Selina Concise Solutions for ICSE Class 10 Physics Chapter 4 Refraction Of Light At Plane Surfaces

EXERCISE - 4 (A)

Question 1. What do you understand by refraction of light?

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

In simple words: When light goes from one clear material (like air) to another (like glass), it bends. This bending is called refraction.

📝 Teacher's Note: Explain refraction using a simple example like a pencil in a glass of water. The pencil looks bent because light bends when it goes from water to air.

🎯 Exam Tip: Remember the key idea: light changes direction when it moves from one transparent medium to another. This is the definition of 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).

Answer:

[Diagram: This diagram shows light bending when it goes from air to glass. The incident ray comes from air, hits the surface, and then bends as the refracted ray inside the glass. The normal line is perpendicular to the surface. The angle of incidence (i) is between the incident ray and the normal. The angle of refraction (r) is between the refracted ray and the normal.]

[Diagram: This diagram shows light bending when it goes from glass to air. The incident ray comes from glass, hits the surface, and then bends as the refracted ray inside the air. The normal line is perpendicular to the surface. The angle of incidence (i) is between the incident ray and the normal. The angle of refraction (r) is between the refracted ray and the normal.]

In simple words: These pictures show how light bends. When light goes from air into glass, it bends towards the normal line. When it goes from glass into air, it bends away from the normal line.

📝 Teacher's Note: Emphasize that the normal is an imaginary line perpendicular to the surface. Students often confuse it with the surface itself. Practice drawing these diagrams in class.

🎯 Exam Tip: Make sure to label all parts correctly: incident ray, refracted ray, normal, angle of incidence (i), and angle of refraction (r). Pay attention to the direction of bending (towards or away from normal).

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?

Answer: The ray of light which is incident normally on a plane glass slab passes undeviated. That is such a ray suffers no bending at the surface because here the angle of incidence is 0°. Thus if angle of incidence \( \angle i = 0^\circ \), then the angle of refraction \( \angle r = 0^\circ \). And the angle of deviation of the ray will also be 0°.

In simple words: If light hits a glass slab straight on (at 90 degrees), it does not bend. So, the angle of refraction is 0 degrees. The light also does not change its path, so the angle of deviation is 0 degrees.

📝 Teacher's Note: Explain that "normally" means the light ray hits the surface at a 90-degree angle. In this special case, light does not bend, it just goes straight through.

🎯 Exam Tip: Remember this special case: when light hits normally (angle of incidence = 0°), both the angle of refraction and the angle of deviation are 0°.

Question 4. What is the cause of refraction of light when it passes from one medium to another?

Answer: The refraction of light (or change in the direction of path of light in other medium) occurs because light travels with different speeds in different media. When a ray of light passes from one medium to another, its direction (except for \( \angle i = 0^\circ \)) changes because of change in its speed.

In simple words: Light bends because its speed changes. Light travels at different speeds in different materials. When its speed changes as it enters a new material, it bends.

📝 Teacher's Note: Use an analogy: Imagine running from a road onto sand. You slow down and change direction. Light does something similar when it changes medium.

🎯 Exam Tip: The main reason for refraction is the change in the speed of light. This is the key point to mention for full marks.

Question 5. 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: Air is a rarer medium while water is denser than air with refractive index of 1.33. Therefore when light ray will travel from air to water it will bend towards the normal.

[Diagram: This diagram shows light hitting the surface between air and water. Part of the light reflects back into the air (reflected ray). The other part enters the water and bends towards the normal (refracted ray). The incident ray comes from air. The normal line is perpendicular to the surface.]

In simple words: When light hits the surface between air and water, some light bounces back (reflection). The rest goes into the water and bends (refraction). Water is "denser" than air, so light bends towards the normal line.

📝 Teacher's Note: Explain that both reflection and refraction happen at the same time when light hits a boundary. The "normal" is an imaginary line at 90 degrees to the surface.

🎯 Exam Tip: In your diagram, show both the reflected ray and the refracted ray. Remember to label all parts and show the bending towards the normal when going from air (rarer) to water (denser).

Question 6. 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, wavelength?

Answer: Speed, intensity and wavelength

In simple words: When light goes from one material to another, its speed changes. Its brightness (intensity) also changes. And the length of its waves (wavelength) changes too. But its color (frequency) stays the same.

📝 Teacher's Note: Explain that frequency is like the "color" of light, which does not change when light moves between different materials. But speed and wavelength do change.

🎯 Exam Tip: Remember that frequency is the only property of light that does NOT change during refraction. Speed, intensity, and wavelength all change.

Question 7. State the snell's laws of refraction of light.

Answer: The Snell's laws of refraction 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 to the sine of the angle of refraction is constant for the pair of the given media.
   \( \frac{\sin i}{\sin r} = {^1\mu_2} \)
   where \( {^1\mu_2} \) is known as the refractive index of the second medium with respect to the first medium.

In simple words: Snell's Law tells us how light bends when it goes from one material to another. It has two main rules. First, the incoming light, the bent light, and a special line (normal) all stay on the same flat surface. Second, there's a fixed ratio between the angles of the incoming and bent light, which helps us find the refractive index.

📝 Teacher's Note: Explain that the "normal" is an imaginary line perpendicular to the surface where light hits. Use a ruler and a glass of water to show how light bends. This makes the concept of refraction clear. Emphasize that the refractive index is a measure of how much light bends when it enters a new medium.

🎯 Exam Tip: Remember both laws. For the second law, write the formula \( \frac{\sin i}{\sin r} = \text{constant} \) and mention that this constant is the refractive index. Use the correct symbols for angles and refractive index.

Question 8. Define the term refraction index of a medium. Can it be less than 1?

Answer: The refractive index of a second medium with respect to a 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.
Refractive index of a medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum.

In simple words: Refractive index tells us how much light slows down when it enters a material. It compares the angle of light coming in to the angle of light bending. It is always more than 1 because light travels fastest in empty space (vacuum) and slows down in any material.

📝 Teacher's Note: Explain refractive index as a "speedometer" for light in different materials. The higher the number, the slower light travels in that material. Use the example of running in water versus running in air to illustrate slowing down.

🎯 Exam Tip: Define refractive index clearly using the ratio of sines. State that it is always greater than 1 and explain why (light is slower in any medium than in vacuum). This shows full understanding.

Question 9. How is the refractive index of a medium related to the speed of light in it?

Answer: A denser medium has a higher refractive index. This means the speed of light in such a medium is lower. In comparison, a medium with a lower refractive index allows light to travel faster.

In simple words: If a material has a high refractive index, light travels slowly through it. If it has a low refractive index, light travels fast. Think of it like walking through thick mud (high index, slow) versus walking on a clear path (low index, fast).

📝 Teacher's Note: Use the analogy of a crowded street versus an empty street. Light moves slower in a "crowded" (denser, higher refractive index) medium. This helps students connect density and refractive index to light speed.

🎯 Exam Tip: Remember the inverse relationship: Higher refractive index means lower speed of light. Lower refractive index means higher speed of light. This is a key concept.

Question 10. A light ray passes from to (a) air, (b) glass. In each case state, how does the speed of light change.

Answer:
(a) Air (its refractive index is less than that of water)
(b) Glass (its refractive index is more than that of water)

In simple words: The question is incomplete. It seems to be asking about light passing from water to air or water to glass. If light goes from water to air, it speeds up because air is "lighter" than water. If it goes from water to glass, it slows down because glass is "heavier" than water.

📝 Teacher's Note: Clarify that the question implies light is moving *from* water *to* air or glass. Explain that air is a "rarer" medium than water, so light speeds up. Glass is a "denser" medium than water, so light slows down. Use the terms "rarer" and "denser" simply.

🎯 Exam Tip: When light goes from a denser medium to a rarer medium (like water to air), its speed increases. When it goes from a rarer medium to a denser medium (like water to glass), its speed decreases. This is the core idea.

Question 11. A light ray in passing from water to a medium (a) speeds up (b) slows down. In each case, give one example of the medium.

Answer:
In case (i) the refractive index of the medium 2 is less than the refractive index of medium 1 because the ray of light speeds up.
While in case (ii) the refractive index of the medium 2 is greater than the refractive index of medium 1 due to because here the ray of light slows down.

In simple words: When light speeds up from water, it means it goes into a material that is "lighter" than water. An example is air. When light slows down from water, it means it goes into a material that is "heavier" than water. An example is glass.

📝 Teacher's Note: This question is similar to the previous one. Reinforce the idea that light speeds up when going from a denser to a rarer medium (e.g., water to air) and slows down when going from a rarer to a denser medium (e.g., water to glass). Give clear examples for each case.

🎯 Exam Tip: For "speeds up," give an example like air (refractive index less than water). For "slows down," give an example like glass (refractive index greater than water). This shows you understand the relationship between refractive index and light speed.

Question 12. How does the speed of light change when it passes from glass to water?

Answer:
Refractive index of water, \( \mu_w = 1.33 \)
Refractive index of air, \( \mu_a = 1.0003 \)
Refractive index of glass, \( \mu_g = 1.5 \)
This implies that \( \mu_a < \mu_w < \mu_g \)
The speed of light decreases when it enters from a rarer medium to denser medium and increases when it enters from a denser medium to rarer medium.
Therefore, the speed of light increases when light ray passes from water to air and the speed of light decreases when light ray passes from water to glass.

In simple words: Glass is "denser" than water (it has a higher refractive index). So, when light goes from glass to water, it moves from a "heavy" material to a "lighter" material. This means the speed of light will increase.

📝 Teacher's Note: Explain that glass is optically denser than water. So, when light moves from glass to water, it is moving from a denser medium to a rarer medium. This causes the light to speed up. Use the given refractive index values to show which medium is denser.

🎯 Exam Tip: Compare the refractive indices of glass (1.5) and water (1.33). Since glass has a higher refractive index, it is optically denser. Light speeds up when going from a denser to a rarer medium. State this clearly for full marks.

Question 13. A ray of light is passing from a transparent medium 1 to another transparent medium 2 (i) Speed up (ii) slows down. In each case, state whether the refractive index of medium 2 is equal to, less than or greater than the refractive index of medium 1.

Answer:
In case (i) the refractive index of the medium 2 is less than the refractive index of medium 1 because the ray of light speeds up.
While in case (ii) the refractive index of the medium 2 is greater than the refractive index of medium 1 due to because here the ray of light slows down.

In simple words: If light speeds up when going from medium 1 to medium 2, then medium 2 is "lighter" than medium 1. So, medium 2 has a smaller refractive index. If light slows down, medium 2 is "heavier" than medium 1. So, medium 2 has a larger refractive index.

📝 Teacher's Note: This question tests the core concept of refractive index and light speed. Emphasize that "speeds up" means going to a medium with a lower refractive index, and "slows down" means going to a medium with a higher refractive index. This is a direct relationship.

🎯 Exam Tip: For (i) "speeds up," state that refractive index of medium 2 is LESS than medium 1. For (ii) "slows down," state that refractive index of medium 2 is GREATER than medium 1. This direct comparison is crucial.

Question 14. What do you understand by the statement the refractive index of glass is 1.5 for white light?

Answer: The refractive index of glass is 1.5 for white light means white light travels in air 1.5 times faster than in glass.

In simple words: When we say the refractive index of glass is 1.5, it means light travels 1.5 times slower in glass than it does in empty space (or air, which is very close to empty space). So, light is faster in air than in glass.

📝 Teacher's Note: Explain that refractive index is a ratio of the speed of light in vacuum (or air) to the speed of light in the medium. So, if it's 1.5, it means light is 1.5 times slower in glass than in air. This is a fundamental definition.

🎯 Exam Tip: The key is to state that light travels 1.5 times faster in air (or vacuum) than in glass. This shows understanding of the definition of refractive index. Mentioning "white light" is important as refractive index can vary slightly for different colors.

Question 15. A monochromatic ray of light passes from air to glass. The wavelength of light in air is \( \lambda \), the speed of light in air is \( c \) and in glass is \( v \). If the absolute 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 \( v \):
Refractive index \( \mu = \frac{\text{Speed of light in air (or vacuum)}}{\text{Speed of light in medium}} \)
\( 1.5 = \frac{c}{v} \)
\( \implies c = 1.5v \)
(b) The wavelength of light in glass:
We know that \( \mu = \frac{\text{Wavelength of light in air}}{\text{Wavelength of light in medium}} \)
Let \( \lambda_g \) be the wavelength of light in glass.
\( 1.5 = \frac{\lambda}{\lambda_g} \)
\( \implies \lambda_g = \frac{\lambda}{1.5} \)

In simple words: (a) The refractive index (1.5) tells us that light travels 1.5 times slower in glass than in air. So, the speed in air (\( c \)) is 1.5 times the speed in glass (\( v \)). (b) The wavelength of light also becomes 1.5 times shorter in glass than in air. So, the wavelength in glass is \( \lambda \) divided by 1.5.

📝 Teacher's Note: Explain that refractive index is a ratio of speeds and also a ratio of wavelengths. When light enters a denser medium, its speed decreases, and its wavelength also decreases, but its frequency remains the same. This is a key concept for understanding light behavior.

🎯 Exam Tip: For part (a), use the formula \( \mu = c/v \) and substitute the given values. For part (b), use the formula \( \mu = \lambda_{\text{air}} / \lambda_{\text{glass}} \). Remember that frequency does not change when light passes from one medium to another.

Question 15. (a) The relation between speed of light in air c and in glass v is given by
\( \frac{c}{v} = \mu \) Or \( c = 1.5 \nu \)
(b) The wavelength of light in glass (\( \lambda' \))
\( \mu = \frac{\lambda}{\lambda'} \implies \lambda' = \frac{\lambda}{1.5} \)

Answer:

(a) The relation between speed of light in air c and in glass v is given by
\( \frac{c}{v} = \mu \) Or \( c = 1.5 \nu \)
(b) The wavelength of light in glass (\( \lambda' \))
\( \mu = \frac{\lambda}{\lambda'} \implies \lambda' = \frac{\lambda}{1.5} \)

In simple words: This question asks about how light changes speed and wavelength when it goes from air into glass. The refractive index (μ) tells us how much light bends. It is a ratio of speeds or wavelengths.

📝 Teacher's Note: Explain that light slows down when it enters a denser medium like glass. This change in speed causes it to bend. The refractive index is a measure of this bending. Use an example of a car moving from a road to mud to show how speed changes direction.

🎯 Exam Tip: Remember the formulas for refractive index in terms of speed (\( \mu = c/v \)) and wavelength (\( \mu = \lambda/\lambda' \)). Write down the correct formula and substitute values carefully.

Question 16. For which colour of white light, is the refractive index of a transparent medium (a) the least (b) the most?

Answer:

(a) The least for red colour and (b) the most for violet colour.

In simple words: White light is made of many colours. Each colour bends differently when it passes through a material. Red light bends the least, and violet light bends the most. So, the refractive index is smallest for red and largest for violet.

📝 Teacher's Note: Explain that different colours of light have different wavelengths. This causes them to bend differently when passing through a prism or lens. This is called dispersion. You can show a rainbow as a natural example of dispersion.

🎯 Exam Tip: Remember the order VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). Violet light bends most, red light bends least. So, refractive index is highest for violet and lowest for red.

Question 17. Name two factors on which the refractive index of a medium depends? State how does it depend on the factors stated by you.

Answer:

The two factors on which the refractive index of a medium depends are:

1. Nature of a medium i.e. its optical density (e.g. \( \mu_g = 1.5, \mu_w = 1.33 \)) - Smaller the speed of light in a medium relative to air, higher is the refractive index of that medium.
2. Physical condition such as temperature - with increase in temperature, the speed of light in medium increases, so the refractive index of medium decreases.

In simple words: The refractive index depends on two main things. First, what the material is made of (like glass or water). Second, how hot or cold the material is. If light moves slower in a material, its refractive index is higher. If the material gets hotter, light moves faster, and the refractive index goes down.

📝 Teacher's Note: Explain optical density as how "thick" a medium is for light. A denser medium slows light more. Use the analogy of running through water versus running through air. Also, explain that temperature makes particles move faster and spread out, making it easier for light to pass through.

🎯 Exam Tip: List both factors: nature of the medium (optical density) and temperature. Clearly state how the refractive index changes with each factor (e.g., higher optical density means higher refractive index; higher temperature means lower refractive index).

Question 18. How does the refractive index of a medium depend on the wavelength of light used?

Answer:

Refractive index of a medium decreases with increase in wavelength of light.
Refractive index of a medium for violet light (least wavelength) is greater than that for red light (greatest wavelength).

In simple words: Different colours of light have different wavelengths. Red light has a long wavelength, and violet light has a short wavelength. The refractive index is higher for shorter wavelengths (like violet) and lower for longer wavelengths (like red). This means violet light bends more than red light.

📝 Teacher's Note: Connect this to the previous question about colours. Explain that wavelength is the "size" of the light wave. Shorter waves interact more with the medium, causing more bending and a higher refractive index. This is why a prism separates white light into colours.

🎯 Exam Tip: Remember the inverse relationship: as wavelength increases, refractive index decreases. Specifically mention that violet light (shortest wavelength) has the highest refractive index, and red light (longest wavelength) has the lowest refractive index.

Question 19. How does the refractive index of a medium depend on its temperature?

Answer:

Refractive index of a medium decreases with the increase in temperature.
With increase in temperature, the speed of light in that medium increases; thus, the refractive index (= velocity of light in vacuum/velocity of light in medium) decreases.

In simple words: When a material gets hotter, its particles move faster and spread out a little. This makes the material less dense. Because it's less dense, light can travel through it faster. Since light travels faster, the refractive index of the material goes down.

📝 Teacher's Note: Explain that temperature affects the density of a medium. A hotter medium is less dense. Less dense means light can pass through more easily and quickly. This is similar to how sound travels faster in hotter air.

🎯 Exam Tip: State clearly: "Refractive index decreases as temperature increases." Also, briefly explain why: higher temperature means faster light speed, which leads to a lower refractive index.

Question 20. A ray of light from air suffers partial reflection and refraction at the boundary of water. In Fig. 4.16, which of the ray A, B, C, D and E is the correct
(i) refracted ray
(ii) partially reflected ray?

[Diagram: This diagram shows a light ray hitting the boundary between air and water. One ray (E) reflects back into the air, and other rays (A, B, C, D) go into the water, bending at different angles.]

Answer:

(i) Ray 'B' is the correct refracted ray as a ray of light bends towards the normal while going from rarer to denser medium.
(ii) Ray 'E' is the partially reflected ray, as reflection of light takes place in the same medium.

In simple words: When light hits water from air, some light bounces back (reflection) and some goes into the water (refraction). The light that bounces back stays in the air. The light that goes into the water bends towards the normal line because water is denser than air. Ray B shows the correct bending into water, and Ray E shows the light bouncing back.

📝 Teacher's Note: Draw a normal line (perpendicular to the surface) at the point where the light hits. Explain that when light goes from a "rarer" medium (like air) to a "denser" medium (like water), it bends *towards* the normal. Reflection means the light bounces off the surface and stays in the same medium.

🎯 Exam Tip: For refraction from rarer to denser medium, the refracted ray bends *towards* the normal. For reflection, the ray stays in the *same* medium. Identify these two key characteristics in the diagram.

Question 21. The diagram alongside shows the refraction of a ray of light from sir to a liquid.
(a) write the values of (i) angle of incidence, (ii) angle of refraction.
(b) use snell's law to find the refractive index of liquid with respect to air.

[Diagram: This diagram shows a light ray going from air to a liquid. The incident ray makes an angle of 30° with the surface in air. The refracted ray makes an angle of 45° with the surface in the liquid.]

Answer:

(a)
(a) Angle of incidence is the angle which the incident ray makes with the normal.
\( \therefore i = 90^\circ - 30^\circ = 60^\circ \)
(b) Angle of refraction is the angle which the refracted ray makes with the normal.
\( \therefore r = 90^\circ - 45^\circ = 45^\circ \)

(b) According to Snell's law:
\( \mu = \frac{\sin i}{\sin r} \)
\( \implies \mu = \frac{\sin 60^\circ}{\sin 45^\circ} \)
\( \implies \mu = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} \)
\( \implies \mu = \frac{\sqrt{3}}{2} \times \sqrt{2} \)
\( \implies \mu = \frac{\sqrt{6}}{2} \)
\( \implies \mu = \frac{2.449}{2} \)
\( \implies \mu = 1.2245 \)

In simple words: First, we need to find the angles from the normal line, not from the surface. The angle of incidence is \( 90^\circ - 30^\circ = 60^\circ \). The angle of refraction is \( 90^\circ - 45^\circ = 45^\circ \). Then, we use Snell's Law, which is a formula that connects these angles to the refractive index. We put the sine values of these angles into the formula to find the refractive index.

📝 Teacher's Note: Emphasize that angles of incidence and refraction are ALWAYS measured from the normal, not from the surface. This is a common mistake. Review basic trigonometry for sine values of common angles like \( 60^\circ \) and \( 45^\circ \).

🎯 Exam Tip: Always calculate the angles of incidence and refraction with respect to the normal first. Then, apply Snell's Law: \( \mu = \frac{\sin i}{\sin r} \). Show all calculation steps clearly to avoid losing marks.

Question 22. The refractive index of water with respect to air is a “w and of glass with respect to air is a g. express the refractive index of glass with respect to water.

Answer: Refractive index of glass with respect to water \( \text{w} \mu \text{g} \) is given by \( \text{w} \mu \text{g} = \frac{\text{a} \mu \text{g}}{\text{a} \mu \text{w}} \)
In simple words: This formula helps us find how much light bends when it goes from water to glass. We use how much it bends from air to glass and from air to water.

📝 Teacher's Note: Explain that refractive index tells us how much light slows down and bends when it enters a new material. Use an example like a spoon looking bent in water.

🎯 Exam Tip: Remember the formula for relative refractive index. Write the symbols correctly: \( \text{w} \mu \text{g} \) means refractive index of glass with respect to water.

Question 23. In fig 4.18, name the ray which represents the correct path of light while emerging out through a glass block.

Answer: The correct ray is B

[Diagram: This diagram shows light entering a glass block from air, refracting, and then emerging back into air. There are four possible paths (A, B, C, D) shown for the emergent ray.]

In simple words: When light enters a glass block, it bends. When it leaves the block, it bends again. The ray B shows the correct path where light bends twice. The emergent ray should be parallel to the incident ray but shifted.

📝 Teacher's Note: Draw this diagram on the board. Show how light bends towards the normal when entering glass and away from the normal when leaving glass. Emphasize that the emergent ray is parallel to the incident ray.

🎯 Exam Tip: For light passing through a glass block, the emergent ray is always parallel to the incident ray. Ray B correctly shows this parallel shift.

Question 24. A ray of light strikes the surface of a rectangular glass block such that the angle of incidence in air is (i) 0°, (ii) 45°. In each case, draw diagram to show the path taken by the ray as it passes through the glass block and emerges from it.

Answer:
(i) Case (i) when angle of incidence is 0°.

[Diagram: This diagram shows a light ray hitting a glass block at 0° (straight on). The ray passes straight through without bending.]

(ii) Case (ii) When angle of incidence is 45°.

[Diagram: This diagram shows a light ray hitting a glass block at 45°. The ray bends towards the normal inside the glass, then bends away from the normal when leaving, becoming parallel to the original ray.]

In simple words: When light hits a glass block straight on (0 degrees), it goes straight through. When it hits at an angle, it bends. It bends when it enters and bends again when it leaves. The final ray comes out parallel to the first ray.

📝 Teacher's Note: Use a laser pointer and a glass slab to show these two cases in class. Students can clearly see the path of light. Explain that light does not bend when it enters straight (normally).

🎯 Exam Tip: Remember two key points: 1) Light does not refract (bend) when it enters a medium perpendicularly (at 0°). 2) When light passes through a glass slab, the emergent ray is parallel to the incident ray.

Question 25. In the adjacent diagram, AO is a ray of light incident on a rectangular glass block.
(a) complete the path of the ray till it emerges out of the block.
(b) In the diagram, mark the angles of incidence (i) and the angle of refractive index of glass related to the angles i and r ?
(c) mark angles 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 complete path of incident ray in glass block is drawn in figure below.

[Diagram: This diagram, which is missing from the source, would show a light ray AO entering a glass block, refracting, and then emerging. The completed path would show the ray bending towards the normal inside the glass and then bending away from the normal when exiting, parallel to the incident ray.]

In simple words: Light bends when it enters the glass block and bends again when it leaves. The diagram would show this bending path.

📝 Teacher's Note: This question requires students to draw the path of light. Guide them to draw the normal at each surface and apply Snell's law (or simply "bend towards/away from normal").

🎯 Exam Tip: When asked to complete a ray diagram for a glass block, always draw the normal at the point of incidence and emergence. Show the bending of light correctly. The emergent ray should be parallel to the incident ray.

Question 26: A ray of monochromatic green light enters a liquid from air, as shown in Fig 4.20. The angle 1 is 45° and angle 2 is 30°.

[Diagram: This diagram shows a light ray entering a liquid from air, bending as it enters. It hits a plane mirror inside the liquid. Angle 1 is the angle of incidence in air, and angle 2 is the angle of refraction in the liquid.]

(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 where ever necessary.
(c) Redraw the diagram if plane mirror becomes normal to the refracted ray inside the liquid. State the principal used.

Answer:
(a) Refractive index of the liquid = \( \frac{\sin i}{\sin r} \)
Here, angle of incidence \( i = 45^\circ \) (angle 1) and angle of refraction \( r = 30^\circ \) (angle 2).
So, refractive index \( \mu = \frac{\sin 45^\circ}{\sin 30^\circ} \)
\( \implies \mu = \frac{1/\sqrt{2}}{1/2} \)
\( \implies \mu = \frac{1}{\sqrt{2}} \times 2 \)
\( \implies \mu = \frac{2}{\sqrt{2}} \)
\( \implies \mu = \sqrt{2} \)
\( \implies \mu = 1.414 \)

(b)

[Diagram: This diagram shows the path of light. A ray enters from air into liquid, bends, hits a mirror, reflects, and then bends again as it leaves the liquid and goes back into the air. All angles are marked.]

The ray reflects from the plane mirror and then refracts back into the air. The angles are marked in the diagram.

(c)

[Diagram: This diagram shows a light ray entering a liquid from air. The mirror is placed straight (normal) to the refracted ray. The light ray hits the mirror at 90 degrees and reflects back along the same path.]

Here the 'Principle of Reversibility' is used. This principle says that if a light ray travels from point A to point B, it can also travel from point B to point A along the exact same path.

📝 Teacher's Note: Explain Snell's Law clearly. Show how light bends when it goes from air to water. Use a pencil in a glass of water as an example. Emphasize that the angle of incidence and refraction are measured from the normal line.

🎯 Exam Tip: For part (a), remember the formula for refractive index: \( \mu = \frac{\sin i}{\sin r} \). For part (b), draw the reflected and refracted rays correctly. For part (c), mention the 'Principle of Reversibility' for full marks.

Question 27: 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 liquid becomes equal to the refractive index of glass. When both have the same refractive index, light passes through them without bending or reflecting, making the glass invisible.

(ii) Light of a single colour (monochromatic light) is used because the refractive index of a medium (like glass or liquid) is different for different colours of light. If white light (many colours) were used, different colours would bend differently, and the glass might still be partly visible.

📝 Teacher's Note: Explain that refractive index tells us how much light bends. If two things have the same refractive index, light doesn't bend when it goes from one to the other. Think of clear glass in clear water – if they were exactly the same, you wouldn't see the glass!

🎯 Exam Tip: For part (i), the key phrase is "refractive index of liquid becomes equal to the refractive index of glass." For part (ii), mention that "refractive index is different for different colours" to get full marks.

Question 28. When a lighted candle is held in front of a thick plane glass mirror, several images can be seen, but the second image is the brightest, give reason.

Answer: When a ray of light from lighted candle falls on the surface of a thick plane glass mirror, a small part of light (nearly 4%) is reflected forming first image which is faint virtual image, while a large part of light (nearly 96%) is refracted inside the glass. This ray is now strongly reflected back by the silvered surface inside the glass. This ray is then partially refracted in air and this refracted ray forms another virtual image. This image is the brightest image because it is due to the light suffering a strong reflection at the silver surface.
In simple words: When light hits a thick mirror, some light reflects from the front (first image, faint). Most light goes inside and hits the silvered back. This light reflects strongly from the back, making the second image very bright.

📝 Teacher's Note: Explain that a thick mirror has two surfaces: the front glass and the silvered back. The silvered back reflects light much better. You can use the example of looking at a shop window at night; you see your faint reflection (front glass) and also the bright reflections of things inside the shop (back surface). This is similar to how the second image is brighter.

🎯 Exam Tip: Mention that the second image is bright because of "strong reflection from the silvered surface" inside the mirror. This is the key point.

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 ________.

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
(c) The refractive index of glass with respect to air is 3/2. The refractive index of air with respect to glass will be 2/3
In simple words: Light slows down when it goes into a thicker material (denser medium). It speeds up when it goes into a thinner material (rarer medium). The refractive index for going one way is the inverse (opposite fraction) of going the other way.

📝 Teacher's Note: Use the analogy of running in water versus running in air. Water is denser, so you run slower. Air is rarer, so you run faster. For refractive index, explain that if glass is 3/2 times "slower" than air, then air is 2/3 times "slower" than glass (which means air is faster). It's like saying if A is twice as tall as B, then B is half as tall as A.

🎯 Exam Tip: Remember that speed "decreases" in denser medium and "increases" in rarer medium. For refractive index, if you reverse the path of light, you just flip the fraction (take the reciprocal).

Multiple Choice Type:

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

Answer: (b) bends towards the normal
Reason: As the speed of light decreases in the denser medium, it bends towards the normal.
In simple words: When light goes from air (thin) into a denser material like water (thick), it slows down. This makes it bend closer to the imaginary line called the normal.

📝 Teacher's Note: Draw a simple diagram on the board showing a light ray entering water from air. Draw the normal line. Show how the ray bends towards the normal. You can compare it to a car moving from a smooth road onto mud at an angle; the wheel that hits the mud first slows down, making the car turn.

🎯 Exam Tip: The key phrase is "bends towards the normal" when light enters a denser medium. Remember this rule for refraction.

Question 2. A light ray does not bend at the boundary in passing from one medium to the other medium if the angle of incident is:
(a) 0° (b) 45°(c) 60°(d) 90°

Answer: (a) 0°
Reason: A ray of light which is incident normally (i.e. at angle of incidence = 0°) on the surface separating the two media, passes undeviated.
In simple words: If light hits a surface straight on (at 0 degrees to the normal), it does not bend. It goes straight through.

📝 Teacher's Note: Explain that "angle of incidence = 0°" means the light ray is exactly perpendicular to the surface, or along the normal line. When light travels along the normal, it doesn't change direction. Use the example of a stick poked straight down into water; it looks straight, not bent.

🎯 Exam Tip: Remember that light does not bend (goes undeviated) only when it hits the surface at an angle of incidence of 0 degrees (normally).

Question 3. The highest refractive index is of:
(a) Glass (b) Water
(c) Diamond (d) Ruby

Answer: (c) Diamond
Reason: As the speed of light in diamond is the least, diamond has the highest refractive index.
In simple words: Diamond slows down light the most compared to other materials. When light slows down a lot, the material has a high refractive index.

📝 Teacher's Note: Explain that a higher refractive index means light bends more when it enters that material, and it also means light travels slower in that material. Diamond has a very high refractive index, which is why it sparkles so much.

🎯 Exam Tip: Diamond has the highest refractive index among common materials. This is a factual recall question. Remember that higher refractive index means slower light speed and more bending.

Numericals:

Question 1. The speed of light in air is \( 3 \times 10^8 \) 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, \( c = 3 \times 10^8 \) m/s
Refractive index of glass, \( \mu = 1.5 \)
Speed of light in glass, \( v = ? \)
We know that,
\( \frac{c}{v} = \mu \)
\( \implies v = \frac{c}{\mu} \)
\( \implies v = \frac{3 \times 10^8 \text{ m/s}}{1.5} \)
\( \implies V = 2 \times 10^8 \text{ m/s} \)
In simple words: We know how fast light travels in air and how much glass slows it down (refractive index). To find light's speed in glass, we divide its speed in air by the refractive index.

📝 Teacher's Note: This is a direct application of the formula for refractive index. Emphasize that refractive index is a ratio of speeds. Make sure students understand the units (m/s) and how to handle powers of 10.

🎯 Exam Tip: Write down the formula \( \mu = \frac{\text{speed of light in vacuum/air}}{\text{speed of light in medium}} \) clearly. Substitute values correctly and show the calculation steps to get full marks.

Question 2. The speed of light in diamond is 125,000 km s\(^{-1}\). What is its refractive index? (speed of light in air = \( 3 \times 10^8 \) m s\(^{-1}\)).

Answer: Given,
Speed of light in diamond, \( v = 125,000 \) km/s \( = 125 \times 10^6 \) m/s (since 1 km = 1000 m)
Speed of light in air, \( c = 3 \times 10^8 \) m/s
Refractive index of diamond, \( \mu = ? \)
We know that,
\( \mu = \frac{c}{v} \)
\( \implies \mu = \frac{3 \times 10^8 \text{ m/s}}{125 \times 10^6 \text{ m/s}} \)
\( \implies \mu = \frac{300 \times 10^6 \text{ m/s}}{125 \times 10^6 \text{ m/s}} \)
\( \implies \mu = \frac{300}{125} \)
\( \implies \mu = 2.4 \)
In simple words: We are given how fast light travels in diamond and in air. To find the refractive index of diamond, we divide the speed of light in air by its speed in diamond. Remember to use the same units for both speeds.

📝 Teacher's Note: Highlight the unit conversion from km/s to m/s. This is a common mistake. Emphasize that refractive index has no units because it's a ratio of two speeds. Explain that a refractive index of 2.4 means light travels 2.4 times slower in diamond than in air.

🎯 Exam Tip: Always convert all speeds to the same unit (usually m/s) before calculating. Write the formula, substitute values, and show the final answer with no units for refractive index.

\( \therefore \mu = \frac{3 \times 10^8 \text{ m/s}}{125 \times 10^6 \text{ m/s}} = 2.4 \)

📝 Teacher's Note: This calculation shows how to find the refractive index. It uses the speed of light in different mediums. Remember, refractive index tells us how much light bends.

🎯 Exam Tip: For such calculations, write the formula first. Then put the correct values. Show your final answer with units.

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?

Answer:
Given: Refractive index of water w.r.t. air is \( _{a}\mu_{water} = \frac{4}{3} \).
Refractive index of air w.r.t. water is:
\( _{water}\mu_{air} = \frac{1}{_{a}\mu_{water}} \)
\( \implies _{water}\mu_{air} = \frac{1}{4/3} \)
\( \implies _{water}\mu_{air} = \frac{3}{4} \)
\( \implies _{water}\mu_{air} = 0.75 \)
In simple words: If light bends a certain way from air to water, it will bend the opposite way from water to air. So, you just flip the fraction.

📝 Teacher's Note: Explain to students that refractive index is reversible. If you know how light bends from medium A to B, you can find how it bends from B to A by taking the reciprocal. Use an example of a spoon in water.

🎯 Exam Tip: Remember the formula: \( \mu_{BA} = \frac{1}{\mu_{AB}} \). Write this formula to get full marks. Show the reciprocal step clearly.

Question 4: A ray of light of wavelength 5400 Å suffer refraction from air to glass. Taking \( _{a}\mu_{g} = 3/2 \), find the wavelength of light in glass.

Answer:
Given: Refractive index of glass w.r.t. air, \( _{a}\mu_{g} = \frac{3}{2} \).
We know that refractive index is also the ratio of wavelengths:
\( _{a}\mu_{g} = \frac{\text{wavelength of light in air}}{\text{wavelength of light in glass}} \)
Substitute the given values:
\( \frac{3}{2} = \frac{5400 \text{ Å}}{\text{wavelength of light in glass}} \)
Now, rearrange to find the wavelength in glass:
\( \text{Wavelength of light in glass} = \frac{2}{3} \times 5400 \text{ Å} \)
\( \implies \text{Wavelength of light in glass} = 3600 \text{ Å} \)
In simple words: When light goes from air to glass, its speed changes. This also changes its wavelength. The refractive index tells us how much the wavelength changes.

📝 Teacher's Note: Explain that the frequency of light does not change when it refracts, but its speed and wavelength do. Use the analogy of cars slowing down when they enter mud, but the number of cars passing a point per second (frequency) remains the same.

🎯 Exam Tip: Remember the formula linking refractive index and wavelength. Write down the formula first. Show all steps clearly, especially the rearrangement of the equation.

Exercise 4(B)

Question 1: What is a prism? With the help of diagram of a prism, indicate its refracting surfaces, refracting edge and base.

Answer:
A prism is a transparent refracting medium bounded by five plane surfaces inclined at some angle.
[Diagram: This diagram would show a triangular prism. It would label the two triangular faces as bases, the three rectangular faces as refracting surfaces, and the lines where the refracting surfaces meet as refracting edges.]
In simple words: A prism is a clear, solid shape, usually like a triangle. It has flat sides that bend light.

📝 Teacher's Note: Show a real prism in class. Let students touch it. Explain its parts using the real object. Emphasize that it's transparent, meaning light can pass through it.

🎯 Exam Tip: Define a prism clearly. For the diagram, draw a simple prism and label its parts: refracting surfaces, refracting edge, and base. This is important for full marks.

Question 2: The diagrams (a) and (b) in Fig. 4.29 below show the refraction of a monochromatic ray of light through a parallel sided glass block and a prism respectively. In each diagram, label the incident, refracted emergent rays and the angle of deviation.

Answer:
(a) Refraction through a parallel sided glass block:
[Diagram: This diagram shows a rectangular glass block. A light ray (incident ray) enters the block, bends (refracted ray), and then exits parallel to the incident ray (emergent ray). The angle of deviation is shown as zero because the emergent ray is parallel to the incident ray. The diagram also labels the angle of incidence (i) and angle of emergence (e).]
(b) Refraction through a prism:
[Diagram: This diagram shows a triangular prism. A light ray (incident ray) enters the prism, bends inside (refracted ray), and then exits (emergent ray). The diagram clearly shows the angle between the incident ray and the emergent ray, which is the angle of deviation. It also shows lateral displacement in the glass block diagram.]
In simple words: These diagrams show how light bends when it passes through a glass block or a prism. We need to mark the path of light and how much it changes direction.

📝 Teacher's Note: Use a laser pointer and a glass slab/prism to demonstrate this in class. Let students observe how the light ray bends. Explain that for a glass slab, the emergent ray is parallel to the incident ray, so deviation is zero. For a prism, there is a clear angle of deviation.

🎯 Exam Tip: For such questions, draw the diagrams neatly. Use a ruler and pencil. Label all parts correctly: incident ray, refracted ray, emergent ray, and angle of deviation. This is key for full marks.

Question 3: Define the term angle of deviation

Answer:
The angle between the direction of the incident ray and the emergent ray is called the angle of deviation.
In simple words: It is how much the light ray changes its direction after passing through a prism or glass.

📝 Teacher's Note: Draw a simple diagram on the board showing an incident ray and an emergent ray from a prism. Extend the incident ray forward and the emergent ray backward to show the angle between them. This visual aid helps students understand.

🎯 Exam Tip: Use the exact definition provided. Make sure to mention both "incident ray" and "emergent ray" in your answer. This is a direct definition question.

Question 4: Complete the following sentence: When light passes from one medium to another, it changes its ________.

Answer:
When light passes from one medium to another, it changes its speed.
In simple words: Light travels at different speeds in different materials. This change in speed makes it bend.

📝 Teacher's Note: Explain that the change in speed is the main reason for refraction. Use the analogy of a car moving from a road to mud – it slows down and changes direction. This helps students visualize the concept.

🎯 Exam Tip: For this blank, "speed" is a key word. You could also say "direction" if the light hits at an angle. But "speed" is the fundamental cause of the bending.

Question 4. 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 incident ray.

In simple words: The angle of deviation tells us how much the light ray bends. It is the angle between the light ray that comes out (emergent ray) and the direction the light ray was going in before it hit the prism (incident ray).

📝 Teacher's Note: Explain that light changes direction when it goes from one material to another, like air to glass. This change in direction is called deviation. Use a ruler in a glass of water to show how light seems to bend.

🎯 Exam Tip: Remember the two key words: "emergent ray" and "incident ray". Write them correctly to get full marks for this fill-in-the-blank question.

Question 5. 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:
In a prism, the ray of light suffers refraction at two faces. The prism produces a deviation at the first surface and another deviation at the second surface. Thus a prism produces a deviation in the path of light.
The value of the angle of deviation (or the deviation produced by a prism) depends on the following four factors:
(a) the angle of incidence (i),
(b) the material of prism (i.e., on refractive index \( \mu \)),
(c) the angle of prism (A),
(d) The colour or wavelength \( ( \lambda ) \) of light used.

In simple words: When light passes through a prism, it bends. This bending is called deviation. It happens because light changes speed when it enters and leaves the glass prism. The amount of bending depends on how the light hits the prism, what the prism is made of, the prism's shape, and the color of the light.

📝 Teacher's Note: Use a prism to show students how white light splits into colors. Explain that each color bends a different amount. This shows that deviation depends on the color of light. Also, show how changing the angle of the light beam changes how much it bends.

🎯 Exam Tip: For full marks, define deviation clearly. Then list at least three of the four factors. Make sure to mention "angle of incidence," "material of prism," and "angle of prism."

Question 6. How does the angle of deviation produced by a prism change with increase in the angle of incidence. Draw a curve showing the variation in the angle of deviation with the angle of incidence at a prism surface.

Answer:
As the angle of incidence increases, the angle of deviation decreases first and reaches to a minimum value \( ( \delta_m ) \) for a certain angle of incidence. By further increasing the angle of incidence, the angle of deviation is found to increase.
Variation of angle of deviation \( ( \delta ) \) with angle of incidence \( (i) \):

[Diagram: This diagram shows a curve plotting the angle of deviation (on the y-axis) against the angle of incidence (on the x-axis). The curve first goes down to a minimum point (labeled \( \delta_m \)) and then goes up.]

In simple words: Imagine you shine a light on a prism. If you slowly change the angle at which the light hits the prism, the light will bend less and less at first. It will reach a point where it bends the least. If you keep changing the angle even more, the light will start bending more again. The diagram shows this bending pattern.

📝 Teacher's Note: This graph is very important. Explain that the minimum deviation point is special. At this point, the light ray inside the prism is parallel to the prism's base. This is a key concept for understanding prisms.

🎯 Exam Tip: Remember the shape of the curve: it goes down, reaches a minimum, and then goes up. Clearly label the axes as "Angle of deviation \( ( \delta ) \)" and "Angle of incidence \( (i) \)". Mark the minimum deviation point \( ( \delta_m ) \).

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.

Answer:
False.
With the increase in the angle of incidence, the deviation produced by a prism first decreases and then increases.
A given prism deviates the violet light most and the red light least.

In simple words: The statement is wrong. How much light bends in a prism depends on the angle it hits the prism. Also, different colors of light bend by different amounts. Violet light bends the most, and red light bends the least.

📝 Teacher's Note: This question checks two important facts about prisms. Emphasize that deviation is NOT constant. It changes with the angle of incidence and also with the color of light. This is why prisms can split white light into a rainbow.

🎯 Exam Tip: State "False" clearly. Then, give both reasons: how deviation changes with the angle of incidence, and how it changes for different colors (violet bends most, red bends least).

Question 8. How does the angle of minimum deviation produces by a prism change with increase in (i) the wavelength of incident light and (ii) the refracting angle of prism?

Answer:
Changes in the angle of deviation as we increase
(i) The wavelength of incident light
As we increase the wavelength, angle of deviation decreases.
(ii) The refracting angle of the prism
The angle of deviation increases with the increase in the angle of prism.

In simple words: (i) Wavelength is like the "size" of the light wave. Longer wavelength light (like red light) bends less. Shorter wavelength light (like violet light) bends more. So, if the wavelength increases, the bending (deviation) decreases. (ii) The refracting angle is the angle at the top of the prism. If this angle gets bigger, the light bends more. So, deviation increases.

📝 Teacher's Note: Connect wavelength to color. Red light has a longer wavelength than violet light. So, red light deviates less. Explain that a wider prism angle means the light travels through more glass, causing more bending.

🎯 Exam Tip: Remember these two relationships: (i) Wavelength increases \( \implies \) deviation decreases. (ii) Refracting angle increases \( \implies \) deviation increases. Write these clearly for each part.

Question 9. Write a relation for the angle of deviation \( ( \delta ) \) for a ray of light passing through an equilateral prism in terms of the angle of incidence \( (i) \) angle of emergence \( (e) \) and angle of prism \( (A) \).

Answer:
The relation between the angle of incident \( (i) \), angle of emergence \( (e) \), angle of prism \( (A) \) and angle of deviation \( ( \delta ) \) for a ray of light passing through an equilateral prism is
\( \delta = (i+e) – A \)

In simple words: This is a formula that connects how much the light bends (deviation) with the angle it enters the prism (incidence), the angle it leaves the prism (emergence), and the top angle of the prism itself.

📝 Teacher's Note: This formula is fundamental for prisms. Make sure students understand what each symbol represents. It's a key equation to remember and apply in numerical problems.

🎯 Exam Tip: Write the formula exactly as \( \delta = (i+e) – A \). Make sure to use the correct symbols for each angle.

Question 10. 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) \( i_2 = i_1 \)
(ii) Angle of deviation is minimum
Explanation: In minimum deviation position, the refracted ray inside the prism travels parallel to it if the prism is equilateral and the angle of incidence is equal to the angle of emergence.

In simple words: (i) When the light ray inside the prism is parallel to its bottom side, the angle at which light enters the prism (i₁) is the same as the angle at which it leaves (i₂). (ii) In this special case, the light bends the least. So, the angle of deviation is at its minimum value.

📝 Teacher's Note: This describes the condition for minimum deviation. It's a very important concept. Explain that when \( i_1 = i_2 \), the light path is symmetrical, and the deviation is the smallest possible. This is often tested.

🎯 Exam Tip: For part (i), state \( i_2 = i_1 \). For part (ii), state "Angle of deviation is minimum." The explanation is also crucial for understanding, so try to remember it.

Question 11: How is the angle of emergence related to the angle of incidence when prism is in the position of minimum deviation? Illustrate your answer with help of a labelled diagram using an equilateral prism?

Answer:
In case of an equilateral prism, when the prism is in the position of minimum deviation \( \delta = \delta_{min} \), the angle of incidence \( i_1 \) is equal to the angle of emergence \( i_2 \).
\( i_1 = i_2 = i \)

[Diagram: This diagram shows a light ray passing through an equilateral prism. It labels the angle of incidence (\( i_1 \)), angle of emergence (\( i_2 \)), and angle of minimum deviation (\( \delta_{min} \)).]

📝 Teacher's Note: Explain that minimum deviation means the light bends the least. At this special point, the light enters and leaves the prism at the same angle. This is a key property of prisms.

🎯 Exam Tip: Remember that for minimum deviation, the angle of incidence (\( i_1 \)) is equal to the angle of emergence (\( i_2 \)). Draw a clear, labelled diagram to get full marks.

Question 12: A light ray of yellow colour is incident on an equilateral glass prism at an angel 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 changes to (a) 30°, (b) 60°, state whether the angle of deviation will be equal to, less than or more than 36° ?

Answer:
(i) As the ray is suffering minimum deviation in an equilateral glass prism so
\( i_1 = i_2 \)
\( \implies i_2 = 48^\circ \)
(ii) If the angle of incidence is changed to
(a) 30°, the angle of deviation will be more than 36°.
(b) 60°, the angle of deviation will be more than 36°.

📝 Teacher's Note: Remind students that the angle of deviation is smallest only at one specific angle of incidence. If you change the angle of incidence, the deviation will increase. Think of it like a U-shaped graph where the bottom of the U is the minimum deviation.

🎯 Exam Tip: For minimum deviation in a prism, the angle of incidence and angle of emergence are always equal. If the angle of incidence changes from this special value, the deviation will always be greater than the minimum deviation.

Question 13: Name the colour of white light which is deviated (i) the most, (ii) the least, on passing through a prism.

Answer:
(i) Violet colour will deviate the most and (ii) Red colour will deviate the least.

📝 Teacher's Note: When white light passes through a prism, it splits into different colours. This is called dispersion. Remember the VIBGYOR order. Violet light bends the most, and red light bends the least.

🎯 Exam Tip: Always remember the order of colours in a spectrum (VIBGYOR). Violet is at one end (most deviation), and Red is at the other (least deviation).

Question 14: Which of the two prism, A made of crown glass and B made of flint glass, deviates a ray of light more?

Answer: B made of flint glass. Because it has higher refractive index.

📝 Teacher's Note: Different types of glass bend light differently. Flint glass is denser than crown glass. Denser materials have a higher refractive index, meaning they bend light more.

🎯 Exam Tip: The material with a higher refractive index will cause more deviation. Flint glass has a higher refractive index than crown glass.

Question 15: How does the angle of deviation depend on the refracting angle of the prism?

Answer: The angle of deviation (\( \delta \)) increases with the increase in the angle of prism (A).

📝 Teacher's Note: The refracting angle (A) is the angle at the top of the prism. If this angle is bigger, the light has to travel through more glass at an angle, so it bends more. Think of it like a thicker lens bending light more.

🎯 Exam Tip: A larger prism angle (A) means a larger angle of deviation (\( \delta \)). This is a direct relationship.

Question 16: An object is viewed through a glass prism with its vertex pointing upwards. It appears to be displaced upward. Explain the reason.

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

[Diagram: This diagram shows light rays from an object O passing through a prism. The rays bend and appear to come from a new position I, which is higher than O.]

Thus, observer sees the object O raised to the position I.

📝 Teacher's Note: When light passes from a denser medium (like glass) to a rarer medium (like air), it bends away from the normal. This bending makes objects appear to be in a different place than they actually are. This is why a coin at the bottom of a glass of water looks higher.

🎯 Exam Tip: The key reason for the upward displacement is the refraction of light as it passes from the prism (denser) to the air (rarer). The light bends away from the normal, making the object appear higher.

Question 17: 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) If the incident ray normal to prism then angle of incidence is 0°.
(b) In this case the angle of refraction from the first face \( r_1 = 0^\circ \).
(c) As the prism is equilateral so \( A = 60^\circ \) and \( r_1 = 0^\circ \). So at the second face of the prism, the angle of incidence will be 60°.
(d) No, the light ray will not suffer minimum deviation. For minimum deviation, the angle of incidence must be greater than 0° and specific to the prism material and angle.

📝 Teacher's Note: "Normally incident" means the light ray hits the surface straight on, at a 90-degree angle to the surface. When light hits a surface normally, it does not bend. For an equilateral prism, the prism angle (A) is 60 degrees. The angle of incidence at the second face is related to A and \( r_1 \).

🎯 Exam Tip: When light is normally incident, both the angle of incidence and angle of refraction are 0°. Remember the prism formula \( A = r_1 + r_2 \). If \( r_1 = 0^\circ \) and \( A = 60^\circ \), then \( r_2 \) (which is the angle of incidence at the second face) will be \( 60^\circ \). Minimum deviation only happens at a specific, non-zero angle of incidence.

Question 18. Fig. 4.30 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:

[Diagram: This diagram shows two identical prisms, A and B, placed side by side. A light ray PQ enters prism A, refracts, then enters prism B, refracts again, and emerges as ray TU, parallel to PQ.]

The light ray PQ enters prism A. It bends towards the normal. Then it travels inside prism A. When it leaves prism A and enters the air gap, it bends away from the normal. This ray then enters prism B. It bends towards the normal again. Finally, it leaves prism B and enters the air. It bends away from the normal. The final ray TU will be parallel to the initial ray PQ. This happens because the prisms are identical and placed parallel. The light ray passes through two parallel surfaces (air-prism A and prism B-air) and then through two more parallel surfaces (prism A-air and air-prism B). The overall effect is a parallel shift of the ray.

📝 Teacher's Note: Explain that when light passes through two identical prisms placed parallel, the final ray is parallel to the first ray. This is like light passing through a glass slab. The light ray shifts but does not change direction.

🎯 Exam Tip: Remember to draw the ray bending towards the normal when entering the prism and away when leaving. The final ray must be parallel to the initial ray for full marks.

Question 19. Fig 4.31 below shows a light ray of single colour incident normally on two prisms A and B. In each case draw the path of the ray of light as it enters and emerges out of the prism. Mark the angle wherever necessary.

[Diagram: This diagram shows two right-angled prisms, A and B. In (i), a light ray hits prism A normally. In (ii), a light ray hits prism B normally. Angles 60° and 30° are marked.]

Answer:
(i) Prism A

[Diagram: This diagram shows a light ray entering prism A normally. It passes straight through the first face. Then it hits the hypotenuse face at 30 degrees to the normal. It refracts out of the prism, bending away from the normal. The angles 30° and 60° are marked.]

When the light ray enters prism A normally (at 90 degrees to the surface), it does not bend. It goes straight. Then it hits the slanted face. The angle of incidence here is 30°. The ray will bend away from the normal as it leaves the prism and enters the air.
(ii) Prism B

[Diagram: This diagram shows a light ray entering prism B normally. It passes straight through the first face. Then it hits the hypotenuse face at 60 degrees to the normal. It refracts out of the prism, bending away from the normal. The angles 30° and 60° are marked.]

When the light ray enters prism B normally, it does not bend. It goes straight. Then it hits the slanted face. The angle of incidence here is 60°. The ray will bend away from the normal as it leaves the prism and enters the air.

📝 Teacher's Note: Remind students that light does not bend when it enters a surface straight on (normally). It only bends when it hits at an angle. The amount it bends depends on the angle and the material.

🎯 Exam Tip: For normal incidence, draw the ray straight. For refraction, show the ray bending away from the normal when going from glass to air. Mark the angle of incidence and refraction clearly.

Question 20. Complete Fig. 4.32 to show the path of the ray of single colour as it enters the prism and emerges out of it. Mark the angles wherever necessary.

[Diagram: This diagram shows a right-angled prism. A light ray hits one face at an angle, with 70° and 45° marked as angles within the prism setup.]

Answer:

[Diagram: This diagram shows a light ray entering a prism. It bends towards the normal when entering. It travels inside the prism. Then it hits the second face and bends away from the normal when leaving. The angles 20°, 70°, and 45° are marked, showing the path of the ray and the angles of incidence and refraction.]

When the light ray enters the prism from the air, it bends towards the normal. This is because light slows down in the prism. Inside the prism, it travels in a straight line. When it leaves the prism and enters the air, it bends away from the normal. This is because light speeds up again in the air. The angles of incidence and refraction are marked to show how the light bends.

📝 Teacher's Note: Use the "FAST to SLOW, bend TOWARDS normal" and "SLOW to FAST, bend AWAY from normal" rule. Air is fast, glass is slow. This helps students remember the direction of bending.

🎯 Exam Tip: Always draw the normal (a line perpendicular to the surface) at the point where the light ray hits. This helps you correctly show the bending of light. Mark all given and calculated angles.

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.
Answer: (c) bends at both the surfaces of prism towards its base.
In simple words: When light goes through a prism, it bends two times. Both times, it bends towards the wider bottom part of the prism.

📝 Teacher's Note: Explain how a prism works by showing a diagram. Tell students light bends when it enters and leaves the prism. This bending always goes towards the base.

🎯 Exam Tip: Remember that light bends twice in a prism. It always bends towards the base of the prism. This is a key point for full marks.

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
Answer: (d) size of prism
In simple words: How much light bends in a prism depends on how light hits it, its color, and what the prism is made of. But it does not depend on how big or small the prism is.

📝 Teacher's Note: Explain that deviation depends on the prism's material, its angle, and the light's color. The physical size of the prism (like its height or width) does not change how much the light bends.

🎯 Exam Tip: For deviation in a prism, remember the key factors: angle of incidence, material, and color of light. The size of the prism is not a factor. This is a common trick question.

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. [Hint: \( \delta_{min} = 2i – A \)]
Answer:
Given,
Angle of incidence, \( i = 48° \)
Refracting angle, \( A = 60° \)
Angle of minimum deviation, \( \delta_{min} = ? \)
As \( \delta_{min} = 2i – A \)
\( \implies \delta_{min} = 2(48°) – 60° \)
\( \implies \delta_{min} = 96° – 60° \)
\( \implies \delta_{min} = 36° \)
In simple words: We are given how light enters the prism (angle of incidence) and the prism's own angle. We use a special formula to find how much the light bends at its least amount (minimum deviation). We put the given numbers into the formula and calculate the answer.

📝 Teacher's Note: This is a direct formula-based question. Make sure students understand what each symbol means: 'i' is angle of incidence, 'A' is prism angle, and '\( \delta_{min} \)' is minimum deviation. Practice similar problems.

🎯 Exam Tip: Write down the given values clearly. Use the correct formula for minimum deviation: \( \delta_{min} = 2i – A \). Show all calculation steps to get full marks. Remember to write the unit, which is degrees (\( ° \)).

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? [Hint: A = 60°, i = (A + \( \delta_{min} \))/2]
Answer:
Given,
Angle of prism, \( A = 60° \)
Angle of minimum deviation, \( \delta_{min} = 36° \)
Angle of incidence, \( i = ? \)
As \( \delta_{min} = 2i – A \)
\( \implies 36° = 2i - 60° \)
\( \implies 2i = 36° + 60° \)
\( \implies 2i = 96° \)
\( \implies i = \frac{96°}{2} \)
\( \implies i = 48° \)
In simple words: Here, we know how much the light bends (minimum deviation) and the prism's angle. We use the same formula as before, but this time we find the angle at which light enters the prism (angle of incidence). We rearrange the formula to find 'i'.

📝 Teacher's Note: This question is the reverse of the previous one. Emphasize how to rearrange the formula \( \delta_{min} = 2i – A \) to find 'i'. Remind students that an equilateral prism always has an angle of 60 degrees.

🎯 Exam Tip: Clearly state the given values. Use the formula \( i = \frac{A + \delta_{min}}{2} \) or rearrange \( \delta_{min} = 2i – A \). Show all steps for calculation. Don't forget the degree symbol (\( ° \)) in the final answer.

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?
Answer:
The refractive index \( \mu \) of a medium is related to the real and apparent depths by the formula:
\( \mu = \frac{\text{Real depth}}{\text{Apparent Depth}} \)
This means if you look at something in water, it looks closer than it actually is. The real depth is how deep it actually is. The apparent depth is how deep it looks to you.
In simple words: When you look at something in water, it seems closer than it really is. The refractive index tells us how much closer it looks. It is found by dividing the real depth (actual depth) by the apparent depth (how deep it seems).

📝 Teacher's Note: Use a real-life example like a coin at the bottom of a glass of water. Ask students if it looks deeper or shallower. Explain that light bends when it goes from water to air, making the coin appear higher.

🎯 Exam Tip: The key formula to remember is \( \mu = \frac{\text{Real depth}}{\text{Apparent Depth}} \). Make sure to write 'Real depth' in the numerator and 'Apparent Depth' in the denominator. This formula is important for understanding how things look in water.

Question 2. Prove that Refractive index = \( \frac{\text{Real depth}}{\text{Apparent depth}} \)
Answer:

[Diagram: This diagram shows an object 'O' at the bottom of a denser medium (like water). Light rays from 'O' travel to the surface and bend as they enter the air. One ray goes straight up (OA). Another ray (OB) bends away from the normal (BC). When we trace the bent ray (BC) backwards, it appears to come from point 'I', which is higher than 'O'. AO is the real depth, and AI is the apparent depth.]

Consider a ray of light incident normally along OA. It passes straight along OAA'.
Consider another ray from O (the object) incident at an angle \( i \) along OB. This ray gets refracted and passes along BC.
On producing this ray BC backwards, it appears to come from the point I, and hence, AI represents the apparent depth, which is less than the real depth AO.
Since AO and BN' are parallel and OB is the transversal,
\( \angle AOB = \angle OBN' \) (Alternate angles)
\( \angle BIA' = \angle CBN \) (Corresponding angles)
In simple words: Imagine an object at the bottom of a pond. Light rays come from it. When these rays leave the water and enter the air, they bend. Because of this bending, our eyes see the object at a shallower place than it actually is. This difference helps us understand the refractive index.

📝 Teacher's Note: This proof explains why things look shallower in water. Draw the diagram on the board and trace the light rays. Explain how the bending of light (refraction) makes the object appear at a different depth. Emphasize the terms 'real depth' and 'apparent depth'.

🎯 Exam Tip: For this proof, it is important to draw a clear diagram. Label the real depth (AO) and apparent depth (AI) correctly. Explain the path of light rays and how they bend. Mentioning alternate and corresponding angles is also key.

In ABAO, Sin i = \( \frac{BA}{OB} \)

In ΔΙΑΒ, Sin r = \( \frac{BA}{IB} \)

We know that refractive index of air w.r.t the medium

\( \mu_{am} = \frac{\sin i}{\sin r} = \frac{BA/OB}{BA/IB} = \frac{IB}{OB} \)

∴ Refractive index of medium w.r.t. air is,

\( \mu_{ma} = \frac{1}{\mu_{am}} = \frac{OB}{IB} \)

Since the point B is very close to point A, i.e the object is viewed from a point vertically above the object.

:: IB = IA and OB = OA.

Hence, \( \mu_{m} = \frac{OA}{IA} = \frac{\text{Real depth}}{\text{Apparent depth}} \)

Question 3: A tank of water is viewed normally from above. State how does the depth of tank appear to change. Draw a labelled ray diagram to explain your answer.

Answer: The depth of the tank appears to be lesser than its real depth. This happens due to the refraction of light from a denser medium (water) to a rarer medium.

[Diagram: This diagram shows light rays bending as they pass from water (denser) to air (rarer), making the bottom of the tank appear shallower than it actually is. It shows real depth and apparent depth.]

In simple words: When you look into a water tank from above, the bottom looks closer than it really is. This is because light bends when it moves from water to air.

📝 Teacher's Note: You can show this easily in class. Put a coin in a glass of water. Look from the top. The coin will seem to be higher than it is. This is a good example of refraction.

🎯 Exam Tip: Remember to mention "refraction of light" and "denser to rarer medium" for full marks. The diagram should clearly show real depth and apparent depth.

Question 4: Water in a pond appears to be only three - quarters of its actual depth. What property of light is responsible for this observation? Illustrate your answer with the help of a ray diagram. How is the refractive index of water calculated from its real and apparent depth?

Answer: 'Refraction of light' is responsible for this property. When light coming from denser medium enters rarer medium, it is bent away from the normal.

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

[Diagram: This diagram shows light rays from an object at the bottom of a pond bending as they exit the water into the air. This makes the object appear shallower (at B') than its actual position (B).]

Refractive index of water = \( \frac{\text{Real depth}}{\text{Apparent depth}} \)

In simple words: Light bends when it goes from water to air. This bending makes the pond look shallower. You can find how much it bends by comparing the real depth to how deep it looks.

📝 Teacher's Note: Explain that light changes speed when it moves from water to air. This change in speed makes the light bend. This bending is called refraction. It makes things in water look closer.

🎯 Exam Tip: The key term is "refraction of light". Make sure to draw a clear diagram showing the bending of light rays and label "Real depth" and "Apparent depth". Also, write the formula for refractive index.

Question 5: Draw a ray diagram to show the appearance of a stick partially immersed in water explain your answer.

Answer: A stick partially immersed in water in a glass container appears bent or raised as shown in figure above. This happens because the rays appear to come from P' (which is the virtual image of the tip P of the stick) due to refraction from denser medium (water) to rarer medium (air) at the surface separating two media.

[Diagram: This diagram shows a stick partially dipped in water. Light rays from the submerged part of the stick bend as they leave the water, making the stick appear bent or broken at the water surface.]

In simple words: When you put a stick in water, it looks bent or broken. This is because light from the part of the stick in water bends when it comes out into the air. Your eyes see the bent light, so the stick looks bent.

📝 Teacher's Note: Bring a glass of water and a pencil to class. Show students how the pencil looks bent when dipped in water. This is a very common and easy-to-understand example of refraction.

🎯 Exam Tip: The main reason is "refraction of light". Your diagram should clearly show the stick, the water level, and how the light rays bend to make the stick look bent or raised.

Question 6: A student puts his pencil into an empty trough and observes the pencil from the position as indicated in Fig. 4.37

[Diagram: This diagram shows a pencil placed inside an empty trough, with an eye observing it from the side. This setup is typically used to demonstrate that an object appears raised when water is added, due to refraction.]

In simple words: This question describes a setup where a student looks at a pencil in an empty bowl. If water were added, the pencil would appear to shift or bend because of how light travels through water and air.

📝 Teacher's Note: This question is likely setting up an experiment. If water is added to the trough, the pencil will appear to be raised or shifted due to refraction. You can ask students what they would observe if water was poured in.

🎯 Exam Tip: For questions like this, if water is added, remember that light bends when it goes from water to air. This bending makes objects look like they are in a different place.

Question 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) Part of the pencil which is immersed in water will look short and raised up.
(ii) The phenomena which is responsible for the above observation is refraction of light.
(iii) The required figure is

[Diagram: This diagram shows a pencil placed partly in water in a trough. Light rays from the submerged part of the pencil bend as they leave the water and enter the air, making the pencil appear bent and raised to the eye.]

In simple words: When you put a pencil in water, it looks broken or bent. This happens because light changes direction when it goes from water to air. This bending of light is called refraction.

📝 Teacher's Note: Show students a real pencil in a glass of water. Ask them to observe how it looks bent. Explain that light bends when it moves from one material to another.

🎯 Exam Tip: Remember the term "refraction of light" for the bending of light. For diagrams, draw the light rays bending away from the normal when going from water to air.

Question 7: A fish is looking at a 1.0m high plant at the edge of the pond. Will the plant appear shorter or taller than its actual height, to the fish. Draw a ray diagram to support your answer.
Answer:
The plant will look taller than its actual height.

[Diagram: This diagram shows a plant at the edge of a pond. A fish (eye of fish) is underwater. Light rays from the plant bend as they enter the water, making the plant appear taller to the fish.]

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

In simple words: When a fish looks at a plant from underwater, the plant looks taller. This is because light bends when it goes from air (above water) into water. The light rays from the plant bend towards the normal, making the plant seem higher up.

📝 Teacher's Note: Explain that light bends when it goes from air to water. This bending makes things look different from what they really are. Use the example of a fish seeing things outside the water.

🎯 Exam Tip: For this question, state "taller" and draw a ray diagram showing light bending towards the normal when entering the denser medium (water) from the rarer medium (air).

Question 8: 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 shift depends and state how does it depend on them.
Answer:
The factors on which the magnitude of shift depends are:
(a) The refractive index of the medium,
(b) The thickness of the denser medium and
(c) The colour (or wavelength) of incident light.

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

In simple words: When you look at something through water, it looks like it has moved up or down. How much it moves depends on three things: how much the light bends in the material (refractive index), how thick the material is (like how deep the water is), and the color of the light.

📝 Teacher's Note: Explain that the "shift" is how much an object appears to move from its real position. Use a glass slab or a thick block of plastic to show how objects behind it appear shifted. Different colors of light bend differently.

🎯 Exam Tip: List the three factors: refractive index, thickness of the medium, and color/wavelength of light. Remember that shift increases with refractive index and thickness, but decreases with wavelength.

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 options
Answer: (a) refraction of light
In simple words: An air bubble inside glass looks closer to the surface than it really is. This happens because light bends when it travels from the glass (denser) to the air (rarer) and then to your eye. This bending is called refraction.

📝 Teacher's Note: Explain that light bends when it goes from a denser material (like glass) to a rarer material (like air). This bending makes things look like they are in a different place.

🎯 Exam Tip: The key concept here is "refraction of light." This is the reason why objects appear shifted or raised when light passes through different mediums.

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 lighten
(d) green light
Answer: (b) violet light
In simple words: When light bends, different colors bend by different amounts. Violet light bends the most. So, the apparent shift (how much something looks moved) is greatest for violet light. This is because the refractive index is highest for violet light.

📝 Teacher's Note: Remind students about the VIBGYOR colors. Violet light bends the most when it passes through a prism or changes medium. This means it has the highest refractive index.

🎯 Exam Tip: Remember that violet light has the highest refractive index and thus experiences the maximum shift or deviation. Red light has the least.

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.
Answer:

In simple words: We know how deep the pond *looks* (apparent depth) and how much light bends in water (refractive index). We need to find the *real* depth. We use a formula that connects these three values.

📝 Teacher's Note: Explain the formula: Refractive Index = Real Depth / Apparent Depth. Make sure students understand the difference between how deep something looks and how deep it actually is.

🎯 Exam Tip: Write down the formula clearly. Substitute the given values correctly. Show your calculation steps to get full marks.

Question 1:

Solution 1:
Apparent depth = 2.7 m
\( \mu_w = \frac{4}{3} \)
Real depth = apparent depth \( \times \mu_w \)
= \( 2.7 \times \frac{4}{3} \)
= 3.6 m

📝 Teacher's Note: This problem shows how things look closer in water. The refractive index tells us how much light bends. It helps us find the real depth of an object.

🎯 Exam Tip: Remember the formula: Real depth = Apparent depth \( \times \) Refractive index. Write down all given values clearly.

Question 2:
A coin is places 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 2:
Refractive index of the water, \( \mu_w = 4/3 \)
Real depth at which the coin is places = 12 cm
Shift in the image = ?
Shift = real depth \( \times (1 - \frac{1}{\mu}) \)
Shift = \( 12 \times (1 - \frac{3}{4}) \)
R = \( \frac{12}{4} \)
= 3cm

📝 Teacher's Note: When you look at a coin in water, it seems higher than it actually is. This "shift" happens because light bends when it goes from water to air. The refractive index tells us how much light bends.

🎯 Exam Tip: The "shift" is the difference between real depth and apparent depth. Make sure to use the correct formula for shift: Real depth \( \times (1 - \frac{1}{\mu}) \).

Question 3:
A postage stamp kept below a rectangular glass block 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 block.

Solution 3:
Refractive index of the glass block, \( \mu_g = 1.5 \)
Shift in the image = 7mm or 0.7 cm
Thickness of glass block or real depth =?
Shift = Real depth \( \times (1 - \frac{1}{\mu}) \)
\( 0.7 = R \times (1 - \frac{1}{1.5}) \)
R = \( \frac{0.7 \times 1.5}{0.5} \)
= 2.1 cm

📝 Teacher's Note: This problem is similar to the coin in water. The glass block makes the stamp appear closer. The shift is given, and we need to find the actual thickness of the glass block (real depth).

🎯 Exam Tip: Pay attention to units. Convert mm to cm if needed. Remember to rearrange the shift formula to find the real depth.

Exercise 4(D)

Question 1:
Explain the term critical angle with the aid of a labelled diagram.

Solution 1:
Critical angle: The angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90° is called the critical angle.

[Diagram: This diagram shows light rays going from a denser medium (Glass) to a rarer medium (Air). The incident ray hits the surface at an angle 'i'. The refracted ray bends away from the normal. When the angle of refraction 'r' becomes 90 degrees, the angle of incidence 'i' is called the critical angle 'i_c'. Another ray shows total internal reflection when 'i' is greater than 'i_c'.]

📝 Teacher's Note: Imagine light trying to leave water and go into air. If it hits the surface at a special angle, it will just skim along the surface of the water, not really coming out. This special angle is the critical angle.

🎯 Exam Tip: For full marks, define critical angle correctly. Mention "denser medium," "rarer medium," and "angle of refraction is 90°." A clear, labelled diagram is also very important.

Question 2:
How is the critical angle related to the refractive index of a medium?

Solution 2:
The critical angle is related to the refractive index of a medium by the relation
\( \mu = \frac{1}{\sin i_c} \)
\( \implies \mu = \text{cosec } i_c \)

📝 Teacher's Note: This formula connects how much light bends (refractive index) with the critical angle. A higher refractive index means light bends more, so the critical angle will be smaller.

🎯 Exam Tip: Remember the formula \( \mu = \frac{1}{\sin i_c} \). This is a key relationship to score marks. Make sure to write it correctly.

Question 3:
State the approximate value of the critical angle for
(a) glass-air surface
(b) water-air surface.

Solution 3:
(a) The critical angle for glass-air surface is
For glass, refractive index \( _a\mu_g = \frac{3}{2} \)
\( \sin i_c = \frac{1}{_a\mu_g} = \frac{2}{3} \)
\( \implies i_c = 42^\circ \)
(b) The critical angle for water-air surface is
For water, refractive index \( _a\mu_w = \frac{4}{3} \)
\( \sin i_c = \frac{1}{_a\mu_w} = \frac{3}{4} \)
\( \implies i_c = 49^\circ \)

📝 Teacher's Note: Different materials have different critical angles. Glass has a smaller critical angle than water. This means light can escape from water into air more easily than from glass into air.

🎯 Exam Tip: Know the approximate critical angles for common materials like glass and water. Remember the formula \( \sin i_c = \frac{1}{\mu} \) and how to use it.

Question 4:
What is meant by the statement the critical angle for diamond is 24°?

Solution 4:
The critical angle for diamond is 24°. This implies that at an incident angle of 24° within the diamond the angle of refraction in the air will be 90°. And if incident angle will be more than this angle then the ray will suffer total internal reflection without any refraction.

📝 Teacher's Note: Diamond has a very small critical angle (24°). This means light inside a diamond hits the surface at an angle greater than 24° very often. When this happens, light cannot escape and bounces back inside. This is why diamonds sparkle so much!

🎯 Exam Tip: Explain what a critical angle of 24° means. Mention that the refracted angle is 90° at this point. Also, state what happens if the angle of incidence is *more* than 24° (total internal reflection).

Question 5: A light ray is incident from a denser medium on the boundary separating it from a rarer medium at an angle of incident equal to the critical angle, what is the angle of refraction for the ray?

Answer: When a ray is incident from a denser medium to a rarer medium at angle equal to critical angle (i=\(i_c\)), the angle of refraction becomes 90°
In simple words: When light goes from a thick material (denser) to a thin material (rarer) at a special angle called the critical angle, it bends so much that it travels along the surface. This means the angle of refraction is 90 degrees.

📝 Teacher's Note: Explain that the critical angle is like a "point of no return" for light. Use an example of looking into water from below. If you look up at a certain angle, you see the surface like a mirror.

🎯 Exam Tip: Remember that when the angle of incidence is the critical angle, the angle of refraction is always 90°. This is a key definition for total internal reflection.

Question 6: Name two factors which affect the critical angle for a given pair of media. State how do the factors affect it.

Answer: The factors which affect the critical angle are:

1. The colour (or wavelength) of light, and
2. The temperature

(i) Effect Of Colour Of Light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus critical angle increases with increase in wavelength of light.
(ii) Effect Of Temperature: The critical angle increases with increase in temperature because on increasing the temperature of medium, its refractive index decreases.
In simple words: The critical angle changes based on the color of light (like red or violet) and how hot or cold the material is. Different colors bend differently. Also, when a material gets hotter, it becomes less dense, which changes how light passes through it.

📝 Teacher's Note: You can explain that different colors of light have different wavelengths. Red light has a longer wavelength than violet light. Also, when a material gets hotter, its particles spread out, making it less dense. This changes its refractive index.

🎯 Exam Tip: For full marks, name both factors (colour/wavelength and temperature). Clearly state how each one affects the critical angle. Remember: longer wavelength (red) means a larger critical angle; higher temperature means a larger critical angle.

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?

Answer: As the wavelength decreases (or increases) refractive index becomes more (or less) and critical angle becomes less (or more).
(i) For red light the critical angle will be more than 45° and
(ii) For blue light the critical angle will be less than 45°.
In simple words: The critical angle changes with the color of light. Red light has a longer wavelength than yellow light, so its critical angle will be bigger. Blue light has a shorter wavelength than yellow light, so its critical angle will be smaller.

📝 Teacher's Note: Remind students of VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). Red light has the longest wavelength, violet has the shortest. The critical angle increases as the wavelength increases. So, red light has a larger critical angle than yellow, and blue light has a smaller critical angle than yellow.

🎯 Exam Tip: Link wavelength to critical angle. Longer wavelength (like red) means a larger critical angle. Shorter wavelength (like blue) means a smaller critical angle. State "more than 45°" for red and "less than 45°" for blue.

Question 8:

(a) What is total internal reflection?
(b) State two conditions necessary for total internal reflection to occur.
(c) Draw diagram to illustrate the total internal reflection
Answer:
(a) Total internal reflection: It is the phenomenon when a ray of light travelling in a denser medium, is incident at the surface of a rarer medium such that the angle of incidence is greater than the critical angle for the pair of media, the ray is totally reflected back into the denser medium.
In simple words: Total internal reflection happens when light tries to go from a thick material (like water) to a thin material (like air). But it hits the surface at a very wide angle. Instead of coming out, it bounces back completely inside the thick material, like a mirror.
(b) The two necessary conditions for total internal reflection are:

1. The light must travel from a denser medium to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the pair of media.

In simple words: For this to happen, light must go from a "denser" (thicker) material to a "rarer" (thinner) material. Also, the light must hit the surface at an angle wider than a special angle called the critical angle.
(c) When incidence angle is more than critical angle i.e., in case of total internal reflection.

[Diagram: This diagram shows a light ray going from glass (denser) to air (rarer). The incident ray hits the surface at an angle greater than the critical angle (\(i > i_c\)), and instead of refracting, it reflects back into the glass as a reflected ray.]

In simple words: The diagram shows light starting in a thick material (glass) and trying to go into a thin material (air). Because the light hits the surface at a very wide angle (more than the critical angle), it bounces back into the glass, like a mirror.

📝 Teacher's Note: Explain total internal reflection using examples like optical fibers (used for internet) or diamonds sparkling. Emphasize that the two conditions are crucial for it to happen.

🎯 Exam Tip: For part (a), define total internal reflection clearly. For part (b), list both conditions correctly. For part (c), if asked to draw, make sure your diagram shows the incident ray, the normal, the reflected ray, and labels for denser/rarer media, and \(i > i_c\).

Question 9: Fill in the blanks to complete the following sentence:

(a) Total internal reflection occurs only when a ray of light passes from a ________ 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 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°.
In simple words: For total internal reflection, light must go from a thick material to a thin material. The critical angle is when light hits the surface at a special angle, and it bends exactly 90 degrees in the thinner material.

📝 Teacher's Note: These blanks test the basic conditions and definition of critical angle and total internal reflection. Make sure students understand "denser" and "rarer" media.

🎯 Exam Tip: Remember the key words: "denser" to "rarer" medium for total internal reflection. For critical angle, it's the angle of "incidence" in the denser medium that causes 90° "refraction" in the rarer medium.

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.

Answer: True.
In simple words: This statement is true. If light hits the surface at an angle wider than the critical angle, it doesn't pass through (refract). Instead, it bounces back completely, which is called total internal reflection.

📝 Teacher's Note: This question checks understanding of total internal reflection. When the angle of incidence is greater than the critical angle, light cannot escape the denser medium; it reflects entirely.

🎯 Exam Tip: This is a direct test of the definition of total internal reflection. If the angle of incidence is greater than the critical angle, refraction stops, and total internal reflection begins. So, the statement is true.

Question 11: The refractive index of air with respect to glass is expressed as \( g\mu_a = \frac{\sin i}{\sin r} \)

(a) Write down a similar expression for \( a\mu_g \) 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) The expression for \( a\mu_g \) in terms of angles i and r is \( a\mu_g = \frac{\sin r}{\sin i} \).
(b) If angle r = 90°, the corresponding angle i is called the critical angle (\( i_c \)).
(c) The physical significance of the angle i (critical angle) in part (b) is that it is the angle of incidence in the denser medium for which the light ray refracts at 90° in the rarer medium. If the angle of incidence is greater than the critical angle, total internal reflection occurs.
In simple words: (a) If \( g\mu_a \) is \( \frac{\sin i}{\sin r} \), then \( a\mu_g \) is the opposite, \( \frac{\sin r}{\sin i} \). (b) When light bends at 90 degrees, the angle it hit the surface at is called the critical angle. (c) This critical angle is important because it tells us when light will stop bending out and start bouncing back inside completely.

📝 Teacher's Note: This question is about Snell's Law and critical angle. \( g\mu_a \) means refractive index of air with respect to glass. \( a\mu_g \) means refractive index of glass with respect to air. These are inverse of each other. When \( r = 90^\circ \), the angle \( i \) is the critical angle.

🎯 Exam Tip: Remember Snell's Law: \( n_1 \sin i = n_2 \sin r \). If \( g\mu_a = \frac{\sin i}{\sin r} \), then \(

Question 12:
The given 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. The critical angle for water-air surface is 48°. (b) Name the phenomenon which the rays A, B and C exhibit.

Answer:
(a)

[Diagram: This diagram shows rays A, B, and C starting from point P. Ray A and B bend away from the normal (refraction). Ray C hits the surface at an angle greater than 48 degrees and reflects back into the water (total internal reflection).]

(b) Rays A and B exhibit the phenomenon of 'refraction of light'.
Rays C exhibits the phenomenon of 'total internal reflection'.

📝 Teacher's Note: Explain that light bends when it goes from water to air. This is called refraction. If the light hits the surface at a very large angle, it bounces back completely. This is total internal reflection.

🎯 Exam Tip: For part (a), draw the rays bending correctly. For part (b), remember "refraction" for bending rays and "total internal reflection" for the ray that bounces back.

Question 13:
In the given 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 \( \angle \)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 \( \angle \) PQN is increased.
(d) The ray PR suffers partial reflection and refraction on the water-air surface. Give reason. 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.

Answer:
(a) Critical angle
In simple words: The critical angle is a special angle. If light hits the surface at this angle, it bends so much that it travels along the surface.
Hint: The angle of incidence in the denser medium for which the angle of refraction in rarer medium is 90° is called the critical angle.

(b) 90°
In simple words: When the angle of incidence is the critical angle, the light ray bends exactly 90 degrees. It travels along the surface between water and air.

(c) Total internal reflection.
In simple words: If the light hits the surface at an angle bigger than the critical angle, it does not leave the water. It bounces back completely, like a mirror. This is called total internal reflection.
Hint: When the angle of incidence is greater than the critical angle, the phenomenon of total internal reflection occurs due to which the ray of light is not refracted but is reflected back in the same medium.

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

[Diagram: This diagram shows ray PQ refracting as QS. Ray PR shows partial reflection and refraction. The refracted ray for PR bends away from the normal. P' is shown as the virtual image of P.]

📝 Teacher's Note: Remind students that light bends when it moves from water to air. This is refraction. If the angle is too big, it bounces back (total internal reflection). If the angle is small, some light reflects, and some refracts.

🎯 Exam Tip: Remember the three possibilities: refraction (bending), total internal reflection (bouncing back), and partial reflection/refraction (some bends, some bounces). Draw the diagrams carefully, showing the bending correctly.

Question 14:
The refractive index of glass is 1.5. From a point P inside a glass block, 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 block.
(b) What is the angle of refraction for the ray PB?
(Take sin 42° = \( \frac{2}{3} \))

Answer:
(a)
\( \mu = 1.5 \)
\( \text{Sin } i_c = \frac{1}{\mu} = \frac{1}{1.5} = 0.667 \)
\( i_c = 41.8^\circ \approx 42^\circ \)
In simple words: First, we find the critical angle for glass. This is the special angle where light starts to bounce back instead of leaving the glass. Here, it is about 42 degrees.

[Diagram: This question requires drawing a diagram showing rays PA, PB, PC. Ray PA (30°) will refract out. Ray PB (42°) will travel along the surface (critical angle). Ray PC (60°) will undergo total internal reflection.]

(b) For ray PB, the angle of incidence is 42°, which is the critical angle.
So, the angle of refraction for ray PB will be 90°.

📝 Teacher's Note: This question helps students understand the critical angle. When the angle of incidence is less than the critical angle, light refracts. When it is equal to the critical angle, it refracts at 90 degrees. When it is more, it reflects totally.

🎯 Exam Tip: Always calculate the critical angle first. Then compare the given angles of incidence with the critical angle to decide if the light will refract or undergo total internal reflection. Remember, at the critical angle, the angle of refraction is 90°.

(b) As angle of incidence inside glass block is 42°
\( \frac{\sin r}{\sin i} = {^a\mu_g} \)
\( \sin r = {^a\mu_g} \times \sin 42^\circ \)
Take \( \sin 42^\circ = \frac{2}{3} \) and \( {^a\mu_g} = \frac{3}{2} \)
\( \sin r = \frac{3}{2} \times \frac{2}{3} = 1 \)

\( \implies r = 90^\circ \)
This also follows from the fact that the ray PB is incident at the critical angle.

[Diagram: This diagram shows a light ray entering a glass block, bending (refracting), and then hitting another surface inside the block. Angles like 30°, 42°, 60° are marked. It shows the path of light from air to glass and back to air.]

In simple words: Light bends when it goes from air to glass. This bending is called refraction. If the light hits the glass at a special angle (called critical angle), it bends so much that it travels along the surface of the glass, making an angle of 90 degrees.

📝 Teacher's Note: Explain Snell's Law (sin i / sin r = constant) simply. Show how light bends when it enters water or glass. Emphasize that the critical angle is a special angle where light almost escapes but instead travels along the surface.

🎯 Exam Tip: Remember the formula for critical angle. Know that when light hits at the critical angle, the refracted ray makes a 90-degree angle with the normal.

Question 15. A ray of light enters a glass slab ABDC as shown in fig. 4.59 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 angle of refraction will be 90° because the ray is incident on the glass at its critical angle.

[Diagram: This diagram shows a glass slab with a circular part. A light ray enters the slab, hits the center of the circular part, and then exits. The critical angle (42°) and the refracted angle (90°) are marked.]

In simple words: When light hits the glass at a special angle called the critical angle (here, 42 degrees), it does not go out of the glass. Instead, it travels along the surface of the glass. This means the angle of refraction is 90 degrees.

📝 Teacher's Note: Explain that when light hits a boundary from a denser medium (like glass) to a rarer medium (like air) at an angle greater than the critical angle, it undergoes Total Internal Reflection. At the critical angle, the refracted ray skims the surface.

🎯 Exam Tip: For full marks, draw the diagram clearly. Mark the critical angle (42°) and the 90° refracted angle correctly. State that the ray is incident at the critical angle, leading to 90° refraction.

Question 16. 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.

Answer: A prism having an angle of 90° between its two refracting surfaces and the other two angles each equal to 45°, is called a total reflecting prism. The light incident normally on any of its faces, suffers total internal reflection inside the prism. Due to this behavior, a total reflecting prism is used to produce following three actions:
(a) To deviate a ray of light through 90°,
(b) To deviate a ray of light through 180°, and
(c) To erect the inverted image without producing deviation in its path.

[Diagram: This diagram shows a total reflecting prism. A light ray (OBJECT PQ) enters the prism, undergoes total internal reflection, and then exits as an inverted image (P'Q'). The angles 45° and 90° are marked inside the prism.]

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

In simple words: A total reflecting prism is a special glass shape with angles 90°, 45°, and 45°. It uses total internal reflection to change the path of light. It can turn light by 90 degrees, turn it by 180 degrees, or flip an upside-down image without changing its direction.

📝 Teacher's Note: Explain that total internal reflection happens when light tries to go from a denser material (like glass) to a lighter material (like air) at a large angle. The light bounces back inside the glass. This prism uses this effect to control light very well.

🎯 Exam Tip: Define the prism with its angles (90°, 45°, 45°). List the three actions clearly. Draw a neat diagram showing one action, like deviating light by 90° or erecting an image, with correct angles.

Question 17. 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: As shown in diagram, a beam of light is incident on face AB of the prism normally so it passes undeviated and strikes the face AC where it makes an angle of 45° with the normal to AC. Because here the incident angle is more than critical angle so rays suffer total internal reflection and reflect at angle of 45°. The beam then strikes face BC, where it is incident normally and so passes undeviated. As a result the incident beam gets deviated through 90°.

[Diagram: This diagram shows a total reflecting prism. A light ray (P) enters face AB straight, hits face AC, reflects totally, and then exits face BC. The angles 45° and 90° are marked, showing how the light ray turns by 90 degrees.]

One instrument in which such a prism is used is a Periscope or Binoculars.

In simple words: A light ray enters the prism straight. Inside, it hits a surface at a special angle (45 degrees). This angle is bigger than the critical angle, so the light bounces back completely inside the prism (total internal reflection). Then, it leaves the prism, but its direction has changed by 90 degrees. This is used in devices like a periscope to see around corners.

📝 Teacher's Note: Use a real periscope or binoculars as an example. Explain how the prism acts like a perfect mirror, but without any silver coating. This makes it very efficient for turning light without much loss.

🎯 Exam Tip: Draw the prism and the path of light clearly, showing the 90-degree deviation. Mark all angles (45° and 90°) correctly. Name an instrument like a periscope or binoculars where this prism is used.

Question 18:

A ray of light OP passes through a right angles 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.

Answer:
(a) The angle of incidence at the face AC is 45° and angle of incidence at the face BC is 0°.
(b) The ray suffers total internal reflection at the face AC.

[Diagram: This diagram shows a right-angled prism with a light ray OP entering it, reflecting inside, and exiting. Angles are marked.]

In simple words: When light hits the face AC, it bounces back completely inside the prism. This is called total internal reflection. When it hits face BC straight on, it passes through without bending.

📝 Teacher's Note: Explain total internal reflection using a simple example like looking into water from below. Show how light bends away from the normal when going from water to air, and if the angle is too big, it reflects back. Emphasize that for normal incidence (0°), light does not bend.

🎯 Exam Tip: Remember the critical angle for glass is about 42°. If the angle of incidence inside the glass is more than 42°, total internal reflection happens. For light hitting a surface at 0° (normally), the angle of incidence is 0°.

Question 19:

In fig. 4.61, a ray of light PA is incident normally on the hypotenuse of an isoceles right angle 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)

[Diagram: This diagram shows an isosceles right-angled prism. A light ray PA enters the hypotenuse normally, reflects twice inside, and emerges. The angles 45° are marked at the reflection points.]

(b) Angle of deviation = 180°
(c) Prism binocular

In simple words: The light ray goes into the prism, bounces off two sides inside, and then comes out. It changes its direction completely, turning around by 180 degrees. This is used in binoculars to make things look closer.

📝 Teacher's Note: Draw the path of light step-by-step on the board. Explain that "normally" means hitting at a 90-degree angle, so it goes straight in. Then, at the other faces, the angle is greater than the critical angle, causing total internal reflection. Show how the final ray is exactly opposite to the initial ray.

🎯 Exam Tip: For part (a), draw the path carefully, showing the two internal reflections. For part (b), remember that a 180° deviation means the light ray turns completely back on itself. For part (c), "binocular" is the key device.

Question 20:

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 can be used to turn a ray of light by 180°. The following diagram can make it further clear.

[Diagram: This diagram shows a right-angled prism. A light ray enters one face, undergoes two total internal reflections, and emerges parallel but opposite to the incident ray, showing a 180° deviation.]

This action of prism is used in binocular.

In simple words: A special glass shape called a total reflecting prism can turn light around completely, by 180 degrees. This is better than a mirror because it reflects all the light. Binoculars use these prisms.

📝 Teacher's Note: Emphasize that a prism reflects light better than a mirror because it uses total internal reflection, which is 100% efficient. Mirrors can lose some light. Show how the prism makes the light turn around completely.

🎯 Exam Tip: The key device is a "total reflecting prism". Make sure your diagram clearly shows the light entering, reflecting twice inside, and exiting with a 180° change in direction. Mention "binocular" as the instrument.

Question 21:

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

Answer:
When total internal reflection occurs from a prism, the entire incident light (100%) is reflected back into the denser medium. Whereas in ordinary reflection from a plane mirror, some light is refracted and absorbed so the reflection is partial.

In simple words: A prism reflects all the light (100%) that hits it from inside. A normal mirror reflects most light, but some light is lost (absorbed or passes through). So, a prism gives a brighter reflection.

📝 Teacher's Note: Explain that "total" means all of it. A mirror is like a shiny surface, but some light always gets lost. A prism, when used for total internal reflection, acts like a perfect mirror, reflecting everything.

🎯 Exam Tip: The main difference is the amount of light reflected. For a prism (total internal reflection), it's 100%. For a plane mirror, it's less than 100% (partial reflection).

Question 22:

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

Answer:
A total reflecting prism gives us an image much brighter than that obtained by using a plane mirror.

In simple words: When you use a total reflecting prism, the picture you see is much brighter. This is because the prism reflects all the light, unlike a mirror which loses some light.

📝 Teacher's Note: Connect this to the previous question. Since total internal reflection reflects 100% of the light, the image formed is brighter and clearer. This is why prisms are often preferred in optical instruments.

🎯 Exam Tip: The key advantage is "brighter image". This is because total internal reflection is more efficient than mirror reflection.

Question 23:

In given 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.
(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) Angle of incidence at face AB = 0° (since it's normal incidence).
Angle of incidence at face AC = 60°.
(b) At face AB: Refraction (no deviation as angle of incidence is 0°).
At face AC: Total internal reflection.
At face BC: Refraction.

[Diagram: The question refers to a figure, but it is not provided in the source. This diagram would show an equilateral prism with a light ray PQ entering face AB normally, reflecting off face AC, and then refracting out of face BC.]

In simple words: The light ray enters the prism straight on (face AB), so it doesn't bend. Then it hits face AC at a big angle and bounces back completely inside the prism. Finally, it hits face BC and bends as it leaves the prism into the air.

📝 Teacher's Note: For an equilateral prism, all angles are 60°. If light enters normally at AB, it goes straight. Then, calculate the angle at AC. Since the angle of incidence (60°) is greater than the critical angle for glass (around 42°), total internal reflection occurs. Finally, at BC, it will refract out.

🎯 Exam Tip: Remember that "normally incident" means the angle of incidence is 0°. For an equilateral prism, the internal angles are 60°. Use geometry to find the angle of incidence at face AC. If this angle is greater than the critical angle, it's total internal reflection. If it's less, it's refraction.

Question 23: [The question text is missing from the source. The diagram above shows a prism ABC with a light ray P-Q entering face AB. Analyze the path of light.]

[Diagram: This diagram shows a prism ABC with angles 90°, 60°, 60°. A light ray P-Q is shown incident on face AB at 90°.]

Answer:

[Diagram: This diagram shows the path of light through the prism ABC. The incident ray enters face AB normally (i=0°), undergoes total internal reflection at face AC (i=60°), and then refracts out of face BC.]

(a) At the face AB, i=0° and at the face AC, i= 60°
(b) At the face AB – refraction,
At the face AC – total internal reflection,
At the face BC – refraction.

📝 Teacher's Note: This problem shows how light behaves inside a prism. Light bends (refracts) when it enters or leaves the glass. If the angle is too big inside the glass, it bounces back completely (total internal reflection).

🎯 Exam Tip: When describing light paths in a prism, always mention if the light is refracting (bending) or undergoing total internal reflection. Clearly state the angle of incidence at each surface where light interacts.

Question 24: Draw a neat labelled ray diagram to show the total internal reflection of a ray of light normally incident on one face of a 30°, 90°, 60° prism.

Answer:

[Diagram: This diagram shows a 30°, 90°, 60° prism ABC. A light ray is incident normally on face AB, undergoes total internal reflection at face AC, and emerges from face BC. Angles are marked.]

📝 Teacher's Note: This diagram shows how a prism can change the direction of light using total internal reflection. This happens when light hits the inside surface of the glass at a large angle and cannot pass through.

🎯 Exam Tip: For drawing diagrams, always use a ruler and pencil. Label all angles, faces (A, B, C), and the path of the light ray clearly. Show the normal lines at the points of incidence and reflection.

Question 25: Two isosceles right -angles glass prisms are placed near each other as shown in Fig. 4.63. Complete the path of the light ray entering the first prism till it emerges out of the second prism.

[Diagram: This diagram shows two isosceles right-angled prisms placed near each other. A light ray enters the first prism.]

Answer:

[Diagram: This diagram shows the complete path of the light ray through two isosceles right-angled prisms. The ray undergoes total internal reflection in both prisms, changing its direction.]

📝 Teacher's Note: This problem shows how prisms can be used to redirect light. Each prism uses total internal reflection to bend the light by 90 degrees. This is like how binoculars work.

🎯 Exam Tip: When completing ray diagrams, remember the laws of reflection and refraction. For total internal reflection, the angle of incidence must be greater than the critical angle. Draw the normals at each surface.

MULTIPLE CHOICE TYPE:

Question 1: The critical angle for glass-air interface is:
(a) 24°
(b) 48°
(c) 42°
(d) 45°
Answer: (c) 42°
In simple words: The critical angle is the special angle at which light stops leaving the glass and starts bouncing back inside. For glass and air, this angle is about 42 degrees.

📝 Teacher's Note: The critical angle is a very important concept for total internal reflection. It depends on the refractive index of the two materials. For common glass, it's around 42 degrees.

🎯 Exam Tip: Remember the critical angle for glass-air interface is 42°. This is a standard value often asked in exams. Make sure to recall it correctly.

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°.
Answer: (d) 90°.
Hint:

[Diagram: This diagram shows a right-angled isosceles prism. A light ray P-Q enters face AB normally, undergoes total internal reflection at face AC, and emerges from face BC, deviating by 90 degrees.]

In simple words: A special type of prism, shaped like a triangle with a 90-degree angle and two 45-degree angles, can bend light by exactly 90 degrees. It does this by bouncing the light completely inside.

📝 Teacher's Note: Right-angled isosceles prisms are used in many optical instruments like binoculars and periscopes. They are very efficient because they use total internal reflection, which means almost no light is lost.

🎯 Exam Tip: Know that a right-angled isosceles prism can deviate light by 90° or 180° depending on how the light enters. For a 90° deviation, the light enters normally on one of the shorter faces.

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°

Answer: (b) 60°
In simple words: An equilateral prism is a special glass shape. All its angles are 60 degrees. When light enters it, it bends and reflects inside. This makes the light change its direction by 60 degrees.

[Diagram: This diagram shows a ray of light entering an equilateral prism. It undergoes total internal reflection inside. The diagram shows how the light ray's path is changed by 60 degrees.]

📝 Teacher's Note: Explain that an equilateral prism has all angles equal to 60 degrees. Show students how light enters, reflects inside, and then leaves. This bending and reflecting changes the light's direction. Use a simple drawing on the board.

🎯 Exam Tip: Remember that a total reflecting equilateral prism deviates light by 60 degrees. This is a key fact to score marks. Make sure to understand the path of light in such a prism.