Goyal Brothers Solutions for ICSE Class 10 Physics Chapter 5 Refraction Of Light

Exercise - 1

Question 1. (a) What do you understand by the term refraction of light?
(b) How does the light deviate when it travels from ?
1. a rarer to a denser medium
2. a denser to a rarer medium?
Answer: (a) Refraction of light : "When light travels from one optical medium to other optical medium, it changes its path, this change in path is called refraction of light".
(b)
(i) It bends towards the normal.
(ii) It bends away from the normal.
In simple words: (a) Refraction means light bends when it goes from one transparent material to another. (b) It bends towards the normal line when entering a thicker medium, and away from it when entering a thinner medium.

Exam Tip: Always remember that the speed of light decreases in a denser medium (bending towards the normal) and increases in a rarer medium (bending away from the normal).

Question 2. (a) State the laws of refraction.
(b) What do you understand by the statement that refractive index of water is 1.33 ?
Answer: (a) Laws of refraction :
(i) Snell's law : For a given pair of optical media and light of a specific color, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant value. This constant is called the refractive index. \[ \mu = \frac{\sin i}{\sin r} \]
(ii) The incident ray, the refracted ray, and the normal at the point of incidence all lie within the same plane.
(b) The statement signifies that the speed of light in vacuum (or air) is 1.33 times greater than its speed when travelling through water.
In simple words: (a) Light obeys two rules: it bends at a constant ratio depending on the materials, and all rays stay on the same flat surface. (b) Water's refractive index of 1.33 means light travels 1.33 times slower in water than it does in air.

Exam Tip: Be sure to write the formula \( \mu = \frac{\sin i}{\sin r} \) clearly when stating Snell's law, specifying that \( i \) is the angle of incidence and \( r \) is the angle of refraction.

Question 3. Describe how will you verify the laws of refraction ?
Answer: **Experiment to Verify the Laws of Refraction:**
1. Secure a clean sheet of white paper onto a flat drawing board.
2. Place a rectangular glass slab in the center and trace its boundary (KLMN) with a pencil.
3. Remove the block. On the boundary line KL, select a point O and construct a perpendicular line \( A'OB' \) to act as the normal.
4. Draw a line AO representing the incident ray, making an angle of incidence \( i \) (e.g., \( 60^{\circ} \)) with the normal.
5. Put the glass slab back exactly on its outlined boundary.
6. Mount two pins, labeled 'a' and 'b', vertically along the line AO, separating them by about 10 cm.
7. Look through the opposite side NM of the slab. Insert two more pins, 'c' and 'd', such that their bases appear aligned in a straight line with the images of pins 'a' and 'b' seen through the glass.
8. Take out the pins and mark their positions with pencil pricks.
9. Remove the glass slab. Connect points c and d with a straight line, extending it to meet the boundary NM at point B.
10. Join points O and B.
- Here, AO represents the incident ray, OB represents the refracted ray, and BC represents the emergent ray.
- Let \( \angle AOA' \) be the angle of incidence \( i \), and \( \angle BOB' \) be the angle of refraction \( r \).
11. With O as the center, draw a circle of a suitable radius that intersects the line AO at point X and the line OB at point Y.
12. Drop perpendiculars from X and Y onto the normal. Let these perpendiculars be XY and X'Y' respectively. Measure their lengths.
13. The ratio \( \frac{XY}{X'Y'} \) represents the refractive index \( \mu \keys{} \), which is found to be a constant (approximately 1.5 for glass).
14. Alternatively, measure \( i \) and \( r \) directly using a protractor and calculate \( \frac{\sin i}{\sin r} \). This ratio remains constant at \( 1.5 \) for different angles of incidence like \( 30^{\circ}, 40^{\circ}, 50^{\circ}, \) etc.
15. This constant ratio confirms Snell's Law (the second law). Since the incident ray AO, the refracted ray OB, and the normal at O are all drawn on the flat sheet of paper, the first law of refraction is also verified.

In simple words: To prove the laws, we trace light going through a glass slab using pins. By measuring how much the light path bends inside the glass compared to outside, we find that the ratio always stays the same, confirming the rules of refraction.

Exam Tip: In experiments verifying Snell's Law, ensure you keep the distance between the pins at least 10 cm apart to reduce parallax errors and get precise alignments.

Question 4. (a) What do you understand by the term lateral displacement ?
(b) State three factors which determine lateral displacement ?
Answer: (a) Lateral Displacement is defined as the perpendicular distance by which an incident light ray shifts from its original path when it emerges parallelly after passing through an optical medium with parallel faces.
(b) The lateral displacement depends on the following three key factors:
1. The angle of incidence of the incoming light ray.
2. The overall thickness of the refracting medium or glass slab.
3. The refractive index of the material (which is determined by both the substance itself and the wavelength of the light used).

In simple words: (a) Lateral displacement is how much a light ray slides sideways from its original path after passing through a flat glass slab. (b) This shift is larger if the slab is thicker, if you shine the light at a sharper angle, or if the glass is denser.

Exam Tip: Keep in mind that lateral displacement is directly proportional to both the thickness of the glass block and the angle of incidence.

Question 5. (a) Why does a stick immersed obliquely in water, appear bent and short ?
Answer: When light rays travel from the submerged portion of the stick (denser medium) to the air (rarer medium), they bend away from the normal upon exiting the water surface. Our eyes perceive these refracted rays as if they are coming from a shallower point, causing the submerged part of the stick to appear raised and shorter than it actually is.

In simple words: When you look at a stick in water, the light coming from it bends as it enters the air. This tricks your eyes into seeing the stick at a shallower depth, making it look bent.

Exam Tip: Always draw arrow heads on the light rays showing direction from the object (inside water) to the observer's eye (in air) to secure full marks.

Question 5. (b) Why does a stamp placed under a glass block, appear raised?
Answer: When light rays originating from the stamp travel from the optically denser glass block into the optically rarer air, they experience refraction and bend away from the normal. When these refracted rays enter our eyes, they are projected backwards to form a virtual image of the stamp at a shallower, elevated position.
In simple words: The glass bends the light rays coming from the stamp underneath it. This makes the stamp's image appear higher up than its actual physical position.

Exam Tip: Be sure to mention that the apparent shift is due to light passing from a denser medium (glass) to a rarer medium (air).

Question 5. (c) Why is twilight formed, before sunrise or sunset ?
Answer: Twilight occurs because of atmospheric refraction of sunlight. Even when the Sun is positioned slightly below the horizon, its light rays enter the Earth's atmosphere obliquely and are bent downwards towards the surface, allowing light to reach our eyes before the actual sunrise and after sunset.
In simple words: Even when the sun is below the horizon, the air layers around the Earth bend its light rays downward so we can still see light before sunrise and after sunset.

Exam Tip: The keyword to include in this explanation is "atmospheric refraction", which is the primary cause of this phenomenon.

Question 5. (d) Why do stars twinkle, but not the planets ?
Answer: Stars appear as extremely distant point sources of light. As their light travels through the Earth's constantly moving atmospheric layers with varying temperatures and densities, the path of the light rays shifts continuously, causing the apparent position and brightness of the star to fluctuate, which results in twinkling. In contrast, planets are much closer and act as extended sources of light. The cumulative variations from different points on a planet average out, making their overall brightness steady.
In simple words: Stars are so far away that they look like tiny points of light, so the moving air easily bends and flickers their light. Planets are closer and larger, so their light averages out and stays steady.

Exam Tip: Contrast the "point source" nature of stars with the "extended source" nature of planets to get full marks on this question.

Question 5. (e) Why do the faces of people sitting around a camp fire appear to shimmer ?
Answer: The air directly above a campfire becomes extremely hot and expands, making it optically rarer than the surrounding cooler air. Since this hot air rises rapidly and mixes turbulently with cold air, its density and refractive index are in constant fluctuation. Consequently, light rays passing through this unstable air suffer continuously varying refraction, making objects on the other side appear to shimmer.
In simple words: The hot air above a fire is constantly moving and changing density. As light passes through this unstable air, it keeps bending in different directions, causing the faces on the other side to look like they are shaking.

Exam Tip: Emphasize that the continuous movement and changing density of hot air cause a constantly fluctuating refractive index, which produces the shimmering effect.

Question 5. (f) Why does a tank filled with water and seen from above appear shallow ?
Answer: Due to refraction, when light rays travel from the water (optically denser) into the air (optically rarer), they bend away from the normal. To an observer above, the refracted rays appear to diverge from a higher point than the actual bottom of the tank, making the apparent depth significantly less than the real depth.
In simple words: Light rays coming from the bottom of a water tank bend as they enter the air, making the bottom appear raised up and the tank look shallower than it really is.

Exam Tip: State the relationship \( \text{Apparent Depth} = \frac{\text{Real Depth}}{\mu} \) to demonstrate a complete understanding of this concept.

Question 5. (g) Why does a fisherman aim his spear at the tail of a fish during spear fishing.
Answer: Due to the refraction of light as it moves from water to air, the fish appears elevated and closer to the surface than its actual position. Since the real depth of the fish is greater than its apparent depth, the fisherman must aim below the apparent image (towards the tail) to hit the fish successfully.
In simple words: Refraction makes the fish look higher up in the water than it actually is. The fisherman aims lower down (at the tail) to make sure the spear hits the real, deeper location of the fish.

Exam Tip: Always mention that the real position of the fish is deeper and slightly behind its apparent image due to refraction.

Question 5. (h) Why is more than one image formed in a thick glass mirror?
Answer: When light falls on a thick glass mirror, a small portion (about 4%) is immediately reflected from the front surface of the glass, forming a faint first image. The remaining 96% enters the glass, refracts, and undergoes a strong reflection at the silvered back surface, creating the brightest second image. The light then undergoes multiple internal reflections between the front and back surfaces, yielding multiple progressively fainter subsequent images.
In simple words: Part of the light reflects off the front glass surface, but most of it reflects off the shiny back surface. As the light bounces back and forth inside the thick glass, it forms several weaker images in a row.

Exam Tip: Note that the second image is always the brightest because it is formed by reflection from the silvered back surface, which reflects nearly 96% of the light.

Question 5. (i) Why does the sun appear bigger during sunrise or sunset?
Answer: During sunrise or sunset, sunlight travels through a much thicker layer of the Earth's atmosphere to reach our eyes. This extensive air layer causes significant refraction and scattering, which optically brings the image of the Sun closer and makes it appear larger to our eyes.
In simple words: At sunrise or sunset, the sun's rays have to travel through a lot more air. This bends the light in a way that acts like a magnifying glass, making the sun look larger.

Exam Tip: Clearly mention the longer optical path of sunlight through the atmosphere at the horizon to justify the apparent increase in size.

Question 6. What is refractive index of a material ? How is it related to (a) real and apparent depth
(b) velocity of light in vacuum or air and the velocity of light in a given medium?
Answer: The refractive index of a medium is defined as the ratio of the speed of light in a vacuum (or air) to the speed of light in that specific medium.
(a) The relationship with real and apparent depth is given by: \[ \text{Refractive Index } (\mu) = \frac{\text{Real Depth}}{\text{Apparent Depth}} \]
(b) The relationship with velocity is: \[ \text{Refractive Index } (\mu) = \frac{\text{Velocity of light in vacuum or air}}{\text{Velocity of light in the medium}} \]
In simple words: Refractive index is a number that tells us how much a material slows down light. It is also the ratio of how deep an object really is compared to how shallow it looks under water or glass.

Exam Tip: Refractive index is a pure ratio of identical physical quantities, so it has no unit. Always state this fact if asked about its unit in exams.

Multiple Choice Questions

Question 1. When a beam of light strikes a glass slab a part of it is :
(a) reflected
(b) absorbed
(c) transmitted
(d) all of the options
Answer: (a) reflected
In simple words: When a light beam hits a glass surface, some of that light immediately bounces off the front surface as a reflection.

Exam Tip: Remember that even transparent glass reflects a small portion (about 4%) of light incident on it.

Question 2. The phenomenon due to which a ray of light deviates from its path while travelling from one optical medium to another optical medium is called :
(a) dispersion
(b) refraction
(c) reflection
(d) diffraction
Answer: (b) refraction
In simple words: Refraction is the scientific name for when light bends and changes direction as it goes from one material into another.

Exam Tip: The main cause of refraction is the change in the speed of light when transitioning from one optical medium to another.

Question 3. When a ray of light travelling in an optically denser medium, emerges into an optically less denser medium it :
(a) deviates towards the normal
(b) deviates away from normal
(c) does not deviate
(d) gets reflected
Answer: (b) deviates away from normal
In simple words: When light leaves a thicker material (like water or glass) and enters a thinner one (like air), it speeds up and bends away from the perpendicular normal line.

Exam Tip: When light travels from a denser to a rarer medium, the angle of refraction (\( r \)) is always greater than the angle of incidence (\( i \)).

Question 4. A ray of light strikes a glass slab at 90°. The angle of incidence is :
(a) 90°
(b) zero
(c) less than 90°, but not zero
(d) none of the options
Answer: (b) zero
In simple words: The angle of incidence is measured from the perpendicular normal line, not the glass surface. A ray that hits straight down at 90° to the glass surface is parallel to the normal, so the angle is 0°.

Exam Tip: Be careful not to confuse the angle with the surface (90°) and the angle of incidence with the normal (which is 0° in this case).

Question 5. Two medium ‘a’ and ‘b’ have same refractive index. A ray of light travelling from medium ‘a ’ to medium ‘b’. will suffer?
(a) refraction at the interfaces
(b) partly suffer reflection at the interfaces
(c) partly gets absorbed in medium ‘b ’
(d) both (b) and (c)
Answer: (b) partly suffer reflection at the interfaces
In simple words: Since both materials have the exact same refractive index, light will not bend when passing between them, but a tiny portion of the light will still reflect at the boundary.

Exam Tip: If two media have the same refractive index, no refraction (bending) occurs, making the boundary invisible to the eye.

Question 6. A ray of light on entering from medium ‘a’ to medium ‘b ’ does not suffer refraction. The angle of incidence in medium ‘a ’ is :
(a) 90°
(b) zero
(c) 45°
(d) 60°
Answer: (b) zero
In simple words: If light enters a new material straight down along the normal (angle of incidence is zero), it goes straight through without bending.

Exam Tip: A ray of light entering normally (\( i = 0^{\circ} \keys{} \)) passes undeviated regardless of the refractive indices of the media.

Question 7. During sun rise or sun set, the sun appears bigger because the rays of light coming from it pass through
(a) larger length of atmosphere
(b) smaller length of atmosphere
(c) the earth gets closer to sun
(d) none of the options
Answer: (a) larger length of atmosphere
In simple words: Near the horizon, sunlight has to travel through a much thicker layer of air, which heavily refracts and magnifies the sun's image.

Exam Tip: The increased optical path length at the horizon leads to maximum refraction and scattering effects.

Question 8. The highest refractive index is of:
(a) glass
(b) water
(c) diamond
(d) cold air
Answer: (c) diamond
In simple words: Diamond slows down light the most out of these choices, giving it the highest refractive index of about 2.42.

Exam Tip: Always remember that diamond has the highest refractive index (2.42) among common materials, which is why it sparkles so brilliantly.

Question 9. During spear fishing a fisherman aims at the :
(a) tail of fish
(b) head of fish
(c) slightly ahead of the head of fish
(d) none of the options
Answer: (a) tail of fish
In simple words: Since refraction makes the fish look shallower and higher than its actual position, the fisherman aims lower and deeper, towards the tail, to hit it.

Exam Tip: Because the apparent position is raised, aiming below the observed image is necessary to strike the real target.

Question 10. When a ray of light enters into another optical medium, its wavelength and velocity change. The material in which wavelength and velocity decrease maximum, when the ray is travelling through air is :
(a) alcohol
(b) diamond
(c) glass
(d) water
Answer: (b) diamond
In simple words: The material with the highest refractive index slows down light the most, which also causes the largest decrease in both its wavelength and speed.

Exam Tip: Wavelength and velocity in a medium are inversely proportional to its refractive index: \( v = \frac{c}{\mu} \) and \( \lambda_m = \frac{\lambda}{\mu} \).

Question 11. A thick glass slab with a silvered side forms multiple images on account of :
(a) reflection of light
(b) dispersion of light
(c) refraction of light
(d) both reflection and refraction of light
Answer: (d) both reflection and refraction of light
In simple words: Multiple images are formed because light is continuously refracted as it enters and leaves the glass slab, and reflected at the silvered and glass surfaces.

Exam Tip: The brightest second image is formed due to strong reflection from the back silvered surface, while other fainter images are caused by repeated internal reflections and refractions.

Numerical Problems on Refraction of Light Through Optical Slabs

Practice Problem 1

Question 1. The velocity of light in air is 3 × 10⁸ ms⁻¹ and in glass is 2 × 10⁸ ms⁻¹ Find the refractive index of glass.
Answer: Given data:
Velocity of light in air (\( v_{\text{air}} \)) = \( 3 \times 10^8 \text{ m s}^{-1} \)
Velocity of light in glass (\( v_{\text{glass}} \keys{} \)) = \( 2 \times 10^8 \text{ m s}^{-1} \)

The formula for the refractive index of glass (\( \mu_g \)) is: \[ \mu_g = \frac{v_{\text{air}}}{v_{\text{glass}}} \] Substituting the given values: \[ \mu_g = \frac{3 \times 10^8}{2 \times 10^8} = 1.5 \]
Thus, the refractive index of glass is \( 1.5 \).
In simple words: To find the refractive index, we divide the speed of light in air by its speed in glass. Since light travels 1.5 times slower in glass, the refractive index is 1.5.

Exam Tip: Always double-check that you cancel the common unit of speed (\( \text{m s}^{-1} \)), as refractive index is a pure ratio with no unit.

Question 2. The velocity of light in air is 3 × 10⁸ ms⁻¹. Calculate the velocity of light in diamond or refractive index 2.5.
Answer: Given data:
Velocity of light in air (\( v_{\text{air}} \)) = \( 3 \times 10^8 \text{ m s}^{-1} \)
Refractive index of diamond (\( \mu_d \)) = \( 2.5 \)

Using the formula: \[ \mu_d = \frac{v_{\text{air}}}{v_{\text{diamond}}} \] We can solve for the velocity in diamond: \[ v_{\text{diamond}} = \frac{v_{\text{air}}}{\mu_d} \] Substituting the given values: \[ v_{\text{diamond}} = \frac{3 \times 10^8}{2.5} = 1.2 \times 10^8 \text{ m s}^{-1} \]
The velocity of light in diamond is \( 1.2 \times 10^8 \text{ m s}^{-1} \).
In simple words: Since light travels 2.5 times slower in diamond than in air, we divide the speed of light in air by 2.5 to find that light travels at \( 1.2 \times 10^8 \text{ m/s} \) in diamond.

Exam Tip: Be sure to write the unit of velocity (\( \text{m s}^{-1} \) or \( \text{m/s} \keys{} \)) in your final numerical answer to avoid losing marks.

Practice Problem 2

Question 1. The angle of refraction in a glass block of refractive index 1.5 is 19°. Calculate the angle of incidence.
Answer: Given data:
Refractive index of glass block (\( \mu \)) = \( 1.5 \)
Angle of refraction (\( r \)) = \( 19^{\circ} \)

Using Snell's Law: \[ \mu = \frac{\sin i}{\sin r} \] Substitute the given values: \[ 1.5 = \frac{\sin i}{\sin 19^{\circ}} \]
\( \implies \sin i = 1.5 \sin 19^{\circ} \keys{} \)
In simple words: Using the law of refraction, the sine of the angle of incidence is calculated by multiplying the refractive index of glass (1.5) by the sine of the angle of refraction (\( 19^{\circ} \)).

Exam Tip: If values of non-standard trigonometric angles like \( \sin 19^{\circ} \) are not provided in the exam paper, it is completely acceptable to leave your final expression in terms of the trigonometric function.

Question 2. Calculate the refractive index of a material, when angle of incidence in air is 50° and angle of refraction in the material is 36°.
Answer: Given parameters:
Angle of incidence (\( i \)) = \( 50^{\circ} \)
Angle of refraction (\( r \)) = \( 36^{\circ} \)

Using the formula for refractive index (R.I.): \[ \text{R.I.} = \frac{\sin i}{\sin r} \] Substituting the given values: \[ \text{R.I.} = \frac{\sin 50^{\circ}}{\sin 36^{\circ}} \]
In simple words: The refractive index is found by dividing the sine of the angle of incidence (\( 50^{\circ} \)) by the sine of the angle of refraction (\( 36^{\circ} \)).

Exam Tip: Ensure you clearly label the angles as \( i \) and \( r \) and write down the general formula for Snell's law before plugging in the values.

Practice Problem 3

Question 1. A coin is placed at a depth of 15 cm in a beaker containing water. The refractive index of water is 4/3, calculate height through which the image of the coin is raised.
Answer: Given data:
Real depth of the coin (\( d_{\text{real}} \)) = \( 15 \text{ cm} \)
Refractive index of water (\( \mu \)) = \( \frac{4}{3} \)

First, we find the apparent depth (\( d_{\text{apparent}} \)): \[ \mu = \frac{d_{\text{real}}}{d_{\text{apparent}}} \] \[ \frac{4}{3} = \frac{15}{d_{\text{apparent}}} \]
\( \implies d_{\text{apparent}} = \frac{15 \times 3}{4} = 11.25 \text{ cm} \)

Next, we calculate the shift in height (height raised): \[ \text{Shift} = d_{\text{real}} - d_{\text{apparent}} \] \[ \text{Shift} = 15 - 11.25 = 3.75 \text{ cm} \]
Therefore, the image of the coin is raised by \( 3.75 \text{ cm} \).
In simple words: Due to the water bending light, the bottom appears shallower at 11.25 cm instead of 15 cm. This makes the coin look like it has been lifted up by 3.75 cm.

Exam Tip: Be careful not to stop at calculating the apparent depth; the question asks for the height through which the image is *raised*, which is the difference between real and apparent depths.

Question 2. The floor of a water tank appears at a depth of 2.5 m. If the refractive index of water is 1.33, find the actual depth of water.
Answer: Given data:
Apparent depth (\( d_{\text{apparent}} \)) = \( 2.5 \text{ m} \)
Refractive index of water (\( \mu \)) = \( 1.33 \)

Using the relationship: \[ \mu = \frac{d_{\text{real}}}{d_{\text{apparent}}} \] We can solve for the actual depth: \[ d_{\text{real}} = \mu \times d_{\text{apparent}} \] \[ d_{\text{real}} = 1.33 \times 2.5 = 3.325 \text{ m} \]
Thus, the actual depth of water in the tank is \( 3.325 \text{ m} \).
In simple words: The tank floor appears to be 2.5 m deep, but since water bends light, the actual physical depth is 1.33 times greater, which is 3.325 m.

Exam Tip: When using the refractive index value 1.33, you can use \( \frac{4}{3} \) to easily carry out the calculation if you prefer exact fractions, which yields \( 3.33 \text{ m} \).

Practice Problem 4

Question 1. A stone placed at the bottom of a water tank appears raised by 80 cm. If the refractive index of water is 4/3, find the actual depth of water in the tank :
Answer: Let the actual depth of water be \( x \text{ cm} \).
Since the stone appears raised by \( 80 \text{ cm} \), the apparent depth is: \[ d_{\text{apparent}} = (x - 80) \text{ cm} \]
Using the formula for refractive index (R.I.): \[ \text{R.I.} = \frac{d_{\text{real}}}{d_{\text{apparent}}} \] Substituting the given values: \[ \frac{4}{3} = \frac{x}{x - 80} \] Cross-multiplying gives: \[ 4(x - 80) = 3x \] \[ 4x - 320 = 3x \]
\( \implies x = 320 \text{ cm} \)

So, the actual depth of water in the tank is \( 320 \text{ cm} \) (or \( 3.2 \text{ m} \)).
In simple words: Since water makes the stone look 80 cm higher than its real place, we can set up an equation using the water's bending index of 4/3. Solving this shows the real depth of the tank is 320 cm.

Exam Tip: Setting up the algebraic equation \( \mu = \frac{x}{x - \text{shift}} \) is the most reliable method for solving problems where the apparent upward displacement is given.

Exercise - 2

Question 1. (a) What do you understand by the following terms.
1. Total intenta! reflection
2. Critical angle ?
(b) Stale two conditions for total internal reflection ?
Answer: (a)
1. When a ray of light travels from an optically denser medium to an optically rarer medium, and the angle of incidence is greater than the critical angle, the ray does not refract but instead gets completely reflected back into the same denser medium, obeying the laws of reflection. This phenomenon is known as total internal reflection.
2. The critical angle is defined as the angle of incidence in the optically denser medium for which the corresponding angle of refraction in the optically rarer medium is exactly 90°.

(b) The two essential conditions required for total internal reflection to occur are:
1. Light must travel from an optically denser medium to an optically rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle for that pair of media.

In simple words: (a) 1. Total internal reflection happens when light inside a material hits the boundary so flatly that it bounces completely backward like a mirror. 2. Critical angle is the special tipping-point angle where the exiting light slides perfectly flat along the surface. (b) For this to occur, light must go from a thicker material to a thinner one, and it must strike at an angle larger than the critical angle.

Exam Tip: When stating the conditions for total internal reflection, always emphasize both points: light must travel from a denser to a rarer medium, and the angle of incidence must exceed the critical angle.

Question 2. (a) What do you understand by the statement, “critical angle for water is 48° ?
(b) Explain, how that refractive index of material is related to the critical angle.
Answer: (a) This statement means that when a light ray travels from water to air at an angle of incidence equal to 48°, the refracted ray will emerge along the boundary of separation, making an angle of refraction of exactly 90° in the air.
(b) The refractive index (\( \mu \)) of a medium is inversely proportional to the sine of its critical angle (\( C \)). It is expressed mathematically as: \[ \mu = \frac{1}{\sin C} \]
In simple words: (a) Saying the critical angle for water is 48° means if you shine a light from under water at exactly 48°, the light will graze flat along the water's surface. (b) The refractive index is simply 1 divided by the sine of this critical angle.

Exam Tip: Be prepared to use the formula \( \mu = \frac{1}{\sin C} \) in numerical problems. Remember that a higher refractive index corresponds to a smaller critical angle.

Question 3. Explain the following :
(a) An empty test tube placed obliquely in water, appears to be filled with mercury.
(b) Bubbles rising up in a fish tank appear silvery.
(c) Air bubbles trapped in a glass paper weight appear silvery.
(d) A crack in window pane appear silvery.
(e) Diamonds sparkle for some time in dark.
(f) The top surface of water contained in a beaker and held above the eye level appear silvery.
Answer: (a) When an empty test tube is placed obliquely in water, light rays travelling through water (denser medium) strike the glass-air boundary of the test tube at an angle of incidence greater than the critical angle of water. These rays undergo total internal reflection and bounce back to our eyes, making the surface of the test tube shine like highly reflective mercury.

(b) As air bubbles rise in a fish tank, light rays traveling through water strike the water-air interface of the bubbles at angles of incidence exceeding the critical angle of water (48°). The rays are totally internally reflected, giving the bubble a bright, silvery appearance.
(c) Similarly, light traveling through a glass paper weight (denser medium) strikes the glass-air boundary of the trapped bubble at an angle greater than the critical angle of glass (42°). Total internal reflection occurs, making the bubble shine like silver.
(d) A crack in a glass window pane traps a thin layer of air. When light travels through glass and meets the glass-air interface at the crack at an angle of incidence greater than the critical angle, it undergoes total internal reflection, making the crack look silvery.
(e) A diamond is cut with many sharp-angled facets so that light entering it faces a very small critical angle (24°). This causes the light to undergo multiple total internal reflections inside the diamond before it can exit, keeping the light trapped and making the diamond sparkle intensely even when viewed in a dimly lit setting.
(f) When water in a beaker is held above eye level, light traveling upwards through the water meets the water-air surface at an angle of incidence greater than the critical angle. The light is totally internally reflected downwards towards our eyes, making the surface look like a polished silver mirror.
In simple words: All these things look silvery or shiny because of total internal reflection. When light inside a denser material hits a layer of air at a very flat angle, it cannot escape and bounces back completely, acting like a highly polished mirror.

Exam Tip: Whenever you are asked to explain why bubbles, cracks, or glass surfaces appear "silvery", "shiny", or "mercury-like", the key scientific concept to name is "total internal reflection" of light.

Question 4. (a) What is a totally reflecting prism ?
(b) By drawing neat diagram explain how totally reflecting prisms are used to turn (i) rays through 90° (ii) rays through 180°.
(c) How is a totally reflecting prism used as an erecting prism ?
Answer: (a) A totally reflecting prism is a right-angled isosceles prism (having angles \( 90^{\circ}, 45^{\circ}, 45^{\circ} \)) designed such that light entering normally through one of its faces strikes the inner surface at an angle of incidence (\( 45^{\circ} \)) greater than the critical angle of glass (\( 42^{\circ} \)), causing the light to undergo total internal reflection.

(b)
(i) To turn rays through 90°: Light is incident normally on one of the perpendicular faces of the prism. It enters undeflected and strikes the hypotenuse at \( 45^{\circ} \), undergoing total internal reflection. The reflected ray exits normally through the other perpendicular face, having turned through exactly 90°.

(ii) To turn rays through 180°: Light enters normally through the hypotenuse face of the prism. It strikes one perpendicular face at \( 45^{\circ} \) (reflecting totally), then strikes the other perpendicular face at \( 45^{\circ} \) (reflecting totally again), and finally exits normally through the hypotenuse, turned through a full 180°.

(c) Used as an erecting prism: An erecting prism is placed so that light from an inverted object enters parallel to the hypotenuse. The rays refract at the first face, strike the hypotenuse at an angle greater than \( 42^{\circ} \), suffer total internal reflection, and then refract out of the second face to emerge as an erect, parallel beam. This flips the image right-side up without changing its lateral position.
In simple words: (a) A totally reflecting prism is a special glass triangle with a 90-degree angle that acts like a highly efficient mirror. (b) By positioning it differently, you can turn a light beam sideways by 90 degrees or turn it completely around by 180 degrees. (c) When used to erect an image, it flips an upside-down image back to upright as light enters and bounces inside the prism.

Exam Tip: Be sure to label the angles (\( 45^{\circ}, 90^{\circ}, 45^{\circ} \)) of the prism and draw arrows indicating normal incidence (\( 90^{\circ} \) to the face) to demonstrate complete ray diagrams in exams.

Question 5. (a) Trace the course of rays through an equilateral glass prism, showing clearly
1. angle of incidence
2. angle of refraction
3. angle of the prism
4. angle of deviation
5. angle of emergence.
(b) On what factors do the angle of deviation in a prism depend?
(c) What do you understand by the term angle of minimum deviation ? In this position how is the angle of incidence related to the angle of emergence ?
Answer: (a) The ray path through an equilateral glass prism is described as follows:
1. Angle of incidence (\( i \)) is the angle between the incident ray and the normal at the first refracting face.
2. Angle of refraction (\( r \)) is the angle between the refracted ray and the normal inside the prism.
3. Angle of the prism (\( A \)) is the angle between the two refracting faces at the apex.
4. Angle of deviation (\( \delta \)) is the angle between the direction of the incident ray produced forward and the emergent ray produced backward.
5. Angle of emergence (\( e \)) is the angle between the emergent ray and the normal at the second refracting face.

(b) The angle of deviation depends on the following four factors:
1. The angle of incidence (\( i \)) of the light.
2. The refracting angle of the prism (\( A \)).
3. The refractive index (\( \mu \)) of the material of the prism.
4. The wavelength (color) of the incident light.

(c) The angle of minimum deviation (\( \delta_m \)) is defined as the unique, lowest value of the angle of deviation achieved when the angle of incidence is adjusted. At this minimum position, the angle of incidence (\( i \)) becomes exactly equal to the angle of emergence (\( e \)), and the refracted ray travels parallel to the base of the prism.
In simple words: (a) When light goes through a prism, it bends twice, causing its final path to turn away from its original path by an angle called the angle of deviation. (b) This bending depends on the angle of incoming light, the prism's glass type, its shape, and the light's color. (c) Minimum deviation is the smallest possible turn the light can make, which happens when the entry and exit angles are exactly equal.

Exam Tip: Remember the critical relation for the minimum deviation position: \( i = e \) and \( r_1 = r_2 \). Mentioning that the refracted ray inside the prism is parallel to its base is a highly valued point by examiners.

Question 6. State four differences between reflection and total internal reflection.
Answer: Differences between Ordinary Reflection and Total Internal Reflection:

In simple words: Ordinary reflection happens on any surface and always wastes some light energy, whereas total internal reflection only happens from inside a denser material and reflects 100% of the light with zero energy loss.

Exam Tip: Highlighting that "100% of light is reflected" and "there is no loss of energy" are the two primary distinct characteristics of total internal reflection compared to ordinary reflection.

Multiple Choice Questions

Question 1. For total internal reflection to take place a ray of light must :
(a) travel from denser to rarer medium
(b) travel from rarer to denser medium
(c) medium does not play any role
(d) none of the options
Answer: (a) travel from denser to rarer medium
In simple words: Total internal reflection can only happen when light is trying to leave a thicker material (like water or glass) and enter a thinner material (like air).

Exam Tip: This is a fundamental prerequisite condition for total internal reflection. Always keep this direction of travel in mind.

Question 2. The critical angle for glass is 42°. The corresponding angle of refraction is :
(a) 0°
(b) 90°
(c) lesser than 90° but more than 42°
(d) no angle of refraction.
Answer: (b) 90°
In simple words: At the critical angle, the refracted light bends so far that it travels perfectly flat along the glass surface, making an angle of 90 degrees with the normal.

Exam Tip: By definition, the angle of refraction in the rarer medium corresponding to the critical angle in the denser medium is always exactly \( 90^{\circ} \).

Question 3. The critical angle for a material X is 45°. The total internal reflection will take place, if the angle of incidence in the denser medium is :
(a) less than 45°
(b) 90°
(c) more than 45°, but not 90°
(d) less than 45°, but not zero degree
Answer: (c) more than 45°, but not 90°
In simple words: Total internal reflection only starts once the entry angle is larger than the critical angle (greater than 45°), but it cannot be 90° because that would mean the light is travelling along the boundary interface rather than being incident on it.

Exam Tip: The angle of incidence must be strictly greater than the critical angle (\( i > C \)) for total internal reflection to occur.

Question 4. Diamonds sparkle more than the glass, because they have :
(a) smaller critical angle than the glass
(b) larger critical angle than the glass
(c) critical angle plays no role
(d) none of the options
Answer: (a) smaller critical angle than the glass
In simple words: Diamond has a very small critical angle of 24°, making it extremely easy for light rays to get trapped and bounce around inside, producing a brilliant sparkle.

Exam Tip: A smaller critical angle increases the likelihood of total internal reflection occurring over a wider range of incident angles, which is why diamonds sparkle intensely.

Question 5. Small air bubbles rising up a fish tank appear silvery when viewed from some particular angle because of the phenomenon of :
(a) reflection
(b) refraction
(c) total internal reflection
(d) dispersion
Answer: (c) total internal reflection
In simple words: When viewed from certain flat angles, light hitting the water-air interface of the bubble bounces completely back to our eyes, making the bubble look like highly reflective silver.

Exam Tip: Remember that any trapped air bubble inside water or glass will appear silvery when viewed at angles exceeding the critical angle due to total internal reflection.

Question 6. An isosceles totally reflecting prism can reflect rays through an angle of :
(a) 60°
(b) 90°
(c) 180°
(d) both (b) and (c)
Answer: (d) both (b) and (c)
In simple words: This prism can be oriented to bend light by either 90 degrees or turn it around completely by 180 degrees.

Exam Tip: Depending on which face the light enters normally (the perpendicular side or the hypotenuse), a right-angled isosceles prism can perform 90-degree or 180-degree deviations.

Question 7. A ray of light is incident on the face of an equilateral prism at angle of 90°. The ray gets totally reflected on the second refracting face. The total deviation produced in the path of ray is :
(a) 60°
(b) 90°
(c) 120°
(d) 180°
Answer: (c) 120°
In simple words: Under these specific angles, the ray suffers a total internal reflection and exits the prism turned away by exactly 120 degrees from its original direction.

Exam Tip: Verify the angles geometrically: for normal incidence, the ray travels straight inside the equilateral prism and hits the second face at \( 60^{\circ} \) (since the prism angle is \( 60^{\circ} \)). This exceeds the critical angle, leading to reflection and a net deviation of \( 120^{\circ} \).

Question 8. A crack in the window pane appears silvery when viewed from some particular angle. This phenomenon due to :
(a) refelction light
(b) refraction of light
(c) total internal reflection of light
(d) dispersion of light
Answer: (c) total internal reflection of light
In simple words: Air inside the crack forms a boundary with glass. When you look at a flat enough angle, light cannot pass through this air boundary and reflects completely, making the crack look silvery.

Exam Tip: The presence of a thin layer of air trapped in glass cracks makes total internal reflection possible at angles greater than \( 42^{\circ} \).

Question 9. When an equilateral prism is in minimum deviation position the angle of incidence is :
(a) greater than the angle of emergence
(b) smaller than the angle of emergence
(c) equal to the angle of emergence
(d) none of the options
Answer: (c) equal to the angle of emergence
In simple words: At the minimum deviation position, the path of light is completely symmetrical, so the entering angle of incidence equals the exiting angle of emergence.

Exam Tip: The condition of symmetry (\( i = e \)) defines the minimum deviation position in a prism.

Question 10. A prism has :
(a) two rectangular and three triangular surfaces
(b) two triangular and three rectangular surfaces
(c) three rectangular and three triangular surfaces
(d) none of the options
Answer: (b) two triangular and three rectangular surfaces
In simple words: A standard triangular prism has two flat triangular faces at the ends and three rectangular sides connecting them.

Exam Tip: A triangular prism is a five-sided solid consisting of three rectangular lateral surfaces and two triangular bases.

Question 11. When a ray of light passes through an equilateral glass prism :
(a) it suffers refraction on the first refracting surfaces
(b) it suffers refraction on both the refracting surfaces
(c) it bends towards the base on both refracting surfaces
(d) both (b) and (c)
Answer: (d) both (b) and (c)
In simple words: As light enters and leaves the glass prism, it gets refracted at both glass-air boundaries, and in both instances, it bends downwards towards the prism's base.

Exam Tip: A ray of light always bends towards the base of the prism during both refraction processes because of the wedge shape of the prism.

Numerical Problems on Lenses

Practice Problems 1

Question 1. A convex lens of focal length 10 cm is placed at a distance of 60 cm from a screen. How far from the lens should be placed an object so as to obtain a real image on the screen? Calculate the magnification of the image and its characteristics.
Answer: Given data:
Focal length of convex lens (\( f \)) = \( +10 \text{ cm} \)
Image distance to screen (\( v \)) = \( +60 \text{ cm} \)

Using the lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Substitute the given values: \[ \frac{1}{60} - \frac{1}{u} = \frac{1}{10} \] Rearranging to solve for \( \frac{1}{u} \): \[ \frac{1}{u} = \frac{1}{60} - \frac{1}{10} \] \[ \frac{1}{u} = \frac{1 - 6}{60} = -\frac{5}{60} = -\frac{1}{12} \]
\( \implies u = -12 \text{ cm} \)

The negative sign indicates that the object must be placed on the left side of the lens, at a distance of \( 12 \text{ cm} \).

Now, we calculate the magnification (\( m \)): \[ m = \frac{v}{u} = \frac{60}{-12} = -5 \]
The characteristics of the image are:
1. The image is real.
2. The image is inverted (as indicated by the negative magnification).
3. The image is magnified, being \( 5 \) times the size of the object.
In simple words: To get a sharp image on a screen 60 cm away, the object must be placed 12 cm in front of the lens. This creates a real, upside-down image that is 5 times larger than the object.

Exam Tip: Remember to use sign conventions correctly: for a convex lens forming a real image, \( f \) and \( v \) are positive, while the object distance \( u \) will naturally solve to a negative value.

Question 2. An object of height 3 cm is placed at a distance of 24 cm from a convex lens of focal length 10 cm, when an image is formed on the screen on the other side of the lens. Calculate
(a) the distance of the screen from the lens
(b) the size of image
(c) the characteristics of image.
Answer: Given data:
Height of the object (\( h_1 \)) = \( 3 \text{ cm} \)
Object distance (\( u \)) = \( -24 \text{ cm} \) (by sign convention)
Focal length of convex lens (\( f \)) = \( +10 \text{ cm} \)

(a) Using the lens formula to find the image distance (\( v \)), which is the distance of the screen: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] \[ \frac{1}{v} - \frac{1}{-24} = \frac{1}{10} \] \[ \frac{1}{v} = \frac{1}{10} - \frac{1}{24} \] \[ \frac{1}{v} = \frac{12 - 5}{120} = \frac{7}{120} \]
\( \implies v = \frac{120}{7} \approx 17.14 \text{ cm} \)
The screen should be placed at a distance of \( 17.14 \text{ cm} \) from the lens on the opposite side.

(b) Using the magnification formula to find the height of the image (\( h_2 \)): \[ m = \frac{v}{u} = \frac{h_2}{h_1} \] \[ \frac{120/7}{-24} = \frac{h_2}{3} \]
\( \implies h_2 = 3 \times \left( -\frac{5}{7} \right) = -\frac{15}{7} \approx -2.14 \text{ cm} \)
The size of the image is \( 2.14 \text{ cm} \) (inverted).

(c) The characteristics of the image are:
1. Real (formed on the other side of the lens).
2. Inverted (as shown by the negative value of \( h_2 \)).
3. Diminished (since \( 2.14 \text{ cm} < 3 \text{ cm} \)).
In simple words: (a) The screen should be set up at a distance of 17.14 cm from the lens. (b) The image formed is 2.14 cm tall. (c) The image is real, upside-down, and smaller than the original 3 cm tall object.

Exam Tip: Keep fractional forms like \( \frac{120}{7} \) in intermediate steps to calculate the height of the image precisely and avoid rounding errors.

Practice Problems 2

Question 1. An object when placed in front of a convex lens forms a real image of 0.5 magnification. If the distance of the image from the lens is 24 cm, calculate focal length of the lens.
Answer: We can interpret this question in two ways depending on whether the 24 cm is the image distance or the object distance:

Case I: If 24 cm is the distance of the real image from the lens (\( v \)):
- Image distance (\( v \)) = \( +24 \text{ cm} \) (positive for a real image)
- Magnification (\( m \keys{} \)) = \( -0.5 \) (negative for a real, inverted image)

Using the magnification formula: \[ m = \frac{v}{u} \] \[ -0.5 = \frac{24}{u} \]
\( \implies u = -48 \text{ cm} \)

Using the lens formula to find the focal length (\( f \)): \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{24} - \frac{1}{-48} = \frac{1}{24} + \frac{1}{48} = \frac{2+1}{48} = \frac{3}{48} \]
\( \implies f = 16 \text{ cm} \)

Case II: If 24 cm is the distance of the object from the lens (\( u \)):
- Object distance (\( u \)) = \( -24 \text{ cm} \)
- Magnification (\( m \)) = \( -0.5 \) (real, inverted image)

Using the magnification formula: \[ m = \frac{v}{u} \] \[ -0.5 = \frac{v}{-24} \]
\( \implies v = +12 \text{ cm} \)

Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{12} - \frac{1}{-24} = \frac{1}{12} + \frac{1}{24} = \frac{2+1}{24} = \frac{3}{24} \]
\( \implies f = 8 \text{ cm} \keys{} \)
In simple words: If 24 cm is the image distance, the object distance is 48 cm, giving a focal length of 16 cm. If 24 cm is the object distance, the image is formed at 12 cm, giving a focal length of 8 cm.

Exam Tip: Remember that a real image is always inverted, so the magnification (\( m \)) must be substituted as negative (\( -0.5 \)) in calculations.

Question 2. A convex lens forms a real image 4 times magnified when placed at a distance of 6 cm from the lens. Calculate the focal length of the lens.
Answer: Given data:
Object distance (\( u \)) = \( -6 \text{ cm} \) (by sign convention)
Magnification (\( m \)) = \( -4 \) (negative for a real, inverted image)

Using the magnification formula: \[ m = \frac{v}{u} \] \[ -4 = \frac{v}{-6} \]
\( \implies v = +24 \text{ cm} \)

Using the lens formula to find the focal length (\( f \)): \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{24} - \frac{1}{-6} \] \[ \frac{1}{f} = \frac{1}{24} + \frac{1}{6} = \frac{1+4}{24} = \frac{5}{24} \]
\( \implies f = \frac{24}{5} = 4.8 \text{ cm} \)
In simple words: Since the object is placed 6 cm away and magnified 4 times, the image is formed at 24 cm on the other side. Using the lens formula, we find the lens has a focal length of 4.8 cm.

Exam Tip: Be sure to write out the full magnification equation \( m = \frac{v}{u} \) and substitute the negative value for real magnification to avoid wrong signs in the lens equation.

Practice Problems 3

Question 1. An object 1.4 cm high when placed in front of a convex lens at a distance of 6 cm, forms a virtual image at a distance of 24 cm from the lens. Calculate (a) the focal length of the lens (b) the size of the image.
Answer: Given data:
Height of the object (\( h_1 \)) = \( 1.4 \text{ cm} \)
Object distance (\( u \)) = \( -6 \text{ cm} \) (by sign convention)
Image distance (\( v \)) = \( -24 \text{ cm} \) (since a virtual image is formed on the same side as the object)

(a) Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Substitute the given values: \[ \frac{1}{f} = \frac{1}{-24} - \frac{1}{-6} \] \[ \frac{1}{f} = -\frac{1}{24} + \frac{1}{6} = \frac{-1 + 4}{24} = \frac{3}{24} = \frac{1}{8} \]
\( \implies f = 8 \text{ cm} \)
The focal length of the convex lens is \( 8 \text{ cm} \).

(b) Using the magnification formula: \[ m = \frac{v}{u} = \frac{h_2}{h_1} \] Substitute the values: \[ \frac{-24}{-6} = \frac{h_2}{1.4} \] \[ 4 = \frac{h_2}{1.4} \]
\( \implies h_2 = 4 \times 1.4 = 5.6 \text{ cm} \)
The size (height) of the virtual image is \( 5.6 \text{ cm} \).
In simple words: (a) Using the lens formula, we find that the focal length of the lens is 8 cm. (b) Since the image distance is 4 times the object distance, the image is enlarged 4 times, making its height 5.6 cm.

Exam Tip: Always remember that virtual images formed by lenses are on the same side as the object, meaning the image distance \( v \) must be substituted as a negative value.

Question 2. A convex lens forms a 2.5 times magnified virtual image when an object is placed at a distance of 8 cm from the lens. Calculate (a) the distance of the image from the lens (b) the focal length of lens.
Answer: Given data:
Magnification (\( m \)) = \( +2.5 \) (positive because a virtual image is always erect)
Object distance (\( u \)) = \( -8 \text{ cm} \) (by sign convention)

(a) Using the magnification formula: \[ m = \frac{v}{u} \] \[ 2.5 = \frac{v}{-8} \]
\( \implies v = 2.5 \times (-8) = -20 \text{ cm} \)
The distance of the virtual image from the lens is \( 20 \text{ cm} \) on the same side as the object.

(b) Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Substitute the values of \( v \) and \( u \): \[ \frac{1}{f} = \frac{1}{-20} - \frac{1}{-8} \] \[ \frac{1}{f} = -\frac{1}{20} + \frac{1}{8} = \frac{-2 + 5}{40} = \frac{3}{40} \]
\( \implies f = \frac{40}{3} \approx 13.33 \text{ cm} \)
The focal length of the convex lens is \( 13.33 \text{ cm} \).
In simple words: (a) Since the lens magnifies the image 2.5 times, the image is formed 20 cm away on the same side as the object. (b) Combining these distances in the lens formula gives a focal length of 13.33 cm.

Exam Tip: A convex lens can form both real (inverted) and virtual (erect) images. Since the question specifies a virtual image, make sure magnification (\( m \)) is taken as positive.

Question 3. An object 1 cm high is placed at a distance of 4 cm from a convex lens of focal length 6 cm. Calculate (a) the position of the image (b) size of a image. State the characteristics of the image.
Answer: Given data:
Height of the object (\( h_1 \)) = \( 1 \text{ cm} \)
Object distance (\( u \)) = \( -4 \text{ cm} \)
Focal length of convex lens (\( f \)) = \( +6 \text{ cm} \)

(a) Using the lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Rearranging for \( \frac{1}{v} \): \[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \] Substitute the values: \[ \frac{1}{v} = \frac{1}{6} + \frac{1}{-4} = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = -\frac{1}{12} \]
\( \implies v = -12 \text{ cm} \)
The image is formed at a distance of \( 12 \text{ cm} \) on the same side of the lens as the object.

(b) Using the magnification formula: \[ m = \frac{h_2}{h_1} = \frac{v}{u} \] \[ \frac{h_2}{1} = \frac{-12}{-4} = 3 \]
\( \implies h_2 = 3 \text{ cm} \)
The size (height) of the image is \( 3 \text{ cm} \).

The characteristics of the image are:
1. The image is virtual.
2. The image is erect.
3. The image is magnified (3 times larger than the object).
In simple words: (a) The image forms 12 cm away on the same side as the object. (b) This virtual image is upright and is 3 times larger than the original object, reaching a height of 3 cm.

Exam Tip: Whenever the object distance is less than the focal length of a convex lens (\( |u| < f \)), the resulting image is always virtual, erect, and magnified.

Practice Problems 4

Question 1. An object 2 cm high is placed at a distance of 25 cm from the optical centre of a concave lens offocal length 15 cm. Calculate (a) the position of the image (b) the size of the image.
Answer: Given data:
Height of the object (\( h_1 \)) = \( 2 \text{ cm} \)
Object distance (\( u \)) = \( -25 \text{ cm} \)
Focal length of concave lens (\( f \)) = \( -15 \text{ cm} \) (focal length of a concave lens is always negative)

(a) Using the lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Rearranging for \( \frac{1}{v} \): \[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \] Substitute the values: \[ \frac{1}{v} = \frac{1}{-15} + \frac{1}{-25} = -\frac{1}{15} - \frac{1}{25} = \frac{-5 - 3}{75} = -\frac{8}{75} \]
\( \implies v = -\frac{75}{8} = -9.375 \text{ cm} \)
The image is formed at a distance of \( 9.375 \text{ cm} \) on the same side as the object.

(b) Using the magnification formula: \[ m = \frac{h_2}{h_1} = \frac{v}{u} \] \[ \frac{h_2}{2} = \frac{-75/8}{-25} = \frac{75}{8 \times 25} = \frac{3}{8} \]
\( \implies h_2 = \frac{3}{8} \times 2 = 0.75 \text{ cm} \)
The size (height) of the image is \( 0.75 \text{ cm} \).
In simple words: (a) For a concave lens, the image is always formed on the same side as the object, which is at a distance of 9.375 cm here. (b) The image is upright and shrunk down to a height of just 0.75 cm.

Exam Tip: Concave lenses only form virtual, erect, and diminished images. Thus, both your image distance (\( v \)) and image size (\( h_2 \)) must solve to be smaller in magnitude than the object's distance and height respectively.

Question 2. A concave lens forms 4 times diminished and virtual image when an object is placed at a distance of 80 cm. Calculate (a) the position of the image (b) the focal length of the lens.
Answer: Given data:
Magnification (\( m \)) = \( +\frac{1}{4} \) (positive because a virtual image is erect)
Object distance (\( u \)) = \( -80 \text{ cm} \)

(a) Using the magnification formula: \[ m = \frac{v}{u} \] \[ \frac{1}{4} = \frac{v}{-80} \]
\( \implies v = \frac{-80}{4} = -20 \text{ cm} \)
The position of the image is at a distance of \( 20 \text{ cm} \) on the same side of the lens as the object.

(b) Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Substitute the values of \( v \) and \( u \): \[ \frac{1}{f} = \frac{1}{-20} - \frac{1}{-80} \] \[ \frac{1}{f} = -\frac{1}{20} + \frac{1}{80} = \frac{-4 + 1}{80} = -\frac{3}{80} \]
\( \implies f = -\frac{80}{3} \approx -26.67 \text{ cm} \)
The focal length of the concave lens is \( -26.67 \text{ cm} \).
In simple words: (a) Since the image is shrunk by 4 times, it forms 20 cm in front of the lens. (b) Inserting this into the lens equation reveals that the lens has a focal length of -26.67 cm.

Exam Tip: Be sure to keep the negative sign for the focal length in your final answer, as the negative sign physically signifies that the lens is concave (diverging).

Question 3. A concave lens has focal length 15 cm. At what distance should the object from the lens be placed, so as to form an image at 10 cm from the lens. Also find magnification of the lens.
Answer: Given data:
Focal length of concave lens (\( f \)) = \( -15 \text{ cm} \)
Image distance (\( v \)) = \( -10 \text{ cm} \) (as a concave lens always forms a virtual image on the same side as the object)

Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Rearranging for \( \frac{1}{u} \): \[ \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \] Substitute the values: \[ \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = -\frac{1}{30} \]
\( \implies u = -30 \text{ cm} \)
The object should be placed at a distance of \( 30 \text{ cm} \) in front of the concave lens.

Now, we calculate the magnification (\( m \)): \[ m = \frac{v}{u} = \frac{-10}{-30} = \frac{1}{3} \approx 0.33 \]
In simple words: To get an image at 10 cm, we must place the object 30 cm in front of the lens. The image is upright and shrunk to one-third of the object's real size.

Exam Tip: Even if the question doesn't explicitly state that the image is virtual, remember that a concave lens can *only* form virtual images. Therefore, you must assign a negative sign to the image distance (\( v \)) from the start.

Practice Problems 5

Question 1. A converging lens has a focal length 40 cm. Calculate its power.
Answer: Given data:
Focal length of converging (convex) lens (\( f \)) = \( +40 \text{ cm} = \frac{40}{100} \text{ m} = +0.4 \text{ m} \)

The formula for the power of a lens (\( P \)) is: \[ P = \frac{1}{f \text{ (in metres)}} \] Substituting the value of focal length: \[ P = \frac{1}{0.4} = +2.5 \text{ D} \]
The power of the converging lens is \( +2.5 \text{ D} \).
In simple words: A focal length of 40 cm is equal to 0.4 meters. Dividing 1 by 0.4 gives us a lens power of +2.5 Dioptres.

Exam Tip: Always convert the focal length from centimeters to meters before calculating power, and remember to express the final power with its unit "D" (Dioptre).

Question 2. A lens which forms a real image has a focal length 8 cm. Calculate its power.
Answer: Since the lens forms a real image, it must be a convex (converging) lens.
Focal length of the lens (\( f \)) = \( +8 \text{ cm} = \frac{8}{100} \text{ m} = 0.08 \text{ m} \)

Using the power formula: \[ P = \frac{1}{f \text{ (in metres)}} = \frac{100}{f \text{ (in cm)}} \] Substituting the given value of focal length: \[ P = \frac{100}{8} = +12.5 \text{ D} \]
The power of the lens is \( +12.5 \text{ D} \).
In simple words: Because this lens forms a real image, it is a convex lens. Converting its 8 cm focal length into meters and dividing 1 by it gives a power of +12.5 Dioptres.

Exam Tip: A real image is always formed by a convex lens, which means the focal length and power will always be positive values.

Practice Problems 6

Question 1. State the nature of the lens and the focal length if its power is +4D.
Answer: Given data:
Power of the lens (\( P \)) = \( +4 \text{ D} \)

- **Nature of the lens:** Since the power of the lens has a positive sign, the lens is convex (converging) in nature.

- **Focal length (\( f \)):** Using the relationship: \[ f = \frac{100}{P} \text{ (in cm)} \] \[ f = \frac{100}{4} = 25 \text{ cm} \] (or \( +0.25 \text{ m} \))
The focal length of the lens is \( 25 \text{ cm} \).
In simple words: Since the power is positive, we know it is a convex lens. Dividing 100 by the power of 4 gives us a focal length of 25 cm.

Exam Tip: The sign of the power tells you the nature of the lens immediately: positive power means a convex lens, while negative power means a concave lens.

Question 2. The number of the glasses of a person is +0.75 D. What is the nature of the lens and what is its focal length ?
Answer: Given data:
Power of the lens (\( P \)) = \( +0.75 \text{ D} \)

- **Nature of the lens:** The positive sign indicates that the person uses a convex (converging) lens to correct their vision.

- **Focal length (\( f \)):** Using the power formula: \[ f = \frac{100}{P} \text{ (in cm)} \] \[ f = \frac{100}{0.75} = \frac{100}{3/4} = \frac{400}{3} \approx 133.33 \text{ cm} \] (or \( 1.33 \text{ m} \))
The focal length of the lens is \( 133.33 \text{ cm} \).
In simple words: The positive power means this person wears convex glasses. By dividing 100 by 0.75, we calculate the focal length to be 133.33 cm.

Exam Tip: Expressing the focal length in both centimeters (\( 133.33 \text{ cm} \)) and meters (\( 1.33 \text{ m} \)) shows a complete and rigorous understanding of the solution.

Practice Problems 7

Question 1. The focal length of a concave lens is 10 cm. Calculate its power.
Answer: Given data:
Focal length of concave lens (\( f \)) = \( -10 \text{ cm} \) (concave lens focal length is negative)
In meters, \( f = -0.1 \text{ m} \)

Using the power formula: \[ P = \frac{1}{f \text{ (in metres)}} = \frac{1}{-0.1} = -10 \text{ D} \]
The power of the concave lens is \( -10 \text{ D} \).
In simple words: Since it is a concave lens, we use a negative focal length of -10 cm (-0.1 m). Dividing 1 by -0.1 gives us a power of -10 Dioptres.

Exam Tip: Always make sure to state the negative sign for both the focal length and the power of a concave lens; omitting the negative sign is a very common point loss on exams.

Question 2. The focal length of the lens of a myopic person is 40 cm. What is the power of the lens ?
Answer: A myopic (near-sighted) person requires a concave (diverging) lens to correct their vision.
Focal length of the concave lens (\( f \)) = \( -40 \text{ cm} = -0.4 \text{ m} \)

Using the power formula: \[ P = \frac{100}{f \text{ (in cm)}} \] \[ P = \frac{100}{-40} = -2.5 \text{ D} \]
The power of the lens is \( -2.5 \text{ D} \).
In simple words: Since myopia requires a concave lens, the focal length is negative (-40 cm). This gives a lens power of -2.5 Dioptres.

Exam Tip: Be sure to link the term "myopic person" with a "concave lens" to justify using a negative sign for the focal length in your calculation.

Practice Problems 8

Question 1. Calculate the focal length of a lens of power -2.75 D.
Answer: Given data:
Power of the lens (\( P \)) = \( -2.75 \text{ D} \)

Using the relationship between focal length and power: \[ f = \frac{100}{P} \text{ (in cm)} \] \[ f = \frac{100}{-2.75} = -\frac{100 \times 100}{275} = -\frac{400}{11} \approx -36.36 \text{ cm} \] (or \( -0.36 \text{ m} \))
The focal length of the lens is \( -36.36 \text{ cm} \) (and the negative sign shows it is a concave lens).
In simple words: Using the power of -2.75 D, we divide -100 by 2.75 to get a focal length of -36.36 cm.

Exam Tip: When simplifying fractions like \( \frac{100}{-2.75} \), convert them to integers (\( -\frac{400}{11} \)) before dividing to ensure your final decimal approximation is highly accurate.

Question 2. The power of a concave lens is -12.5 D. What is the focal length of the lens ?
Answer: Given data:
Power of the concave lens (\( P \)) = \( -12.5 \text{ D} \)

Using the focal length formula: \[ f = \frac{100}{P} \text{ (in cm)} \] \[ f = \frac{100}{-12.5} = -8 \text{ cm} \] (or \( -0.08 \text{ m} \))
The focal length of the concave lens is \( -8 \text{ cm} \).
In simple words: Dividing 100 by the lens power of -12.5 Dioptres gives us a focal length of -8 cm.

Exam Tip: Concave lens powers must always be negative. Double-check that your computed focal length also has a negative sign matching the concave lens properties.

Exercise - 3

Question 1. (a) What do you understand by the term lens ?
(b) What ar the various kinds of lenses ? Draw a neat diagram of each kind.
Answer: (a) A lens is defined as a portion of a transparent refracting medium (such as glass) bounded by two curved surfaces, which are typically spherical.

(b) Lenses are broadly classified into two main types:
(i) **Convex (Converging) Lens:** It is thicker in the middle and tapers towards the edges. Its types include:
- **Biconvex / Double Convex:** Both surfaces curve outwards.
- **Plano-convex:** One surface is flat and the other curves outwards.
- **Concavo-convex:** One surface curves inwards and the other curves outwards.

(ii) **Concave (Diverging) Lens:** It is thinner in the middle and thicker at the edges. Its types include:
- **Biconcave / Double Concave:** Both surfaces curve inwards.
- **Plano-concave:** One surface is flat and the other curves inwards.
- **Convexo-concave:** One surface curves outwards and the other curves inwards.

In simple words: (a) A lens is a curved piece of transparent glass that bends light. (b) Lenses are split into convex (thick in the center, which focuses light) and concave (thin in the center, which spreads light out), with each having three different shapes.

Exam Tip: Be sure to practice drawing clean profiles of all six sub-types of lenses; labeling their relative thickness in the middle versus the edges helps secure full marks.

Question 2. Define the following with respect to converging lens
1. principal axis
2. optical centre
3. first principal focus
4. second principal focus
5. focal length.
Answer: Definitions with respect to a converging (convex) lens:
1. **Principal axis:** It is the straight line passing through the centers of curvature of both spherical surfaces of the lens.
2. **Optical centre:** A special point on the principal axis of the lens such that any incident light ray directed through it passes undeviated (emerging parallel to its original path).
3. **First principal focus:** A point on the principal axis of a convex lens from which rays of light starting become parallel to the principal axis after refraction through the lens.
4. **Second principal focus:** A point on the principal axis where rays of light originally traveling parallel to the principal axis actually converge after undergoing refraction through the lens.
5. **Focal length:** The distance between the optical center and the principal focus of the lens.
In simple words: 1. Principal axis is the central line going straight through the lens. 2. Optical center is the exact middle point where light passes without bending. 3. First focus is where light rays must start to exit parallelly. 4. Second focus is where incoming parallel rays meet. 5. Focal length is the distance from the middle of the lens to its focus.

Exam Tip: For a convex lens, the second principal focus is considered the real focus of the lens, which is why its focal length is always positive.

Question 3. Draw neat diagrams for the formation of images in case of convex lens and state its characteristics when the object is :
1. at infinity
2. between 2F and infinity
3. at 2F
4. in between F and 2F
5. at F.
Answer: The image characteristics for different object positions in a convex lens are:

1. **Object at infinity:**
- *Position of Image:* At the second focus (\( F_2 \))
- *Nature of Image:* Real and inverted
- *Size of Image:* Highly diminished (point-sized)

2. **Object between \( 2F_1 \) and infinity:**
- *Position of Image:* Between \( F_2 \) and \( 2F_2 \)
- *Nature of Image:* Real and inverted
- *Size of Image:* Diminished

3. **Object at \( 2F_1 \):**
- *Position of Image:* At \( 2F_2 \)
- *Nature of Image:* Real and inverted
- *Size of Image:* Same size as the object

4. **Object between \( F_1 \) and \( 2F_1 \):**
- *Position of Image:* Beyond \( 2F_2 \)
- *Nature of Image:* Real and inverted
- *Size of Image:* Magnified

5. **Object at \( F_1 \):**
- *Position of Image:* At infinity
- *Nature of Image:* Real and inverted
- *Size of Image:* Highly magnified
In simple words: As you move the object closer to the lens (from far away up to the focal point), the image moves further away from the lens on the other side, turning from point-sized to extremely large.

Exam Tip: Remember that for all these five positions, the image formed is real and inverted. A virtual, erect image is only formed when the object is placed closer than \( F_1 \).

Question 4. Draw a neat diagram for a simple microscope.
Answer: A simple microscope consists of a single convex lens of short focal length. To use it, the object is placed within the principal focus of the lens (between \( F_1 \) and the optical center \( O \)).

The characteristics of the image formed are:
1. The image is virtual.
2. The image is erect.
3. The image is magnified.

In simple words: A simple microscope works by placing an object very close to a magnifying glass (inside its focal point). This makes the lens project a giant, upright virtual image that you can easily read.

Exam Tip: Be sure to draw the virtual rays (behind the lens) as dashed lines and real rays (passing through the lens) as solid lines to follow correct scientific standards.

Question 5. Draw neat diagrams for the formation of images in case of concave lens and state their characteristics when the object is :
1. at infinity
2. anywhere between infinity and the optical centre
Answer: The image characteristics for a concave lens are:

1. **Object at infinity:**
- *Position of Image:* At the second principal focus (\( F_2 \)) on the same side
- *Nature of Image:* Virtual and erect
- *Size of Image:* Highly diminished (point-sized)

2. **Object anywhere between infinity and the optical centre:**
- *Position of Image:* Between the focus (\( F \)) and optical center (\( O \)) on the same side
- *Nature of Image:* Virtual and erect
- *Size of Image:* Diminished
In simple words: No matter where you place an object in front of a concave lens, the image is always formed on the same side, and it is always upright, virtual, and smaller than the real object.

Exam Tip: Unlike convex lenses, concave lenses can *never* form a real or magnified image. Remembering this rule helps prevent errors in lens selection questions.

Question 6. How will you find the focal length of a convex lens, by using a single pin and a plane mirror ?
Answer: To find the focal length:
1. Place a plane mirror on a horizontal surface, and lay the convex lens flat on top of it.
2. Mount a pin horizontally on a vertical stand, ensuring its tip lies on the principal axis of the lens.
3. Adjust the vertical position of the pin until you see its inverted image. Move the pin until the tip of the pin and the tip of its inverted image coincide perfectly.
4. Eliminate parallax by ensuring that as you move your eye sideways, both the pin and its image move together without any relative shift.
5. Measure the distance from the pin to the optical center of the lens. This distance is equal to the focal length (\( f \)) of the lens.

*Reason:* Light rays originating from the pin at the focus travel parallel to the principal axis after refracting through the lens. These parallel rays hit the plane mirror normally, reflect straight back along their original paths, and re-converge to form an image exactly at the focus.
In simple words: We place a lens on a mirror and move a pin up and down above it until the pin and its upside-down reflection line up perfectly with no relative shift. The distance from the pin to the lens is the focal length.

Exam Tip: In this experiment, the plane mirror acts as a normal reflector. Emphasize that "rays must strike the mirror normally to retrace their path" to secure maximum marks.

Question 7. You are required to form an upright image of an object in case of (a) convex lens, (b) concave lens. What will be the position of the object with respect to the lens in each case ? Support your answer by diagrams and state the characteristics of the image in each case :
Answer: To obtain an upright (erect) image:

(a) **Convex lens:**
- *Object Position:* The object must be placed between the optical center (\( O \)) and the first principal focus (\( F_1 \)).
- *Image Characteristics:* Virtual, erect, and magnified.

(b) **Concave lens:**
- *Object Position:* The object can be placed anywhere between infinity and the optical center of the lens.
- *Image Characteristics:* Virtual, erect, and diminished.
In simple words: (a) A convex lens only makes an upright image if the object is placed extremely close to it (inside the focus), making it look larger. (b) A concave lens always makes an upright image no matter where the object is, but it will always look smaller.

Exam Tip: Highlighting the difference in size is crucial: both lenses form upright, virtual images, but a convex lens magnifies while a concave lens diminishes.

Multiple Choice Questions

Question 1. The point on the principal axis of a convex lens, such that rays of light starting from it on passing through the lens, move parallel to the principal axis is called :
(a) first focal point
(b) second focal point
(c) optical centre
(d) aperture of lens
Answer: (a) first focal point
In simple words: The first focal point is the exact spot on the axis where light rays must start from in order to go straight out parallel after passing through the lens.

Exam Tip: Be sure to distinguish between the first focus (origin of parallel refracted rays) and the second focus (converging point of parallel incident rays).

Question 2. A convex lens can be regarded as a set of prisms and a glass slab, such that refracting angle of the prisms
(a) continuously decreases in outward direction
(b) continuously increases in outward direction
(c) remains same in outward direction
(d) none of the options
Answer: (a) continuously decreases in outward direction
In simple words: A convex lens is like a stack of glass prisms where the prism blocks get flatter and less angled as you move from the center towards the outer edges.

Exam Tip: Prisms with larger angles bend light more, which is why the thick center of a convex lens has the maximum bending capability.

Question 3. A lens forms an inverted image of an object equal to its own size. The object is :
(a) beyond infinity and 2F1
(b) at 2F1
(c) between 2F1 and F1
(d) in between F1 and optical centre
Answer: (b) at 2F1
In simple words: To get an image that is upside-down but exactly the same size as the object, you must place the object at exactly twice the focal distance (2F) on one side of the lens.

Exam Tip: Placing the object at \( 2F_1 \) yields a magnification of exactly \( -1 \), which indicates a real, inverted image of identical height on the opposite side.

Question 4. A convex lens will form a virtual, erect and enlarged image, when the object is :
(a) between 2F1 and F1
(b) 2F1
(c) 2F1 and infinity
(d) F1 and optical centre
Answer: (d) F1 and optical centre
In simple words: A magnifying glass only works when the object is placed very close to the lens, between its focal point and the glass itself, making the image look upright and huge.

Exam Tip: This is the only position for a convex lens that produces a virtual image. It is the core operating principle behind a simple magnifying glass.

Question 5. A concave lens always forms :
(a) real, inverted and enlarged image
(b) virtual, inverted and enlarged image
(c) virtual, erect and diminished image
(d) virtual, erect and enlarged image
Answer: (c) virtual, erect and diminished image
In simple words: Concave lenses are diverging lenses, meaning they can only ever make images that are virtual, upright, and smaller than the real object.

Exam Tip: Remember that a concave lens always behaves the same way regardless of the object's position, producing only virtual, erect, and diminished images.

Questions from ICSE Examination Papers 2003

Question 1. (a) A ray of light, after refraction through a concave lens, emerges parallel to the principal axis. Draw a ray diagram to show the incident ray and its corresponding emergent ray.
(b) The velocity of light in diamond is 21,000 kms-1 What is its refractive index of diamond? (Velocity of light in a air 3 × 108 m/s)
Answer: (a) For a concave lens, a light ray emerging parallel to the principal axis after refraction must have been directed toward the first principal focus (\( F_1 \)) on the other side of the lens.

(b) Given data:
Velocity of light in diamond (\( v \)) = \( 121,000 \text{ km s}^{-1} = 1.21 \times 10^8 \text{ m s}^{-1} \)
Velocity of light in air (\( c \)) = \( 3 \times 10^8 \text{ m s}^{-1} \)

Using the refractive index formula: \[ \mu = \frac{\text{Velocity of light in air}}{\text{Velocity of light in diamond}} \] \[ \mu = \frac{3 \times 10^8}{1.21 \times 10^8} \approx 2.48 \]
In simple words: (a) A ray of light aimed at the focus of a concave lens will bend as it passes through, exiting completely parallel to the central axis. (b) Light travels about 2.48 times slower in diamond than in air, giving diamond a very high refractive index of 2.48.

Exam Tip: (a) Be sure to draw the virtual ray extension behind the concave lens as a dashed line leading directly to the focal point \( F_1 \). (b) Make sure you convert the diamond speed from \( \text{km/s} \) to \( \text{m/s} \) before dividing to ensure both quantities are in identical units.

Question 2. A monochromatic point source of light ‘O ’ is seen through a rectangular glass block ABCD. Paths of two rays, in and outside the block, are shown in figure above.
(a) Does the source monochromatic source appear to be nearer or farther with respect to the surface AB ?
(b) How does the shift in (a) depend up on the thickness (AD) of the glass block ?
(c) Justify your answer in (b) with an appropriate ray diagram.
(d) For the same rectangular block, which colour from the visible spectrum will produce the maximum shift ?
Answer: (a) The point source O appears to be **nearer** with respect to the surface AB.
(b) The upward apparent shift is directly proportional to the thickness of the glass block (AD). This means that as the thickness of the glass block increases, the apparent shift also increases.
(c) Below is the ray diagram justifying this relationship:
When the glass block has a larger thickness AD, the virtual image is formed at I, representing a larger shift (OI). When the thickness is reduced to AE, the virtual image is formed at I', which corresponds to a smaller shift (OI'). Thus, shift is directly proportional to the thickness of the glass block.
(d) **Violet light** will produce the maximum shift. This is because violet light has the shortest wavelength and experience the greatest refractive index in glass, leading to maximum deviation of light rays.
In simple words: (a) The light source looks closer than it really is. (b) The thicker the glass block, the higher up the image is shifted. (c) As shown in the diagram, a thicker block shifts the image up to point I, while a thinner block shifts it less to I'. (d) Violet light is bent the most by glass, so it creates the biggest shift.

Exam Tip: Remember that apparent shift depends on three factors: thickness of the medium, refractive index, and wavelength of the light. Be prepared to state these relationships clearly.

Question 3. A postage stamp appears raised by 7.00 mm when placed under a glass block of refractive index 1.5. Find the thickness of the glass block.
Answer: Let the actual thickness of the glass block (real depth) be \( x \text{ mm} \). Since the stamp appears raised by \( 7.00 \text{ mm} \), the apparent depth is \( (x - 7) \text{ mm} \). Using the relationship: \[ \text{Refractive Index } (\mu) = \frac{\text{Real Depth}}{\text{Apparent Depth}} \] Substituting the given values: \[ 1.5 = \frac{x}{x - 7} \] \[ \frac{3}{2} = \frac{x}{x - 7} \] Cross-multiplying: \[ 3(x - 7) = 2x \] \[ 3x - 21 = 2x \]
\( \implies x = 21 \text{ mm} \) Thus, the thickness of the glass block is \( 21 \text{ mm} \).
In simple words: Since glass bends light, the stamp looks closer to the surface. By setting up the ratio of real depth to apparent depth as 1.5, we calculate that the glass block is exactly 21 mm thick.

Exam Tip: Always define your variables clearly (e.g., let real depth = \( x \)) and express the apparent depth in terms of \( x \) and the shift to solve this type of algebraic problem without errors.

2004

Question 4. (a) What do you understand by the term critical angle ?
(b) Diagram below shows a path of ray AB through an isosceles right angled prism. What is the magnitude of the angle of incidence on (i) face PR (ii) PQ ?
Answer: (a) The critical angle is defined as the angle of incidence in an optically denser medium for which the corresponding angle of refraction in the optically rarer medium is exactly \( 90^{\circ} \). (b) (i) At face PR, the incident ray is perpendicular to the surface. Since the ray is directed along the normal, the angle of incidence is **\( 0^{\circ} \)** (and the angle made with the surface itself is \( 90^{\circ} \)). (ii) At face PQ, the angle of incidence is **\( 45^{\circ} \)**.
In simple words: (a) The critical angle is the tipping-point angle where light refracts flat along the surface of a material. (b) The ray enters face PR straight along the normal, so its angle of incidence is 0°. It then hits face PQ at an angle of 45° with the normal.

Exam Tip: Be careful with the terminology: "normal incidence" means the ray is perpendicular to the boundary surface, which translates to an angle of incidence of exactly \( 0^{\circ} \) with the normal line.

Question 5. The diagram given below shows the position of an object and its image. Copy the diagram and then by drawing two rays locate the position of the lens and its focus, showing clearly the kind of lens used.
Answer: Since the image formed is virtual, erect, magnified, and situated on the same side as the object, the lens must be a **convex (converging) lens**. For a convex lens to produce such an image, the object must be positioned within its focal length (between the first focus \( F_1 \) and the optical center \( O \)). Below is the completed ray diagram:

In simple words: Since the image is larger and upright on the same side, the lens must be a convex lens. Placing the object close to the lens (inside its focus) projects this virtual, magnified image.

Exam Tip: To correctly complete such ray diagrams, use one ray parallel to the principal axis (which refracts through the focus) and a second ray passing straight through the optical center. Extend both backward to locate the virtual image.

Question 6. (a) State the Snell’s law of refraction.
(b) If the velocity of light in air is 3 × 10⁸ m s⁻¹ and refractive index of glass 1-5, calculate the velocity of light in glass.
Answer: (a) Snell's Law states that when a light ray passes from one optical medium to another, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant value for a given pair of media and a given wavelength of light: \[ \frac{\sin i}{\sin r} = \mu \] Where \( \mu \) represents the relative refractive index. (b) Given: - Speed of light in air (\( v_{\text{air}} \)) = \( 3 \times 10^8 \text{ m s}^{-1} \) - Refractive index of glass (\( \mu \)) = \( 1.5 \) The relationship is: \[ \mu = \frac{v_{\text{air}}}{v_{\text{glass}}} \] Rearranging for the speed in glass (\( v_{\text{glass}} \)): \[ v_{\text{glass}} = \frac{v_{\text{air}}}{\mu} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m s}^{-1} \] Thus, the velocity of light in glass is \( 2 \times 10^8 \text{ m s}^{-1} \).
In simple words: (a) Snell's Law says that light bends at a constant ratio of sines when entering a new material. (b) Since light is slowed down 1.5 times by glass, we divide its speed in air by 1.5 to get 200,000,000 meters per second.

Exam Tip: Remember that refractive index is defined in terms of speed of light as the ratio of speed in air/vacuum to speed in the medium, which is always greater than 1 for denser materials.

2005

Question 7. Mention two properties of a wave: one property which varies and the other which remains constant when the wave passes from one medium to another.
Answer: When a wave transitions from one medium into another: 1. The **wavelength** of the wave changes (varies). 2. The **frequency** of the wave remains completely constant.
In simple words: When light enters a new material, its wavelength shrinks or stretches, but its frequency (or color) stays exactly the same.

Exam Tip: Frequency depends strictly on the source of the wave, which is why it never changes during refraction, unlike wavelength and speed which depend on the medium.

Question 8. What is meant by the statement ‘the critical angle for diamond is 24° ? How is the critical angle of a material related to its refractive index ?
Answer: The statement means that when a light ray is traveling inside diamond and strikes the diamond-air boundary at an angle of incidence of \( 24^{\circ} \keys{} \), the refracted ray will emerge along the boundary interface, making an angle of refraction of \( 90^{\circ} \) in the air. The refractive index (\( \mu \)) of a material is related to its critical angle (\( C \)) by the formula: \[ \mu = \frac{1}{\sin C} \]
In simple words: This means if light inside a diamond hits the air boundary at 24°, it will slide completely flat along the surface. The refractive index is simply 1 divided by the sine of this critical angle.

Exam Tip: Highlight that the critical angle is the angle of incidence in the denser medium which corresponds to an angle of refraction of \( 90^{\circ} \) in the rarer medium.

Question 9. The ray diagram given below illustrates the experimental set up for the determination of the focal length of a converging lens using a plane mirror.
(a) State the magnification of the image formed.
(b) Write two characteristics of the image formed.
(c) What name is given to the distance between the object and optical centre in the diagram above ?
Answer: (a) The magnification is \( -1 \), meaning the size of the formed image is exactly equal to the size of the object.
(b) The image formed is **real** and **inverted**.
(c) This distance is called the **focal length** (\( f \)) of the converging lens.
In simple words: (a) The image is exactly the same size as the object. (b) The image is real and upside-down. (c) The distance from the object to the middle of the lens is the focal length.

Exam Tip: In this setup, because the rays are reflected straight back by the plane mirror, the object must be positioned at the principal focus of the convex lens to form a coincident image.

2006

Question 10. An object is placed in front of a convex lens such that the image formed has the same size as that of the object. Draw a ray diagram to illustrate this.
Answer: To obtain a real image of the same size, the object must be placed at a distance equal to twice the focal length (\( 2F_1 \)) from the convex lens. The image is formed at \( 2F_2 \) on the opposite side of the lens. Below is the ray diagram illustrating this:

In simple words: When you place an object at exactly twice the focal distance (2F) in front of a convex lens, its upside-down real image is formed at 2F on the other side, and it is the exact same size.

Exam Tip: Label the object and image positions clearly as \( 2F_1 \) and \( 2F_2 \) respectively, showing the symmetrical ray paths intersecting exactly at \( 2F_2 \).

Question 11. PQ and PR are two light rays emerging from the object P as shown in the figure
(a) What is the special name given to the angle of incidence (∠PQN) of ray PQ ?
(b) Copy the ray diagram and complete it to show the position of the image of the image of the object P when seen obliquely from above.
(c) Name the phenomenon that occurs if the angle of incidence ∠PQN is increased still further.
Answer: (a) The special name given to the angle of incidence \( \angle PQN \) is the **Critical Angle**.
(b) Below is the completed ray diagram showing the virtual image \( P' \) formed above the actual object \( P \):

(c) If the angle of incidence is increased further, the phenomenon of **Total Internal Reflection** will take place.
In simple words: (a) The entry angle of ray PQ is called the critical angle because it bends flat along the surface. (b) Refraction makes the rays bend away from the normal, so the object looks raised up at point P' when viewed obliquely. (c) If you increase the angle further, the light will reflect completely back into the water.

Exam Tip: Be sure to extend the refracted ray from R backwards with a dashed line to locate the virtual image \( P' \) exactly along the normal line passing through the object \( P \).

2007

Question 12. State Snell’s Law of Refraction of light.
Answer: Snell's Law states that for any given pair of media and for light of a specific color, the ratio of the sine of the angle of incidence (\( i \)) to the sine of the angle of refraction (\( r \)) remains a constant: \[ \frac{\sin i}{\sin r} = \text{constant} \] This constant is known as the refractive index of the second medium with respect to the first medium.
In simple words: Snell's law states that light always bends by a constant mathematical ratio of sines when passing from one material to another.

Exam Tip: Write down both the descriptive definition and the mathematical formula \( \frac{\sin i}{\sin r} = \mu \) to ensure you get full marks.

Question 13. An object is placed in front of a converging lens at a distance greater than twice the focal length of the lens. Draw a ray diagram to show the formation of the image.
Answer: When an object is placed beyond twice the focal length (\( 2F_1 \)) of a convex lens, its image is formed on the opposite side of the lens between the principal focus (\( F_2 \)) and twice the focal length (\( 2F_2 \)). The image formed is real, inverted, and diminished. Below is the ray diagram:

In simple words: When an object is placed far away (beyond 2F), the lens creates a real, inverted image that is smaller in size, positioned between F and 2F on the other side.

Exam Tip: Make sure the image formed is visibly smaller (diminished) than the object, and is correctly placed between \( F_2 \) and \( 2F_2 \).

Question 14. Mention one difference between reflection of light from a plane mirror and total internal reflection of light from a prism.
Answer: In the case of a plane mirror, reflection of light occurs at all angles of incidence, but some light energy is always absorbed by the mirror surface. In the case of total internal reflection inside a prism, reflection only occurs when the light strikes the boundary at an angle greater than the critical angle, but 100% of the light is reflected with absolutely no absorption of energy.
In simple words: A plane mirror reflects light at any angle but loses some energy, whereas a prism can only reflect light at angles greater than the critical angle, but it reflects with 100% efficiency.

Exam Tip: Highlight the 100% reflection efficiency (no energy loss) of total internal reflection as the key differentiator in optical instruments.

Question 15. The diagram given below shows a right-angled prism with a ray of light incident on the side AB. (The critical angle for glass is 42°).
1. Copy the diagram and complete the path of the ray of light in and out of the glass prism.
2. What is the value of the angle of deviation shown by the ray ?
Answer: 1. The completed ray diagram is shown below:
The incident ray enters normally through face AB, striking the hypotenuse AC at point R with an angle of incidence of \( 45^{\circ} \). Since \( 45^{\circ} > 42^{\circ} \), the ray undergoes total internal reflection, turns by \( 90^{\circ} \), and exits normally through the bottom face BC.
2. The total angle of deviation produced is exactly **\( 90^{\circ} \)**.
In simple words: 1. The ray enters straight through AB and hits face AC at 45°. Since this is greater than the critical angle, it reflects straight down and exits BC normally. 2. The path of the light ray is turned by exactly 90 degrees.

Exam Tip: Be sure to mark all the angles of incidence and reflection (\( 45^{\circ} \)) at the reflecting hypotenuse face to show why the ray undergoes total internal reflection.

Question 16.
1. With the help of a well-labelled diagram show that the apparent depth of an object, such as a coin, in water is less than its real depth.
2. How is the refractive index of water related to the real depth and the apparent depth of a column of water ?
Answer: 1. Below is the well-labeled diagram showing the apparent depth of a coin placed at the bottom of a water tank:
Due to refraction at the water surface, light rays coming from the actual coin position (A) bend away from the normal, making the coin appear raised at position B.
2. The relation is: \[ \text{Refractive Index of Water } (\mu) = \frac{\text{Real Depth}}{\text{Apparent Depth}} \]
In simple words: 1. Light from the coin bends as it leaves the water, making the coin look higher up than it actually is. 2. Refractive index is calculated by dividing the real depth of the water column by its apparent depth.

Exam Tip: Always show both real depth and apparent depth labels clearly in the diagram, with the apparent depth being shorter than the real depth.

2008

Question 17. (a) (i)A monochromatic beam of light of wavelength \(\lambda\) passes from air into a glass block. Write an expression to show the relation between the speed of light in air and the speed of light in glass.
(ii) As the ray of light passes from air to glass, state how the wavelength of light changes. Does it increase, decreases or remain constant ?
(b) Draw a ray diagram to illustrate the determination of the focal length of a convex lens using an auxiliary plane mirror.
Answer: (a) (i) The relation between the speed of light in air (\( c \)) and in glass (\( v \)) is given by: \[ \frac{c}{v} = \frac{\lambda}{\lambda'} \] Where \( \lambda \) is the wavelength of light in air and \( \lambda' \) is its wavelength in glass. (ii) As light passes from air to glass, its velocity decreases, and its wavelength **decreases** proportionally (while the frequency remains constant). (b) Below is the ray diagram to determine the focal length of a convex lens using an auxiliary plane mirror:
The height of the pin is adjusted vertically until the pin and its real inverted image coincide with no parallax. The average of the distance from the pin to the lens (\( x \)) and the pin to the mirror (\( y \)) gives the focal length of the lens: \[ f = \frac{x + y}{2} \]
In simple words: (a) The ratio of speed of light in air to glass is the same as the ratio of their wavelengths. Wavelength decreases in glass. (b) Adjusting a pin above the lens until it perfectly matches its upside-down reflection gives the focal length by averaging the distances.

Exam Tip: Be sure to explain how the focal length is calculated as the average of the two measured distances to show complete practical precision.

Question 18. (a) (i) Draw a labelled ray diagram to illustrate (1) critical angle (2) total internal reflection, for a ray of light moving from one medium to another.
(ii) Write a formula to express the relationship between refractive index of the denser medium with respect to rarer medium and its critical angle for that pair of media.
(b) (i) The diagram below shows a ray of light incident on an equilateral glass prism placed in minimum deviation position. Copy the diagram and complete it to show the path of the refracted ray and the emergent ray.
(ii) How are angle of incidence and angle of emergence related to each other in this position of the prism ?
Answer: (a) (i) Below are the ray diagrams: 1. **Critical Angle:**
2. **Total Internal Reflection:**

(ii) The relation is: \[ \mu = \frac{1}{\sin C} \]
(b) (i) Below is the completed path of the light ray through the equilateral prism:

(ii) In this position of minimum deviation, the angle of incidence (\( i \)) is exactly equal to the angle of emergence (\( e \)) (\( i = e \)).
In simple words: (a) 1. At the critical angle, refracted light emerges at 90° along the boundary. Beyond this angle, the ray undergoes total internal reflection. 2. Refractive index is \( 1/\sin C \). (b) 1. In minimum deviation, the light ray runs completely parallel to the base inside the prism. 2. The entry angle of incidence equals the exit angle of emergence.

Exam Tip: Remember that in the minimum deviation position, the refracted ray inside a symmetric triangular prism is always parallel to its base.

Question 19. A linear object is placed on the axis of a lens. An image is formed by refraction in the lens. For all positions of the object on the axis of the lens, the positions of the image are always between the lens and the object
(a) Name the lens.
(b) Draw a ray diagram to show the formation of the image of an object placed in front of the lens at any position of your choice except infinity.
Answer: (a) The lens is a **concave lens** (diverging lens). It is the only lens which consistently produces a virtual, erect, and diminished image placed between the lens and the object. (b) Below is the ray diagram:

Question 20. Two isosceles right-angled prisms are placed near each other as shown in the figure. Complete the path of the light ray entering the first isosceles right-angled glass prism till it emerges from the second identical prism.
Answer: Below is the completed path of the light ray through the two right-angled isosceles prisms placed side-by-side:
1. The incident ray G enters the horizontal face AB of the first right-angled isosceles prism normally at H, continuing straight to meet the hypotenuse face AC at point I. 2. At I, the angle of incidence is \( 45^{\circ} \), which is greater than the critical angle of glass (\( 42^{\circ} \)). It undergoes total internal reflection, emerging horizontally and normally from face BC at J. 3. The horizontally traveling ray enters the second identical prism normally through the vertical face DE at point K. 4. It travels straight to the hypotenuse face DF at point L, striking it at \( 45^{\circ} \). It undergoes total internal reflection downwards, exiting normally through the bottom horizontal face EF at point M.
In simple words: The light ray goes straight into the first prism, bounces 90 degrees horizontally off its diagonal face, goes across the gap, and then bounces 90 degrees downwards off the second prism's diagonal face to emerge straight out.

Exam Tip: Carefully align both prisms symmetrically in your sketch, and show normal entry and exit (\( 90^{\circ} \) to the faces) to ensure full marks on ray tracing questions.

2009

Question 21. (a) A ray of light strikes the surface of a rectangular glass block such that the angle of incidence is (i) 0° (ii) 42°. Sketch a diagram to show the approximate path taken by the ray in each case as it passes through the glass block and emerges from it.
(b) State the conditions required for total internal reflection of light to take place.
(c) Copy and complete the following table :
Answer: (a) Below are the ray diagrams: (i) For \( 0^{\circ} \) angle of incidence, the light ray strikes normally and passes completely undeviated through the glass block, emerging straight from the other side.
(ii) For \( 42^{\circ} \) angle of incidence from inside the glass block (which is equal to the critical angle of glass), the refracted ray emerges along the surface of the block at an angle of refraction of \( 90^{\circ} \).
(b) The two conditions required for total internal reflection of light to take place are: 1. The light ray must be traveling from an optically denser medium into an optically rarer medium. 2. The angle of incidence of the light in the denser medium must be strictly greater than the critical angle for that specific pair of media. (c) Below is the completed table:

In simple words: (a) 1. At 0° incidence, light goes straight through without bending. 2. At 42° critical incidence, light bends 90° along the glass surface. (b) Total internal reflection needs light going from dense to rare and hitting at an angle greater than the critical angle. (c) The table shows how a convex lens at F makes a massive real image at infinity, while a concave lens at infinity makes a tiny virtual image at focus.

Exam Tip: Be sure to keep both conditions for total internal reflection memorized, as they are tested frequently as short-answer questions.

Question 22. (a) How does the value of angle of deviation produced by a prism change with an increase in the :
1. value of angle of incidence
2. wavelength of incident light ?
(b) (i) Copy and complete the diagram to show the formation of the image of the subject AB.
(ii) What is the name given to x ?
(c) (i) The diagram below shows a ray of white light PQ cbming from an object P and incident on the surface of a thick glass plane mirror. Copy the diagram and complete it to show the formation of three images of the object P as formed by the mirror.
(ii) Which image will be the brightest image ?
Answer: (a) 1. When the angle of incidence increases, the angle of deviation initially decreases until it reaches its lowest value, known as the angle of minimum deviation. Beyond this position, any further increase in the angle of incidence leads to an increase in the angle of deviation. 2. The angle of deviation is inversely proportional to the wavelength of light. As the wavelength of the incident light increases, the angle of deviation decreases. Consequently, the deviation is maximum for violet light (shortest wavelength) and minimum for red light (longest wavelength). (b) (i) Below is the completed ray diagram showing the virtual, erect, and diminished image \( A_1 B_1 \) formed by the concave lens:
(ii) Point **X** is the **principal focus** of the concave lens, and the distance **OX** represents the **focal length**. (c) (i) Below is the completed ray diagram showing the formation of multiple images by a thick glass mirror. When light ray PQ strikes the glass mirror, a small part reflects at the top surface, forming a faint first image \( P_1 \). The major portion is refracted into the glass, reflects strongly off the silvered bottom surface, and refracts out to form the brightest second image \( P_2 \). Subsequent internal reflections form progressively fainter images such as \( P_3 \).
(ii) The second image (**\( P_2 \)**) is the **brightest image**. This is because it is formed by strong reflection off the silvered bottom surface, which reflects nearly 96% of the incident light, unlike the first image (\( P_1 \)) which is formed by a weak (4%) reflection off the front unsilvered glass surface.
In simple words: (a) 1. Deviation first drops then rises as incidence angle increases. 2. Violet light has shorter wavelength so it is bent the most. (b) A concave lens shrinks the image, and point X is its focus. (c) Out of multiple reflections inside a thick mirror, the second image is the brightest because it comes from the fully silvered back surface.

Exam Tip: Explain the 4% and 96% energy split in ordinary glass reflection to get full marks on thick mirror questions.

2010

Question 23.
(a) (i) What is meant by refraction of light ?
(ii) What is the cause of refraction of light ?
(b) ‘The refractive index of diamond is 2.42 ; What is meant by this statement ?
(c) We can burn a piece of paper by focussing the sun rays by using a particular type of lens.
(i) Name the type of lens used for the above purpose.
(ii) Draw a ray diagram to support your answer.
(d) A ray of light enters a glass slab PQRS, as shown in the diagram. The critical angle of the glass is 42°. Copy this diagram and complete the path of the ray till it emerges from the glass slab Mark the angle in the diagram wherever necessary.
Answer: (a) (i) Refraction of light is the phenomenon where a light ray changes its direction of propagation as it travels from one transparent medium to another.
(ii) Refraction is caused by the change in the velocity of light as it transitions from one optical medium to another. (b) This statement signifies that the velocity of light in diamond is \( \frac{1}{2.42} \) times (or about 41%) the velocity of light in a vacuum or air. (c) (i) A **convex (converging) lens** is used for this purpose. (ii) Below is the ray diagram showing how a convex lens converges parallel rays of sunlight to a single point (the principal focus), concentrating heat energy to burn the paper:
(d) Below is the completed path of the light ray through the glass slab PQRS: The light ray enters normally at PS, going straight to meet PQ at an angle of incidence of \( 48^{\circ} \). Since this exceeds the critical angle (\( 42^{\circ} \)), it undergoes total internal reflection at PQ. The reflected ray then strikes PS at an angle of \( 42^{\circ} \), which is exactly equal to the critical angle, so it refracts and emerges parallel to the face PS.

In simple words: (a) Refraction is light bending when entering a new medium, caused by a speed change. (b) Light is 2.42 times slower in diamond than in air. (c) A convex lens gathers parallel sun rays at its focus to ignite paper. (d) The ray undergoes a bounce inside the glass and then exits flat along the side because it hits at the critical angle.

Exam Tip: Be sure to compute all normal calculations correctly to prove that \( 48^{\circ} > 42^{\circ} \) for total internal reflection in the slab.

Question 24.
(a) A stick partly immersed in water appears to be bent. Draw a ray diagram to show the bending of the stick when placed in water and viewed obliquely from above.
(b) A ray of monochromatic light is incident from air on a glass slab :
1. Draw a labelled ray diagram showing the change in the path of the ray till it emerges from the glass slab.
2. Name the two rays that are parallel to each other. (Hi) Mark the lateral displacement in your diagram.
(c) An erect, magnified and virtual image is formed, when an object is placed between the optical centre and principal focus of a lens.
1. Name the lens.
2. Draw a ray diagram to show the formation of the image with the above stated characteristics. (4)
Answer: (a) Below is the ray diagram showing a stick partly immersed in water appearing to be bent upwards:
(b) (i) Below is the labeled ray diagram showing the passage of light through a glass slab:
(ii) The **incident ray** and the **emergent ray** are parallel to each other. (iii) The perpendicular distance between the produced direction of the incident ray and the emergent ray represents the **lateral displacement** (shown as 'Lateral shift' in the diagram). (c) (i) The lens is a **convex lens**. (ii) Below is the ray diagram showing the formation of a virtual, erect, and magnified image when the object is placed between the optical center and the principal focus:

In simple words: (a) The stick looks bent in water because exiting light rays bend away from the normal. (b) Inside a glass slab, the exiting ray runs parallel to the entering ray but shifted sideways. (c) Placing an object very close to a convex lens magnifies it as an erect virtual image.

Exam Tip: Be sure to draw arrows on your ray diagrams showing the direction of light to secure full marks.

Question 25. (a) (i) Copy the diagram and complete the path of the ray of light through the glass block. In your diagram, mark the angle of incidence by letter “i” and the angle of emergence by the letter “e”
(ii) How are the angle i and ‘e’ related to each other?
(b) A ray of monochromatic light ente -s a liquid from air as shown in the diagram.
(i) Copy the diagram and show in the diagram the path of the ray of light after it strikes the mirror and reenters the medium of air.
(ii) Mark in your diagram the two angles on the surface of separation when the ray of light moves out from the liquid to air.
(c) (i) When does a ray of light falling on a lens pass through it undeviated ?
(ii) Which lens can produce a real and inverted image of an object ?
(d) (i) How is the refractive index of a medium related to its real depth and apparent depth?
(ii) Which characteristic property of light is responsible for the blue colour of the sky ?
Answer: (a) (i) Below is the completed ray diagram showing the path of the light ray through the glass block, with the angle of incidence \( i \) and angle of emergence \( e \) labeled:
(ii) The angle of incidence \( i \) is exactly equal to the angle of emergence \( e \) (\( \angle i = \angle e \)). (b) (i) Below is the completed ray diagram showing the path of the light ray through the liquid, reflecting off the bottom mirror, and refracting back out into the air:
(ii) The angles of incidence and emergence at the boundary are both \( 45^{\circ} \), and the angles of refraction inside the liquid are both \( 30^{\circ} \). (c) (i) A light ray passes undeviated when it is directed through the **optical center** of the lens. (ii) A **convex (converging) lens** is capable of forming a real and inverted image of an object. (d) (i) The relationship is: \[ \text{Refractive Index} = \frac{\text{Real Depth}}{\text{Apparent Depth}} \] (ii) The blue color of the sky is caused by the **scattering of light** (Rayleigh scattering), where shorter wavelengths (blue and violet) are scattered much more intensely by atmospheric particles than longer wavelengths.
In simple words: (a) The entry angle of a ray into a glass block is always equal to the exit angle. (b) When light enters water, it bends, hits the mirror, bounces back, and exits at the exact same angle it started with. (c) Light going through the center of a lens doesn't bend, and only a convex lens can create a real, upside-down image. (d) Water looks shallower than it is because of the refractive index, and the sky looks blue because small dust particles scatter blue light everywhere.

Exam Tip: Always construct perpendicular normal lines at all refraction boundaries in your ray diagrams, and write out all numerical angles explicitly to get full marks on optics questions.

Question 26. (a) (i) State the laws of refraction of light.
(ii) Write a relation between the angle of incidence (i), angle of emergence (e), angle of prism (A) and angle of deviation (d) for a ray of light passing through an equilateral prism.
(b) An object is placed in front of a lens between its optical centre and the focus and forms an erect, virtual, and diminished image.
(i) Name the lens which formed this image.
(ii) Draw a ray diagram to shows the formation of the image.
Answer: (a) (i) **Laws of Refraction:** 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 a given pair of media (Snell's Law). (ii) The relationship is: \[ i + e = A + d \] Where \( i \) is the angle of incidence, \( e \) is the angle of emergence, \( A \) is the angle of the prism, and \( d \) is the angle of deviation. (b) (i) The lens is a **concave lens**. A concave lens is the only lens that forms a virtual, erect, and diminished image for an object placed between its focus and optical center. (ii) Below is the ray diagram showing the formation of the image:

In simple words: (a) The rules of bending say that all light rays stay in the same flat plane, and they bend at a constant mathematical ratio of sines. In a prism, the entering and exiting angles add up to the prism's angle plus the amount the light turned. (b) A concave lens always creates a smaller, upright, virtual image that sits between the lens and the object.

Exam Tip: For the prism formula, always remember that \( i + e = A + d \). For the concave lens diagram, draw virtual ray extensions back to the first focal point as dashed lines to avoid losing presentation marks.

Question 27. (a) (i) Define refractive index of a medium in terms of velocity of light.
(ii) A ray of light moves from a rare medium to a dense medium as shown in the diagram below. Write down the number of the ray which represents the partially reflected ray.
(b) You are provided with a printed piece of paper. Using this paper how will you differentiate between a convex lens and a concave lens ?
(c) A ray of light incident at an angle of incidence ‘i’ passes through an equilateral glass prism such that the refracted ray inside the prism is parallel to its base and emerges from the prism at an angle of emergence ‘e’
(i) How is the angle of emergence ‘e’ related to the angle of incidence ‘i’ ?
(ii) What can you say about the value of the angle of deviation in such a situation ?
Answer: (a) (i) The refractive index of a medium is defined as the ratio of the velocity of light in a vacuum or air to the velocity of light in that specific medium.
(ii) **Ray 2** represents the partially reflected ray. (b) Place each lens approximately 5 cm above the printed paper and view the text through it. If the print appears magnified, the lens is a **convex lens**. If the print appears smaller (diminished), the lens is a **concave lens**. (c) (i) The angle of emergence \( e \) is exactly equal to the angle of incidence \( i \) (\( i = e \)). (ii) Under this symmetrical condition, the angle of deviation achieves its **minimum** value (minimum deviation).
In simple words: (a) Refractive index is a speed ratio showing how much light slows down in a medium. (b) To tell lenses apart using paper, hold them above the text: if the letters look bigger, it is a convex lens; if they look smaller, it is a concave lens. (c) When a light ray inside a prism runs perfectly parallel to the base, the entry and exit angles are equal, and the light suffers the smallest possible bend.

Exam Tip: Be prepared to identify different rays in transition diagrams: the ray in the same medium at the same angle is reflected, while the bent ray in the new medium is refracted.

Question 28. (a) (i) What is meant by the term ‘critical angle’ ?
(ii) How is it related to the refractive index of the medium?
(iii) Does the depth of a tank of water appear to change or remain the same when viewed normally from above ?
(b) A ray of light PQ is incident normally on the hypotenuse of a right angled prism ABC as shown in the diagram given alongside :
(i) Copy the diagram and complete the path of the ray PQ till it emerges from the prism.
(ii) What is the value of the angle of deviation of the ray ?
(iii) Name an instrument where his action of the prism is used.
(c) A converging lens is used to obtain an image of an object placed in front of it. The inverted image is formed between F2 and 2F2 of the lens.
(i) Where is the object placed ?
(ii) Draw a ray diagram to illustrate the formation of the image.
Answer: (a) (i) The critical angle is the angle of incidence in an optically denser medium for which the corresponding angle of refraction in the optically rarer medium is exactly \( 90^{\circ} \).
(ii) The relation is: \[ \mu = \frac{1}{\sin C} \] (iii) The depth of the tank **remains the same** when viewed normally (straight down) from above because the light rays enter the air normally without suffering any refraction. (b) (i) Below is the completed ray diagram showing the path of the light ray PQ through the prism ABC:
(ii) The total angle of deviation produced is exactly **\( 180^{\circ} \)**. (iii) This prism action is used in **binoculars** or a **prism periscope** to invert images. (c) (i) The object must be placed **beyond \( 2F_1 \)** (between \( 2F_1 \) and infinity). (ii) Below is the ray diagram showing the image formation between \( F_2 \) and \( 2F_2 \):

In simple words: (a) Critical angle is the boundary angle of refraction. Refractive index is \( 1/\sin C \). Looking straight down, water doesn't appear raised because the rays don't bend at all. (b) A ray hitting normally on the hypotenuse reflects twice off the perpendicular sides, turning around 180° back out. This is used in binoculars. (c) When the object is placed far away (beyond 2F), the lens creates a smaller real inverted image between F and 2F.

Exam Tip: Be sure to include both internal reflections inside the prism and trace the emergent ray with clear arrowheads to show the full path.

2013

Question 29. (a) A ray of light is moving from a rarer medium to a denser medium and strikes a plane mirror placed at 90° to the direction of the ray as shown in the diagram.
(i) Copy the diagram and mark arrows to show the path of the ray of light after it is reflected from the mirror.
(ii) Name the principle you have used to mark the arrows to show the direction of the ray.
(b) (i) The refractive index of glass with respect to air is 1.5. What is the value of the refractive index of air with respect to glass ?
(ii) A ray of light is incident as a normal ray on the surface of separation of two different mediums. What is the value of the angle of incidence in this case ?
(c) (i) Can the absolute refractive index of a medium be less than one ?
(ii) A coin placed at the bottom of a beaker appears to be raised by 4.0 cm. If the refractive index of water is 4/3, find the depth of the water in the beaker.
(d) An object AB is placed between 2F1 and F1 on the principal axis of a convex lens as shown in the diagram. Copy the diagram and using three rays starting from point A, obtain the image of the object formed by the lens.
Answer: (a) (i) Below is the completed ray diagram showing the light ray reflecting straight back along its incident path after striking the mirror normally:
(ii) The principle used is the **Principle of Reversibility of Light**, which states that the path of a light ray is completely reversible. (b) (i) Given: \( ^a\mu_g = 1.5 \). The refractive index of air with respect to glass (\( ^g\mu_a \)) is: \[ ^g\mu_a = \frac{1}{^a\mu_g} = \frac{1}{1.5} = \frac{2}{3} \approx 0.67 \] (ii) When a ray of light is incident normally on the boundary, its direction is along the normal, so the angle of incidence is exactly **\( 0^{\circ} \)**. (c) (i) No, the absolute refractive index of any medium can never be less than one, because the speed of light in any physical medium is always less than the speed of light in a vacuum (\( c \)). (ii) Let the actual depth of water in the beaker be \( x \text{ cm} \). Since the coin appears raised by \( 4.0 \text{ cm} \), the apparent depth is \( (x - 4) \text{ cm} \). Using the relation: \[ \mu = \frac{\text{Real Depth}}{\text{Apparent Depth}} \] \[ \frac{4}{3} = \frac{x}{x - 4} \] \[ 4(x - 4) = 3x \] \[ 4x - 16 = 3x \]
\( \implies x = 16 \text{ cm} \) The actual depth of water in the beaker is \( 16 \text{ cm} \). (d) Below is the completed ray diagram showing the formation of the magnified, real, and inverted image \( A_1 B_1 \) of the object AB when placed between \( F_1 \) and \( 2F_1 \), using three rays from point A:

In simple words: (a) A light ray hitting a perpendicular mirror bounces straight back, following the rule that light can travel both ways along the same path. (b) The refractive index of air with respect to glass is 0.67, and normal incidence always has a 0° angle. (c) Refractive index is always at least 1. Solving the water ratio shows that a 4 cm rise corresponds to a real depth of 16 cm. (d) For an object between F and 2F, the lens creates a larger, upside-down image beyond 2F on the other side.

Exam Tip: Always show normal incidence with \( i = 0^{\circ} \). For the three-ray tracing diagram, ensure that the parallel ray, focal ray, and optical center ray all meet at a single focus point beyond \( 2F_2 \).

2014

Question 30. (a) Draw the diagram given below and clearly show the path taken by the emergent ray.
(b) (i) A ray of light passes from water to air. How does’ the speed of light change?
(ii) Which colour of light travels fastest in any medium except air? [2]
(c) Name the factors affecting the critical angle for the pair of media.
Answer: (a) Below is the completed ray diagram showing the path of the light ray through the glass slab:
(b) (i) When a ray of light transitions from water (optically denser medium) to air (optically rarer medium), the speed of the light **increases**. (ii) **Red light** travels fastest in any given refracting medium (except air/vacuum) because it suffers the least refraction and has the longest wavelength. (c) The factors that affect the critical angle for a pair of media are: 1. **Color (Wavelength) of Light:** The critical angle is largest for red light and smallest for violet light (since refractive index decreases with increasing wavelength). 2. **Temperature of the Medium:** As the temperature increases, the refractive index of the medium decreases, which causes the critical angle to **increase**.
In simple words: (a) The light ray enters glass at 45°, bends closer to the normal inside, and exits at the exact same angle of 45° parallel to its original path. (b) Light speeds up when entering air from water, and red light travels the fastest inside any medium. (c) The critical angle depends on the wavelength of light (highest for red) and rises when the medium's temperature goes up.

Exam Tip: State that the incident ray and the emergent ray must be parallel to each other, with the emergence angle exactly equal to the angle of incidence.

Question 31. (a) (i) Light passes through a rectangular glass slab and through a triangular glass prism. In what way does the direction of the two emergent beams differ and why?
(ii) Ranbir claims to have obtained an image twice the size of the object with a concave lens. Is he correct? Give a reason for your answer. [4]
(b) A lens forms an erect, magnified and virtual image of an object.
(i) Name the lens.
(ii) Draw a labelled ray diagram to show the formation of the image.
(c) (i) Define the power of a lens.
(ii) The lens mentioned in 2(b) above is of focal length 25 cm. Calculate the power of the lens.
Answer: (a) (i) In a rectangular glass slab, the refracting surfaces are parallel, so the emergent beam emerges parallel to the incident beam, suffering only lateral displacement without any net angular deviation. In a triangular glass prism, the refracting surfaces are inclined at an angle, which causes the emergent beam to undergo both angular deviation (bending towards the base) and dispersion into a spectrum. (ii) No, Ranbir's claim is incorrect. A concave lens can only produce virtual, erect, and **diminished** images. It is physically impossible for a concave lens to produce a magnified image. (b) (i) The lens is a **convex lens**. (ii) Below is the ray diagram showing the formation of an erect, magnified, and virtual image:
(c) (i) The power of a lens is a measure of its ability to bend or deviate light rays passing through it. Mathematically, it is the reciprocal of the focal length of the lens expressed in meters. \[ P = \frac{1}{f \text{ (in metres)}} \] (ii) Given: \( f = +25 \text{ cm} = +0.25 \text{ m} \). The power of the lens (\( P \)) is: \[ P = \frac{100}{f \text{ (in cm)}} = \frac{100}{25} = +4 \text{ D} \] Therefore, the power of the lens is \( +4 \text{ D} \).
In simple words: (a) A glass slab makes light exit parallel to its entry with no deviation, while a prism deviates and splits the light because of its angled surfaces. A concave lens can only shrink images, so Ranbir is incorrect. (b) A convex lens creates a magnified, virtual image when an object is placed closer than its focal point. (c) Power is the reciprocal of focal length in meters. For a 25 cm lens, dividing 100 by 25 gives +4 Dioptres.

Exam Tip: Be sure to distinguish between glass slab lateral displacement and prism angular deviation. State that power of a convex lens is always positive.

2015

Question 32. 1. Name one factor that affects the lateral displacement of light as it passes through a rectangular glass block.
2. The speed of light in glass is 2 × 10⁵ km/s. What is the refractive index of glass?
Answer: 1. The lateral displacement of light is affected by the thickness of the glass block, the angle of incidence, and the refractive index of the glass. 2. Given data: - Speed of light in air (\( c \)) = \( 3 \times 10^8 \text{ m s}^{-1} \) - Speed of light in glass (\( v \)) = \( 2 \times 10^5 \text{ km s}^{-1} = 2 \times 10^8 \text{ m s}^{-1} \) The refractive index of glass (\( \mu_{\text{glass}} \)) is given by: \[ \mu_{\text{glass}} = \frac{\text{Speed of light in air}}{\text{Speed of light in glass}} \] \[ \mu_{\text{glass}} = \frac{3 \times 10^8 \text{ m s}^{-1}}{2 \times 10^8 \text{ m s}^{-1}} = 1.5 \] Therefore, the refractive index of the glass is \( 1.5 \).
In simple words: 1. The thickness of the block affects how much the light is shifted sideways. 2. Since light travels 1.5 times slower in glass than in air, the refractive index of glass is 1.5.

Exam Tip: In speed division calculations, always convert both velocities to the same units (e.g. converting \( 2 \times 10^5 \text{ km/s} \) to \( 2 \times 10^8 \text{ m/s} \)) before performing calculations.

Question 33. (a) (i) Where should an object be placed so that a real and inverted image of the same size as the object is obtained using a convex lens ?
(ii) Draw a ray diagram to show the formation of the image as specified in the part a(i).
Answer: (i) To obtain a real and inverted image of the same size, the object must be placed at a distance equal to twice the focal length (\( 2F_1 \)) in front of the convex lens. (ii) Below is the ray diagram showing the image formed at \( 2F_2 \) on the opposite side of the lens:

In simple words: (i) To get an upside-down real image that is exactly the same size, place the object at 2F. (ii) The ray diagram shows the parallel ray and optical center ray intersecting perfectly at 2F on the opposite side.

Exam Tip: Be sure to label both focal points as \( 2F_1 \) and \( 2F_2 \) on the principal axis to ensure a complete ray diagram representation.

Question 33(b). Jatin puts a pencil into a glass container having water and is surprised o see the pencil in a different state. [4]
(i) What change is observed in the appearance of the pencil?
(ii) Name the phenomenon responsible for the change.
(iii) Draw a ray diagram showing how the eye sees the pencil.
Answer: (i) The pencil appears bent or broken at the water-air interface. It also appears shorter and elevated (raised) inside the water.
(ii) This occurrence is caused by the refraction of light as it moves from water (optically denser medium) to air (optically rarer medium), altering the apparent depth of the submerged portion.
(iii) Below is the ray diagram illustrating how the observer sees the pencil:

In simple words: When a pencil is placed in water, the light coming from the underwater part bends as it enters the air. This tricks our eyes into seeing that part higher up, making the pencil look bent or broken at the surface.

Exam Tip: In your ray diagram, make sure that the backward dashed projections of the refracted rays meet precisely at the virtual image of the pencil tip (shallower position), and use arrowheads to show the direction of light rays from water to the eye.

2016

Question 34. (a) A boy uses blue colour of light to find refractive index of glass. He then repeats experiment using red colour of light. Will the refractive index be same or different in the two cases? Give a reason to support your answer.
(b) Copy the diagram given above and complete the path of ray till it emerges out of prism. The critical angle of glass is 42°. In your diagram mark the angles wherever necessary.
(c) State the dependence of angle of deviation :
1. On refractive index of a material of prism.
2. On the wavelength of light
Answer: (a) The refractive index will be **different** in the two cases. The refractive index of glass varies for different colors of light because speed of light inside the medium depends on its color. Specifically, blue light travels slower than red light in glass, meaning its wavelength is shorter than that of red light. Consequently, blue light has a higher refractive index and suffers greater deviation (bending) than red light. (b) Here is the completed ray diagram:
**Geometric Tracing Steps:** 1. The incident ray PQ strikes face AB normally (\( 90^{\circ} \)), so it enters without bending and meets the base BC at point R. 2. In the right-angled triangle BQR, \( \angle QBR = 60^{\circ} \) and \( \angle BQR = 90^{\circ} \), which gives \( \angle BRQ = 30^{\circ} \). 3. The normal at R is perpendicular to BC. Thus, the angle of incidence at R is \( 90^{\circ} - 30^{\circ} = 60^{\circ} \). 4. Since the angle of incidence (\( 60^{\circ} \)) exceeds the critical angle of glass (\( 42^{\circ} \)), the ray undergoes total internal reflection, reflecting at \( 60^{\circ} \) with the normal (making \( \angle CRS = 30^{\circ} \)). 5. In triangle CRS, \( \angle RCS = 60^{\circ} \) and \( \angle CRS = 30^{\circ} \), meaning \( \angle RSC = 180^{\circ} - (60^{\circ} + 30^{\circ}) = 90^{\circ} \). 6. Thus, the reflected ray strikes face AC normally and emerges undeviated as ray RS. (c) 1. **Refractive Index:** The angle of deviation is directly proportional to the refractive index of the prism material. Prisms with a higher refractive index deviate light rays more than prisms with a lower refractive index. 2. **Wavelength of Light:** The angle of deviation is inversely proportional to the wavelength of light. As wavelength increases, the angle of deviation decreases. Consequently, the deviation is maximum for violet light (shortest wavelength) and least for red light (longest wavelength).
In simple words: (a) Different colors travel at different speeds in glass, so they bend by different amounts; blue light slows down more and bends more than red. (b) Light entering straight into one side reflects off the bottom and shoots straight out the other side normally. (c) A denser glass material bends light more, and shorter wavelengths (like violet) bend much more than longer ones (like red).

Exam Tip: For the prism ray-tracing problem, remember to explicitly calculate and label the angle of incidence as \( 60^{\circ} \) at the base, and show that since \( 60^{\circ} > 42^{\circ} \), total internal reflection must occur.

Question 35. (a)
1. Write a relationship between the angle of incidence and the angle of refraction for a given pair of media.
2. When a ray of light enters from one medium to another medium having different optical densities, it bends, why does this phenomenon occur ?
3. Write a condition where it does not bend when entering a medium of different optical density.
(b) A lens produces a virtual image between the object and the lens.
1. Name the lens
2. Draw a ray diagram to show the formation of image.
Answer: (a) 1. The ratio of the sine of the angle of incidence (\( i \)) to the sine of the angle of refraction (\( r \)) remains constant for any given pair of media. This constant represents the refractive index (\( \mu \)): \[ \frac{\sin i}{\sin r} = \mu \] 2. Refraction occurs because the speed of light changes as it moves from one optical medium into another of a different density, causing the light rays to deviate from their path. 3. A light ray will not bend if it is incident normally (at \( 90^{\circ} \) to the boundary surface, making an angle of incidence \( i = 0^{\circ} \)). (b) 1. The lens used is a **concave lens**. Since a concave lens is the only lens that always produces a virtual, upright, and diminished image positioned between the object and the lens itself, it uniquely fits this description. 2. Below is the ray diagram showing the image formation:

In simple words: (a) Snell's law relates the entry and exit angles. Light bends when entering a new material because its speed changes. It won't bend at all if it enters straight down at a 90-degree angle. (b) A concave lens is the only lens that can produce a smaller, upright, virtual image in front of the object itself.

Exam Tip: Remember that a concave lens always produces virtual, erect, and diminished images. Be sure to use proper sign conventions when drawing or performing calculations on diverging lenses.