CBSE Class 10 Science HOTs Reflection and Refraction Set 08

This CBSE Class 10 Science HOTs page gives 25 higher order thinking questions with answers on Chapter 9, Light – Reflection and Refraction. It combines reasoning about plane, concave and convex mirrors, lenses in different media and refraction with numericals on image formation, magnification, lens power and speed of light, all explained step by step.

Reasoning and Application Based HOTS Questions

Question 1. When we sit in front of a plane mirror and write with our right hand, it appears in the mirror that we are writing with the left hand. Why?
Answer:
A plane mirror forms a laterally inverted image. The left side of the object appears as the right side of the image, and the right side appears as the left side, although the image is erect and of the same size.
So the right hand that writes appears in the mirror as the left hand.
In simple words: A mirror swaps left and right, so your right hand looks like a left hand.

Teacher's Note:
1. Lateral inversion swaps left and right, not up and down.
2. The same reason makes AMBULANCE appear reversed on the front of the van.

Question 2. Explain why, though both a plane mirror and a sheet of paper reflect light but we can see the image of our face in a plane mirror but not in a sheet of paper?
Answer:
A plane mirror has a smooth, polished surface. It gives regular reflection: parallel rays remain parallel after reflection, and they reach the eye in an orderly way to form an image.
A sheet of paper has a rough surface. It gives diffuse (irregular) reflection: parallel rays are scattered in all directions, so no image is formed. The laws of reflection hold at each point, but the normals point in different directions.
In simple words: Smooth surfaces send reflected rays in one direction and give an image; rough surfaces scatter them.

Teacher's Note:
1. Both types of reflection obey the laws of reflection.
2. Scattered reflection is what lets us see non-luminous objects from all directions.

Question 3. When a spherical mirror is held towards the Sun and its sharp image is formed on a piece of carbon paper for some time, a hole is burnt in the carbon paper. Why is a hole burnt in the carbon paper?
Answer:
The mirror is concave. The Sun's rays are almost parallel, so the concave mirror converges them to a very small bright spot at its principal focus.
All the light energy falling on the whole mirror is concentrated on this tiny spot, and the black carbon paper absorbs it. The temperature rises so much that a hole is burnt in the paper.
In simple words: The mirror squeezes a lot of sunlight onto a tiny spot, which becomes very hot.

Teacher's Note:
1. The distance of the spot from the mirror gives the focal length.
2. Never look at the Sun directly or through a mirror.

Question 4. Why can you not use a concave mirror as a rear-view mirror in vehicles?
Answer:
A concave mirror gives a magnified (and, for distant vehicles, inverted and real) image and has a small field of view, so the driver would see only a small part of the road behind.
A convex mirror always gives an erect, diminished image and covers a wide field of view, so it is used as a rear-view mirror.
In simple words: A concave mirror shows too little of the road and gives a confusing image.

Teacher's Note:
1. The two needs are an erect image and a wide field of view.
2. Only the convex mirror gives both.

Question 5. A ray of light travelling in air enters obliquely into water. Does the ray of light bend towards the normal or away from the normal? Why?
Answer:
The ray bends towards the normal.
Air is optically rarer and water is optically denser. The speed of light decreases on entering water, and a ray going from a rarer to a denser medium bends towards the normal.
In simple words: Light slows down in water and so leans towards the normal.

Teacher's Note:
1. Slower medium means the ray bends towards the normal.
2. Reverse the path (water to air) and it bends away from the normal.

Question 6. Why does a beam of light bend when it enters glass at an angle? Why does it not bend if it enters the glass at right angles?
Answer:
The speed of light in glass is less than that in air. When a beam enters obliquely, one edge of the beam reaches the glass first and slows down while the other edge is still travelling faster in air, so the beam turns (refraction).
When the beam enters at right angles (along the normal), the whole beam enters the glass at the same time and slows down together. Nothing turns it, so the beam goes straight (the angle of refraction is 0°).
In simple words: Light bends only when different parts of the beam change speed at different moments.

Teacher's Note:
1. The change of speed is the cause of refraction.
2. For normal incidence the direction does not change but the speed does.

Question 7. Explain why, a real image can be projected on a screen but a virtual image cannot?
Answer:
In a real image the reflected or refracted rays actually meet at a point, so light energy is really present there and a screen placed at that point shows the image.
In a virtual image the rays only appear to come from a point; they do not actually meet. No light reaches that point, so it cannot be received on a screen.
In simple words: Real images are made of light that really arrives; virtual images only seem to be there.

Teacher's Note:
1. Real image: rays actually meet. Virtual image: rays only appear to meet.
2. A virtual image can still be seen by the eye or a camera looking into the mirror or lens.

Question 8. Which causes more bending (or more refraction) of light rays passing through it: a convex lens of long focal length or a convex lens of short focal length? Why?
Answer:
A convex lens of short focal length causes more bending.
A short focal length means the lens is more strongly curved (thicker in the middle). Its rays converge at a point very close to the lens, so they are turned through larger angles. Its power P = 1/f is also greater.
In simple words: The shorter the focal length, the more strongly the lens bends light.

Teacher's Note:
1. Power and bending go together: more power means more bending.
2. Link focal length, curvature and power in your answer.

Question 9. Which has more power: a thick convex lens or a thin convex lens, made of the same glass? Give reason for your choice.
Answer:
The thick convex lens has more power.
A thicker lens has more curved surfaces, so it bends the rays more and has a shorter focal length. Since P = 1/f, a smaller f gives a larger power.
In simple words: A fatter lens bends light more, so it is more powerful.

Teacher's Note:
1. Same glass means the same refractive index, so curvature decides the power.
2. Use P = 1/f to link the two ideas.

Question 10. A concave mirror and a convex lens are held separately in water. What changes, if any, do you expect in the focal length of either?
Answer:
Concave mirror: No change. Reflection does not depend on the medium, and f = R/2 depends only on the radius of curvature.
Convex lens: The focal length increases. In water the refractive index of glass relative to the surrounding medium becomes smaller (1.5/1.33 instead of 1.5), so the lens bends light less and its focal length becomes larger.
In simple words: The mirror is unaffected, but the lens becomes weaker in water.

Teacher's Note:
1. The bending depends on the refractive index of the lens material relative to the surrounding medium.
2. This is why a glass lens is weaker in water than in air.

Question 11. A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent if instead of water we use liquids like kerosene or turpentine? Support your answer with reason.
Answer:
No, the pencil will not appear bent to the same extent.
The apparent bending is due to refraction of light at the surface of the liquid. The amount of bending depends on the refractive index of the liquid (water 1.33, kerosene 1.44, turpentine 1.47). Liquids with a larger refractive index bend the light more, so the pencil appears more bent in kerosene or turpentine than in water.
In simple words: Different liquids bend light by different amounts, so the pencil looks bent differently.

Teacher's Note:
1. Refractive index measures how much a medium bends light.
2. Name the reason as refraction at the air-liquid interface.

Question 12. A convex lens is made of a material of refractive index n₁. When placed in a medium of refractive index n₂, it behaves as a diverging lens. How are n₁ and n₂ related?
Answer:
A lens bends light according to the refractive index of its material relative to the surrounding medium.
A convex-shaped lens acts as a diverging lens only when the surrounding medium is optically denser than the lens material. Hence n₂ > n₁.
(For example, an air bubble in water has a convex shape but diverges light.)
In simple words: A lens thicker in the middle diverges light when the medium around it is denser than the lens.

Teacher's Note:
1. Compare the refractive index of lens material with that of the medium.
2. If n₂ = n₁ the lens has no effect on the rays.

Question 13. A convex lens made of material of refractive index n₂ is kept in a medium of refractive index n₁. A parallel beam of light is incident on the lens. Compare the path of rays of light emerging from the convex lens if (i) n₁ < n₂ (ii) n₁ = n₂ (iii) n₁ > n₂
Answer:
(i) n₁ < n₂: the lens is optically denser than the medium, so it behaves as a normal convex lens and the emerging rays converge to a point (real focus).
(ii) n₁ = n₂: there is no refraction at the surfaces, so the rays pass through undeviated (the lens becomes invisible, with infinite focal length).
(iii) n₁ > n₂: the lens is optically rarer than the medium, so the convex lens behaves like a concave lens and the emerging rays diverge.
In simple words: Compare the lens with its surroundings: denser converges, equal does nothing, rarer diverges.

Teacher's Note:
1. The effect of a lens depends on the ratio of the two refractive indices.
2. Part (ii) is the idea behind a glass rod becoming invisible in a liquid of the same refractive index.

Question 14. A man standing in front of a spherical mirror, finds his image having a very small head, a fat body and legs of normal size. What type of mirror are used in the three parts?
Answer:
Very small head: convex mirror, because it makes the image diminished.
Fat body: concave mirror, because it makes an enlarged (magnified) image.
Legs of normal size: plane mirror, because it forms an image of the same size.
In simple words: Convex makes things small, concave makes them big, and plane keeps the size.

Teacher's Note:
1. Link each effect to magnification: less than 1, more than 1 and equal to 1.
2. Funny mirrors in fairs are built this way.

Question 15. How do you distinguish between a plane mirror, a concave mirror and a convex mirror without touching them?
Answer:
Hold each mirror at a short distance from your face (or an object) and look at the image formed.
Plane mirror: the image is erect and of the same size as the face.
Concave mirror: the image is erect and enlarged (magnified) when the face is close to the mirror.
Convex mirror: the image is erect and diminished, and the mirror shows a larger field of view.
In simple words: Same size means plane, bigger means concave, smaller means convex.

Teacher's Note:
1. Keep the face within the focal length of the concave mirror to see an erect image.
2. Compare the size of the image with that of the object.

Question 16. Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?
Answer:
The window pane is at a finite distance, so its image is formed beyond the focus (at 15 cm). The building is much farther away, so its rays are nearly parallel and the image forms closer to the lens, at the focus.
Sudha must move the screen towards the lens.
The sharp image of a very distant object is at the principal focus, so the focal length is a little less than 15 cm; it is approximately 15 cm.
In simple words: For a farther object the image moves closer to the lens, so the screen is moved in.

Teacher's Note:
1. As the object moves away, the image moves towards the focus.
2. The image of a very distant object lies at the focus.

Question 17. A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Answer:
Yes, the statement is correct.
Magnified virtual image: object between the optical centre and the focus, i.e. at a distance less than 20 cm from the lens.
Magnified real image: object between F and 2F, i.e. at a distance between 20 cm and 40 cm from the lens. (At 40 cm the real image is of the same size.)
In simple words: A convex lens magnifies when the object is inside 2F: virtual within F, real between F and 2F.

Teacher's Note:
1. The virtual image is erect and the real image is inverted.
2. Use the magnifying glass for the virtual image and the projector for the real image.

Numerical Based HOTS Questions

Question 18. An object of 5 cm is placed 20 cm away from converging lens of focal length 10 cm. Draw the ray diagram to produce the image and find the position, size and the nature of the image formed.
Answer:
Given: h = 5 cm, u = −20 cm, f = +10 cm.
\( \dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{10}-\dfrac{1}{20}=\dfrac{1}{20} \), so v = +20 cm (on the other side of the lens, at 2F).
m = v/u = 20/(−20) = −1, so h′ = −5 cm: the image is real, inverted and of the same size as the object.

In simple words: The object at 2F gives a real, upside-down image of the same size at 2F on the other side.

Teacher's Note:
1. Object at 2F gives image at 2F with m = −1.
2. Show F and 2F on both sides in the ray diagram.

Question 19. An object 3 cm in height is placed 20 cm from convex lens of focal length 12 cm. Find the nature, position and height of the image.
Answer:
Given: h = 3 cm, u = −20 cm, f = +12 cm.
\( \dfrac{1}{v}=\dfrac{1}{12}-\dfrac{1}{20}=\dfrac{5-3}{60}=\dfrac{1}{30} \), so v = +30 cm.
m = v/u = 30/(−20) = −1.5, so h′ = −1.5 × 3 = −4.5 cm.
The image is real, inverted and enlarged, 30 cm from the lens on the other side, and 4.5 cm tall.
In simple words: The image is 30 cm away, upside down and 1.5 times larger (4.5 cm).

Teacher's Note:
1. The object lies between F and 2F, so the image is beyond 2F.
2. A negative image height means an inverted image.

Question 20. Seema went to the science museum and saw her image when she was standing around 2 m away from the mirror. She was surprised to see her image which is just half of her actual height. What would be the nature and focal length of the mirror? Seema conducted another activity using an object of height 12 cm and a convex mirror of focal length 30 cm. She placed the object at a distance of 30 cm in front of the mirror. What is likely to be the height of the image formed?
Answer:
Part 1. The image is erect and half her height, so m = +1/2. A mirror that always gives a diminished erect image is a convex mirror.
u = −200 cm and m = −v/u = 1/2, so v = +100 cm.
\( \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{100}-\dfrac{1}{200}=\dfrac{1}{200} \), so f = +200 cm = 2 m.
Part 2. f = +30 cm, u = −30 cm, h = 12 cm: \( \dfrac{1}{v}=\dfrac{1}{30}+\dfrac{1}{30}=\dfrac{1}{15} \), so v = +15 cm.
m = −v/u = −15/(−30) = +0.5, so h′ = 0.5 × 12 = 6 cm (erect, virtual).
In simple words: Half-sized erect image means a convex mirror of focal length 2 m; the 12 cm object gives a 6 cm image.

Teacher's Note:
1. For a convex mirror, f is positive and u is negative.
2. A positive magnification below 1 means an erect, diminished, virtual image.

Question 21. A student has two lenses A and B which have focal length 50.0 cm and 20 cm respectively. (A) (i) State whether lenses A and B are thicker in the middle or at the edges. (ii) Which lens bends the light rays more strongly? (B) Where should the student place the object to get an image at a distance of 30 cm behind the lens B? (C) What would be the combined power of two lenses A and B?
Answer:
(A)(i) Both focal lengths are positive, so both are convex lenses. They are thicker in the middle than at the edges.
(ii) Lens B (f = 20 cm) bends the rays more strongly, because a shorter focal length means a greater power.
(B) For lens B: f = +20 cm and v = +30 cm. \( \dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=\dfrac{1}{30}-\dfrac{1}{20}=-\dfrac{1}{60} \), so u = −60 cm: the object must be 60 cm in front of lens B.
(C) P_A = 1/0.5 = +2 D, P_B = 1/0.2 = +5 D. For lenses in contact the powers add: P = 2 + 5 = +7 D (equivalent focal length ≈ 14.3 cm).
In simple words: Both are convex; the 20 cm lens is stronger; keep the object at 60 cm; together they have +7 D.

Teacher's Note:
1. Use P = 1/f with f in metres.
2. The combined power of lenses in contact is the sum of their powers.

Question 22. A concave mirror of focal length f produces an image n times the size of an object. What would be the object distance for which the image is real?
Answer:
For a real image the magnification is negative: m = −v/u = −n, so v = nu.
Put v = nu in the mirror formula: \( \dfrac{1}{nu}+\dfrac{1}{u}=\dfrac{1}{f} \), so \( \dfrac{n+1}{nu}=\dfrac{1}{f} \).
Hence u = \( \dfrac{(n+1)f}{n} \). Because f is negative for a concave mirror, u is negative; the object is at a distance \( \dfrac{n+1}{n}|f| \) in front of the mirror.
Check: for n = 2 and f = −10 cm, u = −15 cm and v = −30 cm, giving m = −2.
In simple words: For a real image n times as big, the object distance is (n+1)/n times the focal length.

Teacher's Note:
1. Substitute v = nu (same sign for a real image) into the mirror formula.
2. The result confirms the object lies between F and C for n > 1.

Question 23. If the light from the Sun subtend an angle θ at the pole of a concave mirror of focal length f, where will it converge? Also find the diameter of the image. Draw the necessary ray diagram.
Answer:
The Sun is very far away, so its rays are almost parallel and the image is formed at the principal focus of the concave mirror, at a distance f from the pole.
Rays from the two opposite edges of the Sun make an angle θ with each other. All rays from one point of the Sun are parallel and meet at one point of the focal plane, so the two edges give image points separated by about f × θ (arc length = radius × angle, with θ in radians).
Diameter of the image ≈ fθ, a tiny bright spot (real, inverted and highly diminished).

In simple words: The Sun's image is a small bright spot at the focus, about f times θ wide.

Teacher's Note:
1. Use the small angle approximation: tan θ ≈ θ for θ in radians.
2. The image of a very distant object always forms at the focus.

Question 24. An object is held at a distance of 60 cm from a convex mirror of focal length 20 cm. At what distance from the mirror, should a plane mirror be held so that images in the two mirrors coincide?
Answer:
Convex mirror: u = −60 cm and f = +20 cm. \( \dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{60}=\dfrac{4}{60} \), so v = +15 cm. The image is 15 cm behind the convex mirror.
Let the plane mirror be at distance d in front of the convex mirror, on the object side. The object is (60 − d) from the plane mirror, so its image is (60 − d) behind it. Measured from the pole of the convex mirror (positive = behind it), the image is at 60 − 2d.
For the two images to coincide: 60 − 2d = 15, so d = 22.5.
The plane mirror should be held 22.5 cm from the convex mirror (between the object and the convex mirror).
In simple words: The convex mirror's image is 15 cm behind it; a plane mirror 22.5 cm in front of it puts its image at the same spot.

Teacher's Note:
1. A plane mirror forms its image as far behind as the object is in front.
2. Take distances from the same reference point (the pole of the convex mirror).

Question 25. How much time will light take to cross 2 mm thick glass pane if refractive index of glass is 3/2?
Answer:
Speed of light in glass v = c/n = (3 × 10⁸)/(3/2) = 2 × 10⁸ m/s.
Thickness d = 2 mm = 2 × 10⁻³ m.
Time t = d/v = (2 × 10⁻³)/(2 × 10⁸) = 10⁻¹¹ s.
In simple words: Light takes 10⁻¹¹ seconds to cross the 2 mm glass pane.

Teacher's Note:
1. Convert millimetres to metres before dividing.
2. Light travels slower in glass than in air by the factor n.