Selina Concise Class 10 Physics Solution Chapter No. 4- ‘Refraction of Light at Plane Surfaces’ For ICSE Board Students.
Selina Concise Class 10 Physics Chapter 4 Refraction of Light at Plane Surfaces Exercise All Questions and Answers by Physics Teacher here in this post.
Exercise 4 (A)
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.
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) air to glass.
(ii) 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 angle of incident is zero as the incident ray of light is normal to the plane. Hence, angle of refraction is also zero and the angle of deviation will also be zero.
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 speed of ray changes as the medium changes because change in the density of the medium.
Question: 5
What is the cause of refraction of light when it passes from one medium to another?
Solution: Refraction of light causes as the speed of light changes with change in the medium because of difference in density. As the speed of ray changes the angle of the path of ray changes.
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:
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 changes from incident ray as a ray travels from one medium to another.
Question: 8
State the Snell’s laws of refraction of light.
Solution: Snell’s law of refraction states that,
The incident ray, the refracted ray and point incident lies in the same plane.
The ratio of sine of the angle of incident I to the sine to angle of refraction r is constant called as refractive index.
Question: 9
Define the term refractive index of a medium. Can it be less than 1?
Solution: Refractive index of second medium to the first can be given by the ratio of sine of angle of incident to the sine of angle of refraction. It can never be less than 1 for a transparent medium.
Question: 10
(a) Compare the speeds of light of wavelength 4000
(b) How is the refractive index of a medium related to the speed of light in it and in vacuum or air?
Solution:
(a) The speed of light will be the same as speed of light depends upon the medium not the wavelength.
(b) The speed of light is highest in vaccum or air and it decreases as the light passes through denser medium.
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: Speed of light increases as the light passes to rarer medium of air from denser medium of water. The speed of light decreases as the light passes through lighter medium of water to denser medium of 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) As the light ray speeds up the ray passes from denser medium of water to rarer medium like air.
(b) As the light ray slows down as the ray passes from water to another more dense material such as glass.
(i) The refractive index is less than that of water as the speed of light ray increases.
(ii) The refractive index is greater than that of water as the speed of light ray decreses.
Question: 13
What do you understand by the statement ‘the refractive index of the glass is 1.5 for white light’?
Solution: If the refractive index of glass is 1.5. we can state that the light ray travels 1.5 times faster in air.
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) If the refractive index of glass is 1.5 and speed in air and glass is c and v respectively. Then
Relation between c and v is given by
Refractive index = c/v
Therefore, c = 1.5v
(b) The wavelength in glass
Refractive index = wavelength in air / wavelength in glass
Wavelength in glass = 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: Yes, the refractive index will change as the speed of red light is greater than speed of blue light in the glass medium. Hence the refractive index will decrease as the light is changed from the blue to red.
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 for red colour is least in the transparent medium.
(ii) The violet colour shows the most refractive index in the transparent medium.
(b) Red colour travels fastest in any medium other than air as the wavelength of red colour is longest.
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 refractive index depends upon the nature of medium as the speed of lightindifferent medium are different, which affects the refractive index. And the physical conditions like the temperature, pressure also affects the refractive index of a medium as the increase in temperature cause increase in the speed of light which decreases the refractive index of a particular medium.
Question: 18
How does the refractive index of a medium depend on the wavelength of light used?
Solution: The refractive index is inversely proportional to the wavelength of light used. Hence the refractive index decreases with the increase in the wavelength.
Question: 19
How does the refractive index of a medium depend on its temperature?
Solution: refractive index is inversely proportional to the temperature as the speed of light increase wit increase in the temperature of the medium. Hence, refractive index decreases for a medium when the temperature of medium increases.
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) When the refractive index of liquid becomes equal to the refractive index of the glass the glass piece can not be seen.
(b) As the refractive index in glass or liquid are different for different wavelength. Hence, the light of single wavelength that is 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:
The relation between the angle of incident and the angle of refraction is Refractive index=sin i/sin r
Question: 23
The refractive index of water with respect to air is aµw and of glass with respect to air is aµg. Express the refractive index of glass with respect to water.
Solution: The refractive index of glass with respect to water
wµg= aµg / aµw
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:
Lateral displacement is the perpendicular distance between the path of emergent ray to the direction of incident ray.
Question: 25
A ray of light strikes the surface at a rectangular glass slab such that the angle of incidence is (i) 0o, (ii) 45o. 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) 0o
(ii) 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:
Question: 27
A ray of green light enters a liquid from air, as shown in the figure. The angle 1 is 45o and angle 2 is 30o.
(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 liquid
Given, Angle od incident i = 45°
angle of refraction r = 30°
refractive index = sin i / sin r
= sin 45° / sin 30°
= 1/√(2)/1/2
= √(2)
= 1.414
(b)
(c)
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 light emitted by illuminated object falls on thick plane mirror most of the light gets refracted and little amount of it gets reflected which forms the light image. Now this refracted light gets reflected strongly from the silvered surface of mirror. This ray gets refracted a little in air and forms another virtual image. This image is brightest as the light is reflected strongly by 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) Decreases
(b) Increases
(c) 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: (b) 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) 0°
Question: 3
The highest refractive index is of:
a.) Glass
b.) Water
c.) Diamond
d.) Ruby
Solution: (c) Diamond
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 = 3 x 108 m/s
Refactive index = 1.5
Speed of light in glass = speed of light in air / refractive index
= 3 x 108 m/s / 1.5
= 2 x 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 air = 3 x 108 m/s
Speed of light in diamond = 125000 km/s = 1.25 x 108 m/s
Refractive index = speed of light in air / speed of light in diamond
= 3 x 108 m/s / 1.25 x 108 m/s
= 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
Refractive index of water w.r.t. air = 4/3
Refractive index of air w.r.t. water = 1/ Refractive index of water w.r.t. air
= 1/ 4/3
= ¾
= 0.75
Question: 4
A ray of light of wavelength 5400
Å Suffers refraction from air to glass. Taking refractive index = 3/2, find the wavelength of light in glass.
Solution: Given
Wavelength of light in air = 5400 Ao
refractive index = 3/2
refractive index = wavelength in the air / wavelength in glass
wavelength in glass = 5400 x 2/3
wavelength in glass = 3600 Ao
Exercise 4 (B)
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 transparent medium bounded by five plane surfaces with a triangular cross section is called as prism.
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)
Question: 3
Define the term angle of deviation.
Solution: The angle between the of direction incident ray and the emergent ray is called as angle of deviation.
Question: 4
Complete the following sentence:
Angle of deviation is the angle which the ________ ray makes with the direction of ________ ray.
Solution: Emergent ray, 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: Prism has two surface which refracts light which are inclined. The prism produces deviation at first surface and then the second surface produces another deviation.
The value of the angle of deviation depends upon
(i) Angle of incident
(ii) The angle of prism
(iii) The wavelength of light used
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.
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) Deviation produced by prism is directly proportional to the refractive index of prism. Hence, more the refractive index higher the deviation angle.
(ii) Based on the refractive index of prism the deviation produced by the different wavelength of light. Higher the wavelength lower refraction. Hence, a prism deviates violet the most and red the 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) Minimum angle of deviation is inversely proportional to wavelength. Hence, increase in wavelength decreases the angle of deviation.
(ii) Angle of deviation is directly proportional to refracting angle.
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:
Relation between 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) is given by
= (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:
(a) The relation between the angle of emergence ‘i2‘ to the angle of incidence ‘i1‘ is the
i2 = i1
(b) In this case 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:
Angle of incident and angle of refraction is same.
Question: 13
A light ray of yellow colour is incident on an equilateral glass prism at an angle of incidence equal to 48o and suffers minimum deviation by an angle of 36o. (i) What will be the angle of emergence? (ii) If the angle of incidence is changed to (a) 30o, (b) 60o, state whether the angle of deviation will be equal to less than or more than 36o.
Solution:
(i) In equilateral glass prism the ray suffers minimum deviation
I1 = I2
I2 = 48°
(ii) (a) if the angle of incident changes to 30°, then the angle of deviation will be more than 36°
(b) If the angle of incident changes to 60°, then the angle of deviation will be less than 36°.
Question: 14
Name the colour of white light which is deviated (i) the most, (ii) the least, on passing through a prism.
Solution: The violet colour gets deviated 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 prism made of flint glass has more refractive index than prism made of crown glass. Hence prism B deviates a ray more.
Question: 16
How does the angle of deviation depend on refracting angle of the prism?
Solution: The refracting angle of prism is directly proportional to the angle of deviation.
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:
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) The angle of incident on the first face of the prism is 0°.
(b) The angle of refraction on the first face is 0°.
(c) As the prism is equilateral glass prism angle of prism is 60°. hence, the angle of incident at the second prism face is 60°.
(d) There will be no minimum deviation for a ray.
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: (c) 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: (d) size of prism
Numericals
Question: 1
A ray of light incident at an angle 48° on a prism of refracting angle 60° suffers minimum deviation. Calculate the angle of minimum deviation.
Solution: Given
Angle of incident, I = 48°
Refracting angle, A = 60°
Minimum deviation = 2i – A
= 2 (48) – 60
= 96 – 60
= 36°
Question: 2
What should be the angle of incidence for a ray of light which suffers a minimum deviation of 36° through an equilateral prism?
Solution: Given
Angle of prism, A = 60°
Minimum deviation = 36°
Minimum deviation = 2i – A
36° = 2i – 60°
(36° + 60°)/2 = i
48° = i
Hence, angle of incident is 48°.
Exercise 4 (c)
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 between refractive index based on real and apparent depth is
Refractive index = 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:
The apparent depth is less than real depth as the refractive index of water is more than air when the ray travel from the water to eye the light get refracted accordingly.
Real depth / apparent depth = refractive index.
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.
(b) refractive index = 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 stick appears slightly raised due to change in the medium. As the refractive index alters the ray of light depending on the medium.
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 will appear taller than original tree to the fish due to refraction of light.
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:
(i) The part of pencil in the water will appear short.
(ii) Due to refraction of light pencil will appear short.
(iii)
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 factor on which the magnitude of shift depends upon are refractive index of medium, the thickness of denser medium and wavelength of light used.
The shift is directly proportional to the refractive index and thickness of the medium and inversely proportional to 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) 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: (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
Refractive index = 4/3
Apparent depth = 2.7m
Real depth = apparent depth x refractive index
Real depth = 2.7 x 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 = 4/3
Real depth = 12 cm
Shift = real depth x (1-1/refractive index)
Shift = 12 x (1 – ¾)
Shift = 12/4
Shift = 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
Shift = 7mm = 0.7 cm
Refractive index = 1.5
Shift = real depth x (1- 1/ refractive index)
0.7 = R x (1 – 1/ 1.5)
R = (0.7 X 1.5) / 0.5
R = 2.1 cm
Exercise 4 (D)
Question: 1
Explain the term critical angle with the aid of a labelled diagram.
Solution: When the angle of refraction become 90 the incident angle that causes it is called as critical angle.
Question: 2
How is the critical angle related to the refractive index of a medium?
Solution: The relation between critical angle and refractive index of a medium is
Refractive index = 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 to air surface
Refractive index = 3/2
Sin C = 1 / refractive index = 2/3
C = 42°
(b) The critical angle for water to air surface
Refractive index = 4/3
Sin C = 1/ refractive index = ¾
C = 49°
Question: 4
What is the meant by the statement ‘the critical angle for diamond is 24°?
Solution: The critical angle for diamond 24o. Hence, if the incident angle 24°, the angle of refraction in the air will be 90o within diamond. Hence, 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 90o.
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 affecting the critical angle are the wavelength of light and the temperature. The critical angle is directly proportional to the value of wavelength. The critical angle is directly proportional to the 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 is directly proportional to the critical angle. Hence, for red light the critical angle will be more than 45o and for blue light the critical angle will be less than 45o.
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) A ray traveling from denser medium and it incident on the surface of rarer medium at an angle grater than the critical angle for the pair of media, the ray is totally reflected back into denser medium without any refraction. This phenomenon is known as total internal reflection.
(b) Two condition necessary for total internal reflection are light must travel from denser medium to rarer medium and angle of incident must be greater than the critical angle.
(c)
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) Denser, rarer
(b) incident, refraction, 90o
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 gμa = sin i / sin r
(a)Write down a similar expression for aμg in terms of angle i and r.
(b) If angle r = 90o, what is the corresponding angle i called?
(c) What is the physical significance of the angle i and part (b)?
Solution:
(a) Refractive index of glass to air = sin r / sin i
(b) As angle of refraction is 90o. Then the incident angle is called as critical angle.
(c) Total internal reflection happens when angle of incident 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:
The conditions for total internal reflection are light must travel from a denser medium to rarer medium, angle of incident must be greater than critical angle.
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:
Ray A and B shows refraction of light.
Ray c shows 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) 900
(c) If critical angle is increased total internal reflection occurs.
(d) As the angle of incident is less than critical angle the ray gets partially reflected and partially refracted.
(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°, 42° and 60° 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:
(c) as angle of incident is 42o
sin I / sin r = refractive index
3/2 = sin 42 / sin r
Sin r = 3/2 x 2/3
Sin r = 1
r = 90o
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:
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: The prism having two equal refracting sides which have 90o angle between them is called as total reflecting prism. It can deviate ray by 90o and 180o and it can invert the image without deviation.
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:
This prism shows the deviation of light through 90o.
This type of prism are 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) angle of incident at face AC is 45o. Angle of incident at face BC is 0o.
(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) angle of deviation is 180o.
(c) it is used in prism binoculars.
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:
Face AB shows refraction
Face AC shows total internal reflection
Face BC shows refraction.
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: Total reflecting prism.
It is used in binoculars.
Question: 25
Mention one difference between the reflection of light from a plane mirror and total internal reflection of light from a prism.
Solution: Total internal reflection reflects complete intensity of light on the other hand reflection from plane mirror is partial as some of the light gets refracted and absorbed.
Question: 26
State one advantage of using a total reflecting prism as a reflector in place of a plane mirror.
Solution: Total reflecting prism gives much brighter long lasting image while plane mirror gives less bright image which fades away gradually.
Multiple Choice Type:
Question: 1
The critical angle for the glass-air interface is:
a.) 24°
b.) 48°
c.) 42°
d.) 45°
Solution: (c) 42°
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: (d) 90°
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: (b) 60°
