answers are accurate.

List of subtopics covered in Selina Solutions Concise Physics Class 10 Chapter 4 Refraction of Light at Plane Surfaces

List of Exercises

Key Features of Selina Solutions Concise Physics Class 10 Chapter 4 Refraction of Light at Plane Surfaces

• Selina Solutions are written in a lucid manner for students
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• The accurate answers of Selina help students to understand each and every concept clearly
• Exercise questions, multiple-choice and numerical problems at the end of the chapter help students comprehend the concepts without error

The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2020-21 will be updated soon.