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Refraction of Light at Plane Surfaces: ICSE Class 10 Physics
Get summaries, questions, answers, solutions, notes, extras, workbook solutions, PDF and guide of chapter 4 Refraction of Light at Plane Surfaces: ICSE Class 10 Physics which is part of the syllabus of students studying under the Council for the Indian School Certificate Examinations board. These solutions, however, should only be treated as references and can be modified/changed.
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Summary
Light usually travels straight. When light moves from one clear material to another, like air to water, it bends. This is refraction. It happens because light’s speed changes. Light is fastest in air, slower in water, and even slower in glass.
If light slows down in a material, it’s optically denser. If it speeds up, the material is optically rarer. Glass is optically denser than air. “Optical density” is about light’s speed, not weight.
Rules govern light bending. The incoming ray, bent ray, and the normal (a line at 90 degrees to the surface) are on one flat surface. A number called refractive index links the angles of incoming and bent light for two materials.
Refractive index shows how much a material bends light. It compares light’s speed in air to its speed in the material. Higher refractive index means more slowing and bending. When light refracts, its speed and wavelength change. Its color (frequency) does not change.
Light doesn’t always bend. It passes straight if it hits a surface at 90 degrees. It also goes straight if both materials have the same refractive index, so light’s speed is unchanged. Light’s path can be reversed; a bent ray sent backward follows its original path.
A glass slab makes light bend when entering and leaving. The outgoing light is parallel to the incoming light but shifted sideways. This is lateral displacement. A prism, with triangular ends, bends light towards its thicker base. The total bend is the angle of deviation.
Refraction changes how we see things. A pool bottom looks shallower. A pencil in water appears bent. This is because light from these objects bends when going from water to air to our eyes.
Light can get trapped inside a denser material. When light moves from a denser material (like glass) to a rarer one (like air), it bends away from the normal. If it hits the boundary at a large enough angle, called the critical angle, it reflects entirely back into the denser material. This is total internal reflection. For this, light must go from denser to rarer, and its angle must be greater than the critical angle.
Total internal reflection is useful. Prisms use it to turn light by 90 degrees in periscopes or 180 degrees in binoculars. This reflection is very bright. Diamonds sparkle due to this effect..
Workbook solutions (Concise/Selina)
Exercise A
MCQ
1. In the diagram shown below, which one of the following statements is true ?
(a) C is incident ray and A is reflected ray
(b) A is incident ray and D is refracted ray
(c) A is incident ray and B is refracted ray
(d) B is incident ray and C is refracted ray
Answer: (c) A is incident ray and B is refracted ray
2. 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
Answer: (b) bends towards the normal
3. A light ray does not bend at the boundary while passing from one medium to the other medium if the angle of incidence is:
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Answer: (a) 0°
4. The diagram given below shows light travelling from air into glass. Four angles A, B, C and D are shown. Which of the following formulae is used to calculate the refractive index µ of the glass ?
(a) µ = sin A / sin C
(b) µ = sin C / sin A
(c) µ = sin B / sin C
(d) µ = sin B / sin A
Answer: (c) µ = sin B / sin C
5. The incorrect statement among the following is :
(a) When a ray of light gets refracted from a denser to a rarer medium, the speed of light decreases.
(b) The frequency of light does not change on refraction
(c) The speed of light V in a medium is related to its wavelength λ and frequency f as V = fλ
(d) The incident ray, the refracted ray and the normal at the point of incidence all lie in one plane.
Answer: (a) When a ray of light gets refracted from a denser to a rarer medium, the speed of light decreases.
6. In refraction of light, the relation between incident angle i and refracted angle r is :
(a) i > r
(b) r > i
(c) i = r
(d) there is no fixed relation
Answer: (d) there is no fixed relation
7. The intensity of the refracted light is always ………… the intensity of incident light :
(a) greater than
(b) equal to
(c) less than
(d) none of the above
Answer: (c) less than
8. The ray AO is shown in the figure given below. The path of refracted ray and emergent ray will be:
(1) the refracted ray will bend towards normal
(2) the refracted ray will bend away from normal
(3) the emergent ray will bend towards normal
(4) the emergent ray will bend away from normal
(a) (1) and (3)
(b) (1) and (4)
(c) (2) and (3)
(d) (2) and (4)
Answer: (b) (1) and (4)
9. What is the deviation of a ray of light which is incident normally on the surface of two separating media :
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Answer: (a) 0°
10. If refractive indices of medium 1 and medium 2 are the same, the speed of light will :
(a) increase
(b) decrease
(c) remain the same
(d) none of the above
Answer: (c) remain the same
11. When light passes from a denser medium to a rarer medium, its wavelength :
(a) decreases
(b) increases
(c) remains the same
(d) none
Answer: (b) increases
12. Which of the following is correct when a ray of light passes from a rarer medium to a denser medium :
(1) speed will decrease
(2) frequency remains unchanged
(3) wavelength will decrease
(4) intensity will decrease
(a) (1)
(b) (1) and (4)
(c) (1), (2) and (4)
(d) all of the above
Answer: (d) all of the above
13. The refractive index of water is 4/3 and that of glass is 3/2. The refractive index of glass with respect to water is:
(a) 9/8
(b) 2
(c) 8/9
(d) 1
Answer: (a) 9/8
14. The optical density of turpentine is higher than that of water, while its mass density is lower than water. The figure given below shows a layer of turpentine floating over water taken in a container. For which one of the four rays incident on turpentine in the path shown in the figure is correct ?
(a) Ray 1
(b) Ray 2
(c) Ray 3
(d) Ray 4
Answer: (b) Ray 2
15. With respect to light passing from air into water, which of the following statements is correct ?
(a) Angle of incidence is equal to the angle of refraction
(b) Angle of incidence is always greater than the angle of refraction
(c) Angle of incidence is always less than the angle of refraction
(d) Angle of incidence has no effect on the angle of refraction
Answer: (b) Angle of incidence is always greater than the angle of refraction
Very Short Answer Type Questions
1. 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 ?
Answer: (i) For a ray of light incident normally on a plane glass slab, the angle of refraction will be 0°.
(ii) The angle of deviation for the ray will be 0°.
2. An obliquely incident light ray bends at the surface due to change in speed, when passing from one medium to another. The ray does not bend when it is incident normally. Will the ray have different speed in the other medium ?
Answer: Yes, even when incident normally and not bending, the ray will have a different speed in the other medium if the media are different, as the speed of light changes but the direction of light does not change in case of normal incidence.
3. 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 the incident ray : speed, intensity, frequency, and wavelength ?
Answer: When a ray of light passes from medium 1 to medium 2, the quantities of the refracted ray that will differ from that of the incident ray are speed, intensity, and wavelength. The frequency of light does not change on refraction.
4. A light ray passes from water to (i) air, and (ii) glass. In each case, state how does the speed of light change.
Answer: (i) When a light ray passes from water to air, the speed of light increases because air is optically rarer than water.
(ii) When a light ray passes from water to glass, the speed of light decreases because glass is optically denser than water.
5. (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 white light travels fastest in any medium except air ?
Answer: (a) (i) For white light, the refractive index of a transparent medium is the least for red light.
(a) (ii) The refractive index is the most for violet light.
(b) Red light travels fastest in any transparent medium except air (or vacuum), because in a given medium, the speed of red light is maximum.
6. How does the refractive index of a medium depend on the wavelength of light used ?
Answer: The refractive index of a medium decreases with an increase in the wavelength of light used.
7. How does the refractive index of a medium depend on its temperature ?
Answer: The refractive index of a medium decreases with an increase in its temperature.
8. 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.
Answer: The refractive index of glass with respect to water (wµg) can be expressed as wµg = aµg / aµw.
9. 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 …..
Answer: (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.
Short Answer Type Questions
1. What do you understand by refraction of light ?
Answer: Refraction of light is understood as the change in the direction of the path of light when it passes from one transparent medium to another transparent medium. The refraction of light is essentially a surface phenomenon.
2. What is the cause of refraction of light when it passes from one medium to another ?
Answer: The cause of refraction of light when it passes from one medium to another is the change in speed of light in going from one medium to another. This change in speed causes its direction (or path) to change.
3. State the Snell’s laws of refraction of light.
Answer: Snell’s laws of refraction of light are:
(1) The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
(2) 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. Mathematically, sin i / sin r = constant (1µ2).
4. Define the term refractive index of a medium. Can it be less than 1 ?
Answer: The refractive index of a 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. Alternatively, the absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium.
No, the refractive index of a transparent medium cannot be less than 1 because the speed of light in any medium is always less than that in vacuum (i.e., V < c). Obviously µ = 1 for air or vacuum.
5. (a) Compare the speeds of light of wavelength 4000Å (i.e. violet light) and 8000Å (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?
Answer: (a) The speeds of light of wavelength 4000Å (violet light) and 8000Å (red light) in vacuum are the same. The speed of light of all colours is the same in air (or vacuum). The ratio of their speeds is 1:1.
(b) The refractive index of a medium (µ) is related to the speed of light in it (V) and in vacuum or air (c) by the formula: µ = c/V.
6. 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 medium is equal to, less than or greater than the refractive index of water.
Answer: (a) If a light ray passing from water to a medium speeds up:
(i) An example of such a medium is air, as air is optically rarer than water.
(ii) The refractive index of this medium (air) is less than the refractive index of water.
(b) If a light ray passing from water to a medium slows down:
(i) An example of such a medium is glass, as glass is optically denser than water.
(ii) The refractive index of this medium (glass) is greater than the refractive index of water.
7. What do you understand by the statement ‘the refractive index of glass is 1.5 for white light’ ?
Answer: The statement ‘the refractive index of glass is 1.5 for white light’ means that the ratio of the speed of white light in vacuum (or air) to the speed of white light in glass is 1.5. This implies that white light travels 1.5 times faster in vacuum (or air) than it does in glass, or the speed of light in glass is 1/1.5 (or 2/3) times the speed of light in vacuum.
8. 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 glass.
Answer: (a) The relationship between c and V, given that the refractive index of glass (µ) is 1.5, is µ = c/V, so 1.5 = c/V, or c = 1.5 V.
(b) The wavelength of light in glass (λ’) is related to the wavelength in air (λ) by λ’ = λ/µ. So, the wavelength of light in glass is λ/1.5.
9. A boy uses blue colour of light to find the refractive index of glass. He then repeats the experiment using red colour of light. Will the refractive index be the same or different in the two cases ? Give a reason to support your answer.
Answer: The refractive index will be different in the two cases.
Reason: The speed of blue light in glass is less than that of red light. Since the refractive index of glass is µ = (speed of light in air c) / (speed of light in glass V), and V is smaller for blue light than for red light, the refractive index for blue light (µblue) will be greater than the refractive index for red light (µred).
10. Name two factors on which the refractive index of a medium depends ? State how does it depend on the factors stated by you.
Answer: Two factors on which the refractive index of a medium depends are:
(i) The colour or wavelength of light: The refractive index of a medium decreases with the increase in wavelength of light. For visible light, the refractive index is maximum for violet colour and minimum for red colour.
(ii) Physical condition such as temperature: With an increase in temperature, the refractive index of the medium decreases.
(Another factor is the nature of the medium: Less the speed of light in a medium as compared to that in air, more is the refractive index of the medium.)
11. Light of a single colour is passed through a liquid having a piece of glass suspended in it. On changing the temperature of liquid, at a particular temperature the glass piece is not seen.
(i) When is the glass piece not seen ?
(ii) Why is the light of a single colour used ?
Answer: (i) The glass piece is not seen when the refractive index of the liquid becomes equal to the refractive index of the glass at that particular temperature.
(ii) Light of a single colour is used because the refractive index of a medium (both glass and liquid) is different for the light of different colours. Using a single colour ensures that the condition of equal refractive indices can be precisely met for that specific wavelength.
12. 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 brightest. Give reason.
Answer: When an illuminated object is held in front of a thick plane glass mirror, a small fraction of light (nearly 4%) is reflected from the front unsilvered surface of the glass, forming a faint first virtual image. A larger fraction of light (nearly 96%) is refracted into the glass, strikes the silvered surface at the back, and is then strongly reflected. This strongly reflected ray, upon emerging from the front surface after refraction, forms the second virtual image. This second image is the brightest because it is formed due to the light suffering a strong first reflection at the silvered surface (PN). Subsequent images are formed by multiple reflections within the glass and are of gradually decreasing brightness.
Long Answer Type Questions
1. Draw diagrams to show the refraction of light from (i) air to glass, and (ii) glass to air. In each diagram, label the incident ray, refracted ray, the angle of incidence (i) and the angle of refraction (r).
Answer: (i) For refraction of light from air to glass (rarer to denser medium), the light ray bends towards the normal, so the angle of refraction (r) is less than the angle of incidence (i).
(ii) For refraction of light from glass to air (denser to rarer medium), the light ray bends away from the normal, so the angle of refraction (r) is greater than the angle of incidence (i).
2. A light ray suffers reflection and refraction at the boundary in passing from air to water. Draw a neat labelled ray diagram to show it.
Answer: When a light ray passes from air to water, it strikes the boundary separating the two media. A part of the light is reflected back into the air, obeying the laws of reflection (angle of incidence equals angle of reflection). The remaining part of the light enters the water and undergoes refraction. Since water is optically denser than air, the refracted ray bends towards the normal, and the angle of refraction is less than the angle of incidence.
3. In the figure, 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 ?
Answer: (a) (i) To complete the diagram for the reflected ray B, it should be drawn in the air such that the angle of reflection (angle between reflected ray B and the normal) is equal to the angle of incidence (angle between incident ray A and the normal).
(a) (ii) For the refracted ray C in water, since water is denser than air, the ray will bend towards the normal. Thus, the angle of refraction r (angle between refracted ray C and the normal in water) will be less than the angle of incidence i.
(b) The angles of incidence i and refraction r are related by Snell’s law: sin i / sin r = µw, where µw is the refractive index of water with respect to air.
4. The diagram below shows the refraction of a ray of light from air to a 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.
Answer: (a) (i) For the ray of light from air to liquid shown, the angle of incidence is 60°.
(a) (ii) The angle of refraction is 45°.
(b) Using Snell’s law, the refractive index of the liquid with respect to air (µ) is given by:
µ = sin (angle of incidence i) / sin (angle of refraction r)
µ = sin 60° / sin 45°
µ = (√3/2) / (1/√2)
µ = √3 / √2, which is approximately 1.22.
5. 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.
Answer: Due to refraction of light at two parallel surfaces of a parallel-sided glass block, the angle of emergence is equal to the angle of incidence, so the emergent ray and the incident ray are parallel, but they are not along the same line. The emergent ray is laterally displaced from the path of the incident ray. The perpendicular distance between the path of the emergent ray and the direction of the incident ray is called lateral displacement.
6. A ray of light strikes the surface of a rectangular glass slab such that the angle of incidence in air is (i) 0°, (ii) 45°. 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.
Answer: (i) When the angle of incidence in air is 0°, the ray of light is incident normally on the surface of the glass slab. In this case, the ray passes undeviated through the slab. The angle of refraction at each surface is also 0°.
(ii) When the angle of incidence in air is 45°, the ray of light is incident obliquely on the surface of the glass slab. The ray enters from air (rarer medium) to glass (denser medium), so it slows down and bends towards the normal. It travels inside the glass in a straight path. At the other surface, the ray enters from glass (denser medium) to air (rarer medium), so it speeds up and bends away from the normal. The emergent ray is parallel to the incident ray but is laterally displaced.
7. In the adjacent diagram, AO is a ray of light incident on a rectangular glass slab.
(a) Complete the path of the ray till it emerges out of the slab.
(b) In the 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 the 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.
Answer: (a) The ray AO enters the glass slab and bends towards the normal, travelling along a path OB inside the slab. At the second interface, it emerges out of the slab along a path BC, bending away from the normal. The emergent ray BC is parallel to the incident ray AO.
(b) The angle of incidence (i) is the angle between the incident ray AO and the normal at the point of incidence O. The angle of refraction (r) is the angle between the refracted ray OB and the normal at the point of incidence O. The refractive index of glass (µ) is related to the angles i and r by Snell’s law: µ = sin i / sin r.
(c) The angle of emergence (e) is the angle between the emergent ray BC and the normal at the point of emergence B. The angle of emergence e is equal to the angle of incidence i (e = i).
(d) The incident ray AO and the emergent ray BC are parallel to each other.
(e) The lateral displacement is the perpendicular distance between the emergent ray BC and the original path of the incident ray AO if it had passed undeviated.
8. A ray of green light enters a liquid from air, as shown in the figure. Angle 1 is 45° and angle 2 is 30°.
(a) Find the refractive index of the 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.
Answer: (a) From the given figure and information, a ray of light enters a liquid from air.
- The angle of incidence, i = Angle 1 = 45°
- The angle of refraction, r = Angle 2 = 30°
To find the refractive index of the liquid (μ), we use Snell’s law of refraction. Snell’s law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, which is the refractive index of the second medium with respect to the first.
Formula:
μ = sin i / sin r
Calculation:
- Substitute the given values of the angles into the formula:
μ = sin 45° / sin 30° - Use the standard values for the sine of these angles:
sin 45° = 1/√2
sin 30° = 1/2 - Calculate the refractive index:
μ = (1/√2) / (1/2)
=> μ = (1/√2) × 2
=> μ = 2/√2
=> μ = √2
=> μ ≈ 1.414
Therefore, the refractive index of the liquid is √2 or approximately 1.414.
(b) The refracted ray inside the liquid strikes the plane mirror at an angle of incidence of 30°. It reflects off the mirror at an angle of reflection of 30°. This reflected ray then travels to the liquid-air interface, where it is incident at an angle of 30°. It then refracts into the air, bending away from the normal at an angle of refraction of 45°.
(c) If the plane mirror becomes normal to the refracted ray, the ray will strike the mirror at an angle of incidence of 0° and will be reflected back along the same path. It will then retrace its path through the liquid and emerge from the liquid into the air along the original path of incidence. The principle used is the principle of reversibility of the path of light, which states that the path of a light ray is reversible.
Numerical Questions
1. The speed of light in air is 3 × 10⁸ m s⁻¹. Calculate the speed of light in glass. The refractive index of glass is 1.5.
Answer: Given:
Speed of light in air, c = 3 × 10⁸ m s⁻¹
Refractive index of glass, µ = 1.5
We know the relationship between refractive index (µ), speed of light in air (c), and speed of light in a medium (V) is:
µ = c / V
To find the speed of light in glass (V), we rearrange the formula:
V = c / µ
Substituting the given values:
V = (3 × 10⁸ m s⁻¹) / 1.5
=> V = 2 × 10⁸ m s⁻¹
Thus, the speed of light in glass is 2 × 10⁸ m s⁻¹.
2. The speed of light in diamond is 125,000 km s⁻¹. What is its refractive index ? (Speed of light in air = 3 × 10⁸ m s⁻¹).
Answer: Given:
Speed of light in diamond, V = 125,000 km s⁻¹
Speed of light in air, c = 3 × 10⁸ m s⁻¹
First, we need to ensure the units for speed are consistent. Let’s convert the speed of light in diamond from km/s to m/s.
V = 125,000 km/s × 1000 m/km
=> V = 125,000,000 m/s
=> V = 1.25 × 10⁸ m/s
Now, we use the formula for refractive index (µ):
µ = Speed of light in air (c) / Speed of light in diamond (V)
µ = (3 × 10⁸ m s⁻¹) / (1.25 × 10⁸ m s⁻¹)
=> µ = 3 / 1.25
=> µ = 2.4
The refractive index of diamond is 2.4.
3. The refractive index of water with respect to air is 4/3. What is the refractive index of air with respect to water?
Answer: Given:
Refractive index of water with respect to air, ₐµ_w = 4/3
According to the principle of reversibility of the path of light, the refractive index of a medium 1 with respect to medium 2 is the reciprocal of the refractive index of medium 2 with respect to medium 1.
This can be written as: ₂µ₁ = 1 / ₁µ₂
Applying this principle to find the refractive index of air with respect to water (wµₐ):
wµₐ = 1 / ₐµ_w
=> wµₐ = 1 / (4/3)
=> wµₐ = 3/4
=> wµₐ = 0.75
The refractive index of air with respect to water is 3/4 or 0.75.
4. A ray of light of wavelength 6600 Å suffers refraction from air to glass. Taking ₐµ_g = 3/2, find the wavelength of light in glass.
Answer: Given:
Wavelength of light in air, λ_air = 6600 Å
Refractive index of glass with respect to air, ₐµ_g = 3/2
When light travels from air to another medium, its wavelength changes. The relationship is given by:
λ_medium = λ_air / µ_medium
To find the wavelength of light in glass (λ_glass):
λ_glass = λ_air / ₐµ_g
=> λ_glass = 6600 Å / (3/2)
=> λ_glass = 6600 × (2/3) Å
=> λ_glass = 2200 × 2 Å
=> λ_glass = 4400 Å
The wavelength of light in glass is 4400 Å.
Exercise B
MCQ
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 the prism towards its base
(d) bends at both the surfaces of the prism opposite to its base.
Answer: (c) bends at both the surfaces of the prism towards its base
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
Answer: (d) size of prism
3. In the figure given ahead, the correct statement is :
(a) ABED and ABC are the refracting surfaces.
(b) line AD is the refracting edge.
(c) angle a is the angle of deviation
(d) AB is the base of the prism
Answer: (b) line AD is the refracting edge.
4. The angle of prism A is given as :
(a) r₁ – r₂
(b) i₁ + i₂
(c) r₁ + r₂
(d) r₁/r₂
Answer: (c) r₁ + r₂
5. The angle of deviation becomes minimum when :
(a) i₁ = r₁
(b) i₁ = r₂
(c) i₁ = i₂
(d) r₂ = i₂
Answer: (c) i₁ = i₂
6. In position of minimum deviation, the refracted ray inside the prism is ………… to its base:
(a) normal
(b) inclined at 45°
(c) parallel
(d) none of the above
Answer: (c) parallel
7. Mention the incorrect statement:
(a) The angle of deviation produced by a prism depends on the colour or wavelength of the light used.
(b) The angle of incidence also affects the angle of deviation produced by the prism.
(c) In a prism as well as a parallel-sided glass slab, the emergent ray is parallel to the incident ray with a lateral displacement.
(d) The angle of deviation of prism is also affected by the material of prism.
Answer: (c) In a prism as well as a parallel-sided glass slab, the emergent ray is parallel to the incident ray with a lateral displacement.
8. A ray of light is incident normally on the face of an equilateral glass prism. The angle of refraction from the first face of prism is:
(a) 45°
(b) 90°
(c) 60°
(d) 0°
Answer: (d) 0°
9. Assertion (A): If the angles of the base of a prism are equal, then in the position of minimum deviation, the refracted ray will pass parallely to the base of the prism.
Reason (R): For minimum deviation, angle of incidence is equal to the angle of emergence.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (a) both A and R are true and R is the correct explanation of A
Very Short Answer Type Questions
1. Complete the following sentence: Angle of deviation is the angle which the ………… ray makes with the direction of ………… ray.
Answer: Angle of deviation is the angle which the emergent ray makes with the direction of the incident ray.
2. 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 colours of light.
Answer: The statement is false.
3. 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 prism?
Answer: (i) With an increase in the wavelength of incident light, the angle of minimum deviation produced by a prism decreases.
(ii) With an increase in the refracting angle of the prism, the angle of minimum deviation produced by a prism increases.
4. Write a relation for the angle of deviation (δ) for a ray of light passing through an equilateral prism in terms of the angle of incidence (i₁), angle of emergence (i₂), and angle of prism (A).
Answer: The relation for the angle of deviation (δ) is δ = (i₁ + i₂) – A.
5. Name the colour of white light which is deviated (i) the most, and (ii) the least, on passing through a prism.
Answer: (i) Violet light is deviated the most.
(ii) Red light is deviated the least.
6. Which of the two prisms, A made of crown glass and B made of flint glass, deviates a ray of light more?
Answer: Prism B, made of flint glass, deviates a ray of light more.
Short Answer Type Questions
1. Define the term angle of deviation.
Answer: The angle of deviation is the angle between the direction of the incident ray (produced forward) and the emergent ray (produced backward) when a light ray passes through a prism.
2. How does the deviation produced by a prism depend on (i) the refractive index of its material, and (ii) the wavelength of incident light?
Answer: (i) The deviation produced by a prism increases with an increase in the refractive index of its material. A prism with a higher refractive index produces greater deviation than a prism with a lower refractive index, for a given angle of incidence.
(ii) The deviation produced by a prism decreases with an increase in the wavelength of incident light. For visible light, the refractive index of the material of a prism is maximum for violet colour and minimum for red colour. Consequently, a given prism deviates violet light the most and red light the least.
3. 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. (i) How is the angle of emergence ‘i₂’ related to the angle of incidence ‘i₁’? (ii) What can you say about the angle of deviation in such a situation?
Answer: (i) In this situation, the angle of emergence ‘i₂’ is equal to the angle of incidence ‘i₁’.
(ii) In such a situation, the angle of deviation is minimum.
4. A light ray of yellow colour is incident on an equilateral glass prism at an angle of incidence equal to 48° and suffers minimum deviation by an angle of 36°. (i) What will be the angle of emergence? (ii) If the angle of incidence is changed to (a) 30°, (b) 60°, state in each case whether the angle of deviation will be equal to, less than or more than 36°?
Answer: (i) The angle of emergence will be 48°.
(ii) If the angle of incidence is changed to (a) 30°, the angle of deviation will be more than 36°. If the angle of incidence is changed to (b) 60°, the angle of deviation will be more than 36°.
5. How does the angle of deviation depend on the refracting angle of the prism?
Answer: The angle of deviation (δ) increases with an increase in the refracting angle of the prism (A).
Long Answer Type Questions
1. What is a prism? With the help of a diagram of the principal section of a prism, indicate its refracting surfaces, refracting angle and base.
Answer: A prism is a transparent medium bounded by five plane surfaces with a triangular cross section.
In the principal section of a prism, which is a triangle (e.g., ABC), the two rectangular surfaces (e.g., ABED and ACFD) which are polished act as the refracting surfaces. The angle between these two refracting surfaces (e.g., ∠BAC) is the refracting angle, or angle of prism, denoted by A. The rectangular surface opposite to the refracting edge (e.g., BCFE) is the base of the prism, which is usually grounded or made rough.
2. Diagrams (a) and (b) in the figure below show the refraction of a ray of light of single colour through a prism and a parallel sided glass slab respectively.
(i) In each diagram, label the incident, refracted, emergent rays and the angle of deviation.
(ii) In what way the direction of the emergent ray in the two cases differ with respect to the incident ray? Explain your answer.
Answer: (i) For a prism (diagram a), the incident ray enters the first refracting surface, bends towards the normal (if entering a denser medium), travels inside as the refracted ray, strikes the second refracting surface, and then refracts again (bending away from the normal if emerging into a rarer medium) to exit as the emergent ray. The angle of deviation is the angle between the direction of the produced incident ray and the direction of the produced emergent ray.
For a parallel-sided glass slab (diagram b), the incident ray enters the first surface, refracts, travels inside as the refracted ray, strikes the second parallel surface, and refracts again to exit as the emergent ray. The emergent ray is parallel to the incident ray but laterally displaced. The angle of deviation (angular) is zero.
(ii) In the case of a prism, the refraction of light occurs at two inclined faces, so the emergent ray is not parallel to the incident ray, but it is deviated towards the base of the prism.
In the case of a parallel-sided glass slab, the refraction of light occurs at two parallel faces, so the emergent ray is parallel to the incident ray with a lateral displacement.
3. What do you understand by the deviation produced by a prism? Why is it caused? State three factors on which the angle of deviation depends.
Answer: The deviation produced by a prism is the angle between the direction of the incident ray (produced forward) and the emergent ray (produced backward) when a light ray passes through the prism.
It is caused because a ray of light suffers refraction at two inclined faces of the prism. In each refraction, the ray bends towards the base of the prism. Three factors on which the angle of deviation depends are:
(1) The angle of incidence (i).
(2) The material of the prism (i.e., its refractive index μ).
(3) The angle of the prism (A).
4. (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.
(b) Using the curve in part (a) above, how would you infer that for a given prism, the angle of minimum deviation δmin is unique for light of a given wavelength?
Answer: (a) As the angle of incidence increases, the angle of deviation produced by a prism first decreases, reaches a minimum value (δmin) for a certain angle of incidence, and then on further increasing the angle of incidence, the angle of deviation begins to increase.
(b) Using the i-δ curve, it can be inferred that for a given prism, the angle of minimum deviation δmin is unique for light of a given wavelength because there is only one lowest point on the curve. This means only one horizontal line can be drawn parallel to the i-axis that touches the curve at its minimum point, corresponding to a single value of δmin. This unique minimum deviation occurs at a specific angle of incidence where the angle of incidence equals the angle of emergence.
5. 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.
Answer: When a monochromatic ray passes through a prism and suffers minimum deviation, the refracted ray inside the prism is parallel to its base (if the prism is equilateral or isosceles). In this position, the angle of emergence (i₂) is equal to the angle of incidence (i₁).
6. An object is viewed through a glass prism with its vertex pointing upwards. Draw a ray diagram to show the formation of its image as seen by the observer on the other side of the object.
Answer: When an object is viewed through a glass prism with its vertex pointing upwards, the rays of light from the object, after passing through the prism, bend towards the base of the prism. The emergent rays appear to come from a virtual image that is shifted upwards, towards the vertex of the prism, relative to the object’s actual position.
7. 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?
Answer: (a) The angle of incidence on the first face of the prism is 0°.
(b) The angle of refraction from the first face of the prism is 0°.
(c) For an equilateral prism, the angle of prism A = 60°. Since r₁ = 0°, and A = r₁ + r₂, the angle of incidence at the second face, r₂, will be 60°.
(d) No, the light ray will not suffer minimum deviation by the prism.
8. Figure 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.
[Hint : The emergent ray out of the prism B will be parallel to the incident ray PQ]
Answer: The completed diagram will show the ray PQ entering prism A, undergoing refraction, then entering prism B, undergoing further refraction, and finally emerging from prism B. The emergent ray from prism B will be parallel to the incident ray PQ.
Numericals
1. A ray of light incident at an angle of incidence 48° on a prism of refracting angle 60° suffers minimum deviation. Calculate the angle of minimum deviation.
[Hint : δ_min = 2i – A]
Answer: Given:
- Angle of incidence (i) = 48°
- Refracting angle of the prism (A) = 60°
The condition is that the ray suffers minimum deviation.
For minimum deviation, the relationship between the angle of minimum deviation (δ_min), the angle of incidence (i), and the angle of the prism (A) is given by:
δ_min = 2i – A
Substituting the given values in the formula:
δ_min = 2(48°) – 60°
=> δ_min = 96° – 60°
=> δ_min = 36°
Therefore, the angle of minimum deviation is 36°.
2. What should be the angle of incidence for a ray of light which suffers minimum deviation of 36° through an equilateral prism ?
[Hint : A = 60°, i = (A + δ_min)/2]
Answer: Given:
- Angle of minimum deviation (δ_min) = 36°
- The prism is an equilateral prism, which means its refracting angle (A) is 60°.
For a ray of light suffering minimum deviation, the angle of incidence (i) is related to the angle of the prism (A) and the angle of minimum deviation (δ_min) by the formula:
i = (A + δ_min) / 2
Substituting the given values in the formula:
i = (60° + 36°) / 2
=> i = 96° / 2
=> i = 48°
Therefore, the angle of incidence for the ray of light should be 48°.
Exercise C
MCQs
1. Refractive index of a liquid can be found using :
(a) µ = real depth x apparent depth
(b) µ = apparent depth/real depth
(c) µ = real depth/apparent depth
(d) none of the above
Answer: (c) µ = real depth/apparent depth
2. 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.
Answer: (a) refraction of light
3. 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) green light.
Answer: (b) violet light
4. The shift by which the object appears to be raised depends on :
(a) the refractive index of the medium
(b) the thickness of denser medium
(c) the wavelength of incident light
(d) all of the above
Answer: (d) all of the above
5. An object when placed in a rarer medium and is viewed from a denser medium appears to be at a _________ distance than its real distance.
(a) smaller
(b) greater
(c) same
(d) cannot say
Answer: (b) greater
6. A glass slab is placed over a piece of paper on which VGYR is printed with letter V in violet, G in green, Y in yellow and R in red colour. The letter of colour _________ appears to be the most raised.
(a) red
(b) green
(c) yellow
(d) violet
Answer: (d) violet
7. Assertion (A): The stars twinkle while the planets do not.
Reason (R): Stars are more distant than the planets.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (b) both A and R are True and R is not the correct explanation of A
8. Assertion (A): An empty test tube placed in water in a beaker with its mouth outside the water surface appears silvery when viewed from a suitable direction.
Reason (R) : The substance in water appears silvery due to refraction of light.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (d) assertion is true but reason is false
9. Assertion (A): A ray of light travelling from a rarer medium to a denser medium, slows down and bends away from the normal.
Reason (R) : The speed of light is higher in a rarer medium than in a denser medium.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (c) assertion is false but reason is true
Short Answer Type Questions
1. How is the refractive index of a medium related to the real and apparent depths of an object in that medium?
Answer: The refractive index of a medium with respect to air (aµm) is related to the real and apparent depths by the formula: aµm = real depth / apparent depth.
2. An object placed in one medium when seen from the other medium, appears to be vertically shifted. Name two factors on which the magnitude of the shift depends and state how does it depend on them.
Answer: The shift by which the object appears to be raised depends on the following factors:
(1) The refractive index of the medium: Higher the refractive index of the medium, more is the shift.
(2) The thickness of the denser medium: For a given medium, shift is directly proportional to the thickness of medium. Thicker the medium, more is the shift.
(3) The colour (or wavelength) of incident light: The shift decreases with the increase in the wavelength of the light used. Since µV > µR, therefore the shift is more for violet light than for red light in a given medium.
Long Answer Type Questions
1. Prove that, refractive index = real depth / apparent depth
Answer: Consider a point object O kept at the bottom of a transparent medium, such as water or glass, separated from air by the surface PQ. A ray of light OA, starting from the object O, is incident on the surface PQ normally, so it passes undeviated along the path AA’. Another ray OB, starting from the object O, strikes the boundary surface PQ at B and suffers refraction. Since the ray travels from a denser medium to a rarer medium, it bends away from the normal N’BN.
When viewed by the eye, the ray BC appears to be coming from a point I which is the virtual image of O, obtained on producing A’A and CB backwards. Thus, the object at O will appear to be at I, which is at a lesser depth (AI) than its actual depth (AO).
For the incident ray OB, the angle of incidence is i = ∠OBN’ and the angle of refraction is r = ∠CBN.
Since AO and BN’ are parallel and OB is a transversal line, ∠AOB = ∠OBN’ = i.
Similarly, IA’ and BN are parallel and IC is the transversal line, so ∠BIA’ = ∠CBN = r.
In the right-angled triangle BAO, sin i = BA/OB.
In the right-angled triangle IAB, sin r = BA/IB.
For refraction from the medium to air, by Snell’s law, the refractive index of air with respect to the medium (mµa) is given by:
mµa = sin i / sin r = (BA/OB) / (BA/IB) = IB/OB
The refractive index of the medium with respect to air (aµm) is the reciprocal of mµa.
aµm = 1 / mµa = OB / IB
When the object is viewed from a point vertically above it, point B is very close to point A. Therefore, we can approximate IB ≈ IA and OB ≈ OA.
Hence, aµm = OA / IA.
Since OA is the real depth and IA is the apparent depth,
Refractive index = real depth / apparent depth.
2. A tank of water is viewed normally from above. (a) State how does the depth of the tank appear to change. (b) Draw a labelled ray diagram to explain your answer.
Answer: (a) When a tank of water is viewed normally from above, it appears to be shallower than its actual depth. An object placed in a denser medium, when viewed from a rarer medium, appears to be at a depth lesser than its real depth. This is because of the refraction of light.
(b) The ray diagram explains this phenomenon. Consider a point object O at the bottom of the tank. A ray of light OA starting from O travels normally to the surface and passes undeviated. Another ray OB from O strikes the water-air surface at B and refracts away from the normal, travelling along BC. When viewed from above, the ray BC appears to come from a point I, which is the virtual image of O. This image I is formed at a lesser depth (apparent depth) than the actual depth of the object O.
3. Water in a pond appears to be only three-quarter 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 ?
Answer: (a) The property of light responsible for this observation is the refraction of light. When light travels from a denser medium (water) to a rarer medium (air), it bends away from the normal. This makes the bottom of the pond appear raised, and thus the pond appears shallower than it actually is. For water, the refractive index is 4/3, which is why the apparent depth is three-fourths of the real depth.
The illustration is shown in a ray diagram where an object O at the real depth of the pond is viewed from air. Rays of light from O, after refracting at the water surface, appear to diverge from a virtual image I, which is at the apparent depth.
(b) The refractive index of water is calculated from its real and apparent depths using the formula:
Refractive index = Real depth / Apparent depth
4. Draw a ray diagram to show the appearance of a stick partially immersed in water. Explain your answer.
Answer: When a straight stick is placed obliquely in water, the portion of the stick under water appears to be shortened and raised up. This is due to the refraction of light from water to air.
The rays of light coming from the tip P of the stick, when they cross the water-air interface, bend away from the normal. These rays appear to be coming from a point P’, which is the virtual image of the point P. The same is true for every point of the stick inside the water. Thus, the immersed part of the stick PO appears to be P’O. This makes the immersed part of the stick appear raised and therefore bent at the point O on the surface of the water. The entire stick XOP appears as XOP’.
5. A fish is looking at a 1.0 m high plant at the edge of a pond. Will the plant appear to the fish shorter or taller than its actual height ? Draw a ray diagram to support your answer.
Answer: The plant will appear taller than its actual height to the fish.
An object placed in a rarer medium (air) when viewed from a denser medium (water) appears to be at a greater distance than its real distance. In the ray diagram, the top of the plant is at A and is viewed from inside the water by the fish at O. The rays of light from A bend towards the normal as they enter the water at P. To the observer (the fish) in the water, these rays appear to come from a point Q, which is higher than the point A. Therefore, the plant appears taller.
6. A student puts his pencil into an empty trough and observes the pencil from the position as indicated in the figure. (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 stated observation. (iii) Complete the diagram showing how the student’s eye sees the pencil through water.
Answer: (i) When water is poured into the trough, the part of the pencil immersed in water will appear to be raised and bent at the surface of the water.
(ii) The phenomenon that accounts for this observation is the refraction of light.
(iii) To complete the diagram, rays of light are drawn from a point on the pencil tip inside the water. These rays travel from the denser medium (water) to the rarer medium (air). At the water surface, the rays bend away from the normal. When these refracted rays are extended backwards, they appear to meet at a point above the actual position of the pencil tip. This point is the virtual image of the pencil tip. The eye sees this raised virtual image, making the pencil appear bent.
Numericals
1. A water pond appears to be 3 m deep. If the refractive index of water is 4/3, find the actual depth of the pond.
Answer: Given:
Apparent depth = 3 m
Refractive index of water (μ) = 4/3
We know the formula relating refractive index, real depth, and apparent depth:
Refractive index (μ) = Real depth / Apparent depth
To find the actual depth (Real depth), we can rearrange the formula:
Real depth = Refractive index (μ) × Apparent depth
Substituting the given values:
Real depth = (4/3) × 3 m
=> Real depth = 4 m
Thus, the actual depth of the pond is 4 m.
2. A coin is placed at the bottom of a beaker containing water (refractive index = 4/3) at a depth of 16 cm. By what height the coin appears to be raised when seen from vertically above?
Answer: Given:
Real depth of the coin = 16 cm
Refractive index of water (μ) = 4/3
The height by which the coin appears to be raised is the shift. First, let’s calculate the apparent depth.
Apparent depth = Real depth / Refractive index (μ)
Substituting the given values:
Apparent depth = 16 cm / (4/3)
=> Apparent depth = 16 × (3/4) cm
=> Apparent depth = 12 cm
Now, we can find the shift:
Shift = Real depth – Apparent depth
Shift = 16 cm – 12 cm
=> Shift = 4 cm
Alternatively, using the direct formula for shift:
Shift = Real depth × (1 – 1/μ)
Shift = 16 × (1 – 1/(4/3))
=> Shift = 16 × (1 – 3/4)
=> Shift = 16 × (1/4)
=> Shift = 4 cm
Thus, the coin appears to be raised by a height of 4 cm.
3. A postage stamp kept below a rectangular glass slab of refractive index 1.5 when viewed from vertically above it, appears to be raised by 7.0 mm. Calculate the thickness of the glass slab.
Answer: Given:
Shift = 7.0 mm
Refractive index of the glass slab (μ) = 1.5
The thickness of the glass slab is its real depth. Let the thickness (real depth) be ‘t’.
We know the formula for shift:
Shift = Real depth × (1 – 1/μ)
Substituting the given values and ‘t’ for real depth:
7.0 mm = t × (1 – 1/1.5)
We can write 1.5 as 3/2. So, 1/1.5 is 2/3.
7.0 = t × (1 – 2/3)
=> 7.0 = t × (1/3)
Now, we can solve for t:
t = 7.0 × 3
=> t = 21.0 mm
The thickness can also be expressed in cm:
t = 2.1 cm
Thus, the thickness of the glass slab is 21.0 mm or 2.1 cm.
Exercise D
MCQs
1. At angle of incidence equal to the critical angle, the angle of refraction is :
(a) 0°
(b) 30°
(c) 90°
(d) 180°
Answer: (c) 90°
2. The condition required for total internal reflection to take place is :
(a) light ray passes from a rarer to denser medium.
(b) light ray passes from a denser to rarer medium.
(c) angle of incidence is less than critical angle.
(d) none of the above.
Answer: (b) light ray passes from a denser to rarer medium.
3. A scientist while performing an experiment desires that no incident light escapes from the top of a glass block. He varies the angle X continuously and finally manages to achieve his objective. Which of the following options describes the correct answer?
(a) Decreased Total internal reflection
(b) Decreased Total internal refraction
(c) Increased Total internal refraction
(d) Increased Total internal reflection
Answer: (d) Increased Total internal reflection
4. When a ray of light travels from substance A to substance B and reaches their boundary, it suffers total internal reflection. Out of the following possibilities, which one is correct?
(a) i < C optical density of A is less than B.
(b) i = C optical density of A is more than B.
(c) i > C optical density of A is less than B.
(d) i > C optical density of A is more than B.
Answer: (d) i > C optical density of A is more than B.
5. An equilateral prism can be used to deviate a ray of light through :
(a) 30°
(b) 60°
(c) 75°
(d) 90°
Answer: (b) 60°
6. Which of the following figures will depict deviation of ray of light A through 180°
Answer: (4)
7. The diagram below shows a light source P embedded in a rectangular glass block ABCD of critical angle 42°. The ray PQ would :
(a) pass undeviated
(b) suffer refraction
(c) suffer reflection
(d) suffer total internal reflection
Answer: (d) suffer total internal reflection
8. In the process of total internal reflection, …….. of energy (or intensity) of light is reflected back.
(a) 70%
(b) 80%
(c) 100%
(d) 50%
Answer: (c) 100%
9. Out of violet, blue, yellow and red, the critical angle would be the least for …….. light for a pair of media
(a) violet
(b) blue
(c) blue
(d) yellow
Answer: (a) violet
Very Short Answer Type Questions
1. How is the critical angle related to the refractive index of a medium?
Answer: The critical angle C is related to the refractive index µ of a medium by the relation sin C = 1/µ.
2. State the approximate value of the critical angle for (a) glass-air surface (b) water-air surface.
Answer: The approximate value of the critical angle for (a) glass-air surface is 42° and (b) water-air surface is 49°.
3. 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?
Answer: The angle of refraction for the ray is 90°.
4. 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, and (ii) blue light?
Answer: For (i) red light, the critical angle will be more than 45°, and for (ii) blue light, it will be less than 45°.
5. Which colour of light has a higher critical angle? Red light or green light.
Answer: Red light has a higher critical angle than green light.
6. Fill in the blanks to complete the following sentences:
(a) Total internal reflection occurs only when a ray of light passes from a ………… medium to a ………… medium.
(b) Critical angle is the angle of ………… in denser medium for which the angle of ………… in rarer medium is ………… .
Answer: (a) Total internal reflection occurs only 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 90°.
7. 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.
Answer: True.
Short Answer Type Questions
1. What is meant by the statement ‘the critical angle for diamond is 24°’?
Answer: The statement ‘the critical angle for diamond is 24°’ means that at an angle of incidence of 24° in diamond (denser medium), the angle of refraction in air (rarer medium) is 90°. If the angle of incidence in diamond is more than 24°, light will suffer total internal reflection.
2. Name two factors which affect the critical angle for a given pair of media. State how do the factors affect it.
Answer: The critical angle for a given pair of media depends on their refractive indices. But the refractive index of a medium is affected by the following two factors:
(1) the colour (or wavelength) of light
(2) temperature.
The refractive index of a transparent medium decreases with the increase of wavelength of light, therefore the critical angle for a pair of media is the least for violet light and the most for red light i.e. the critical angle increases with the increase in wavelength of light. On increasing the temperature of a medium, its refractive index decreases, therefore the critical angle increases with the increase in temperature.
3. 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 the angles i and r.
(b) If angle r = 90°, what is the corresponding angle i called?
(c) What is the physical significance of the angle i in part (b)?
Answer: (a) A similar expression for aµg in terms of the angles i (in air) and r (in glass) is aµg = sin i / sin r.
(b) If angle r = 90°, the corresponding angle i (in glass) is called the critical angle C.
(c) The physical significance of the angle i (critical angle C) in part (b) is that if the angle of incidence in the denser medium (glass) is greater than this angle C, light will suffer total internal reflection.
4. The table below gives refractive indices of some substances with respect to air.
| Substance | Water | Turpentine | Glass | Diamond |
| Refractive Index (µ) | 1.33 | 1.47 | 1.58 | 2.41 |
(a) Which medium will have the lowest critical angle and which medium will have the highest critical angle with respect to air?
(b) If the temperature of all the substances is raised by 30°C, what will happen to their critical angle?
Answer: (a) Diamond will have the lowest critical angle and water will have the highest critical angle with respect to air.
(b) If the temperature of all the substances is raised by 30°C, their refractive index will decrease, and therefore their critical angle will increase.
5. Mention one difference between reflection of light from a plane mirror and total internal reflection of light from a prism.
Answer: In reflection of light from a plane mirror, some light is refracted and absorbed, so the reflection is not 100%, whereas in total internal reflection of light from a prism, the entire incident light (100%) is reflected back into the denser medium.
6. State one advantage of using a total reflecting prism as a reflector in place of a plane mirror.
Answer: An advantage of using a total reflecting prism as a reflector in place of a plane mirror is that the image obtained by the use of a total reflecting prism is much brighter than that obtained by using a plane mirror, and the brightness of the image formed by a total reflecting prism always remains unchanged.
Long Answer Type Questions
1. Explain the term critical angle with the aid of a labelled diagram.
Answer: When a ray of light passes from a denser medium to a rarer medium, at a certain angle of incidence i = C, the angle of refraction becomes 90°, i.e., at i = C, r = 90°. The angle C is called the critical angle. Thus, critical angle is the angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90°.
2. (a) What is total internal reflection?
(b) State two conditions necessary for total internal reflection to occur.
(c) Draw diagrams to illustrate total internal reflection.
Answer: (a) When a ray of light travelling in a denser medium, is incident at the surface of a rarer medium at an 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 called total internal reflection.
(b) There are two necessary conditions for total internal reflection:
(1) The light must travel from a denser to a rarer medium.
(2) The angle of incidence must be greater than the critical angle for the pair of media.
(c)
3. The figure below shows 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.
Answer: For ray A, since the angle of incidence (48°) is equal to the critical angle, the refracted ray will go along the surface of water and air. For ray B, since the angle of incidence is less than the critical angle, it will be refracted into air, bending away from the normal.
A ray will suffer total internal reflection when it travels from water to air and its angle of incidence is greater than the critical angle (48°).
4. The figure 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.
Answer: (a) Ray A will be refracted away from the normal. Ray B, incident at the critical angle (48°), will be refracted along the surface. Ray C, incident at an angle greater than the critical angle, will undergo total internal reflection.
(b) Ray A exhibits refraction. Ray B exhibits refraction at the critical angle (grazing emergence). Ray C exhibits total internal reflection.
5. 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 for 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 the image of the object P by the letter P’ when seen vertically from above.
Answer: (a) The special name given to the angle of incidence ∠PQN for the ray PQ is the critical angle.
(b) The angle of refraction for the refracted ray QS is 90°.
(c) If the angle of incidence ∠PQN is increased, total internal reflection occurs.
(d) The ray PR suffers partial reflection and refraction on the water-air surface because its angle of incidence is less than the critical angle.
(e) The refracted ray for the incident ray PR will bend away from the normal as it passes from water to air. The image P’ will appear at a lesser depth than the actual object P.
6. The refractive index of glass is 1.5. From a point P inside a glass slab, draw rays PA, PB and PC incident on the glass-air surface at an angle of incidence 30°, 42° and 60° respectively.
(a) In the diagram show the approximate direction of these rays as they emerge out of the slab.
(b) What is the angle of refraction for the ray PB? (Take sin 42° = 2/3)
Answer: (a) For an angle of incidence of 30° (which is less than the critical angle for glass, approximately 42°), the ray PA will refract into the air, bending away from the normal. For an angle of incidence of 42° (which is the critical angle, as sin C = 1/1.5 = 2/3, so C = 42°), the ray PB will refract along the glass-air surface (grazing emergence). For an angle of incidence of 60° (which is greater than the critical angle), the ray PC will undergo total internal reflection.
(b) Given:
Refractive index of glass, μg = 1.5
We know:
sin(ic) = 1 / μ
Substituting the values:
sin(ic) = 1 / 1.5
= 0.667
So,
ic = 41.8° ≈ 42°
Now, applying:
sin(r) / sin(i) = a / μg
Therefore,
sin(r) = (a / μg) × sin(i)
Substitute i = 42°:
sin(r) = (a / μg) × sin(42°)
Given:
sin(42°) = 2/3
and
a / μg = 3/2
Substituting these values:
sin(r) = (3/2) × (2/3)
= 1
As sin(r) = 1, the angle of refraction r = 90°.
7. A ray of light enters a glass slab ABDC as shown in the figure 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.
Answer: The ray strikes AC at O. Since the angle of incidence at O on the circular surface AC is 42°, which is the critical angle, the ray will travel along the surface AC. When it reaches point C, it will strike the surface CD. If CD is perpendicular to AC, the ray will emerge undeviated. (The exact path after C depends on the geometry of the slab at C which isn’t fully specified beyond the circular arc AC). Assuming the ray travels along AC and then meets the boundary CD, if the incidence is normal, it passes undeviated. If the incidence is oblique, it refracts.
8. What is a total reflecting prism? State three actions that it can produce. Draw a diagram to show one such action of the total reflecting prism.
Answer: A total reflecting prism is a prism having an angle of 90° between its two refracting surfaces and the other two angles each equal to 45°. It is called a total reflecting prism because the light incident normally on any of its faces, suffers total internal reflection inside the prism. Three actions it can produce are:
(a) to deviate a ray of light through 90°,
(b) to deviate a ray of light through 180°, and
(c) to erect an inverted image without producing deviation in its path.
One such action, deviating a ray of light through 90°, is shown in the figure.
9. 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.
Answer: A total reflecting prism can be used to turn a ray of light through 90° as shown in the figure. In this setup, a beam of light is incident normally at the face AB. It passes undeviated into the prism and strikes the face AC at an angle of incidence equal to 45°. Since this is greater than the critical angle for glass-air (approx. 42°), the beam suffers total internal reflection and emerges from face BC, having been deviated by 90°.
One instrument in which such a prism is used is a periscope.
10. A ray of light XY passes through a right angled isosceles prism as shown below in the figure.
(a) What is the angle through which the incident ray deviates and emerges out of the prism?
(b) Name the instrument where this action of prism is put into use.
(c) Which prism surface will behave as a mirror?
Answer: (a) The angle through which the incident ray deviates and emerges out of the prism is 90°.
(b) This action of the prism is used in a Periscope.
(c) The surface AB of the prism behaves as a mirror.
11. Draw a diagram of a right angled isosceles prism which is used to make an inverted image erect.
Answer: A right angled isosceles prism used to make an inverted image erect is shown in the figure below.
12. In the figure, a ray of light PQ is incident normally on the hypotenuse of an isosceles right angled prism ABC.
(a) Complete the path of the ray PQ till 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.
Answer: (a) The ray PQ incident normally on face AC passes undeviated. It strikes face AB at an angle of incidence of 45° and undergoes total internal reflection. Then it strikes face BC at an angle of incidence of 45° and undergoes total internal reflection. Finally, it emerges normally from face AC.
(b) The angle of deviation of the ray PQ is 180°.
(c) This action is used in a Binocular.
13. In the figure, 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.
Answer: (a) At the face AB, the angle of incidence i = 0°. At the face AC, the angle of incidence i = 60°.
(b) At the face AB, the phenomenon is refraction (passes undeviated). At the face AC, the phenomenon is total internal reflection (since 60° > 42°). At the face BC, the phenomenon is refraction (as it emerges into air).
14. Draw a neat labelled ray diagram to show total internal reflection of a ray of light incident normally on one face of a 30°, 90°, 60° prism.
Answer: A ray of light incident normally on the face BC (opposite to the refracting angle 30°) of a 30°, 90°, 60° prism passes undeviated as QR into the prism and it strikes the face AC at an angle of incidence 60° at the point R. Since this angle is greater than the critical angle (42°), the light ray QR suffers total internal reflection at the glass-air interface. The reflected ray RS then strikes face AB at 30°, which is less than critical angle, so it refracts out.
15. Copy the diagram given below and complete the path of the light ray till it emerges out of the prism. The critical angle of glass is 42°. In your diagram mark the angles wherever necessary.
Answer: The ray PQ is incident normally on face AB, so it passes undeviated. It then strikes face AC at an angle of incidence of 60°. Since 60° > 42° (critical angle), the ray undergoes total internal reflection. The reflected ray then strikes face BC normally (angle of incidence 0°), so it emerges undeviated.
16. 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.
Answer: A total reflecting prism (specifically a right-angled isosceles prism) can be used to turn a ray of light through 180°.
This device is used in a binocular.
17. Two isosceles right-angled glass prisms P and Q are placed near each other as shown in the figure. Complete the path of the light ray entering the prism P till it emerges out of the prism Q.
Answer: The light ray enters prism P, undergoes total internal reflection at its hypotenuse, emerges, enters prism Q, undergoes total internal reflection at its hypotenuse, and then emerges from prism Q, parallel to its initial path but displaced.
18. Complete the path of ray PQ through the glass prism ABC shown in the figure till it emerges out of the prism. Given the critical angle of the glass is 42°.
Answer: The ray PQ enters normally so passes undeviated. It strikes face AC at an angle of incidence of 30°. Since 30° < 42° (critical angle), the ray refracts out of face AC, bending away from the normal.
19. The diagram given below shows a light source P embedded in a rectangular glass block ABCD of critical angle 42°. Complete the path of ray PQ till it emerges out of the block. [Write necessary angles]
Answer: The ray PQ strikes the surface at Q with an angle of incidence of 42°. Since this is the critical angle, the ray will travel along the surface of the block (grazing emergence).
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