Exercise — Light - Reflection and Refraction

17 questions✓ Free · step-by-step

11 markNCERT Cl-10 Science, Exercise Q1
Which one of the following materials cannot be used to make a lens? (a) Water (b) Glass (c) Plastic (d) Clay
Hint. A lens has to let light pass through it and bend as it does.
Step 1 — Recall what a lens needs to do. A lens must be transparent, so that light can pass through and refract at its surfaces.

Step 2 — Check each material. Water, glass and plastic are all transparent; clay is opaque and blocks light rather than transmitting it.

✦ Answer: (d) Clay.
Where students slip. Picking water because it seems like an unusual lens material — water is genuinely transparent and can form a lens (e.g. a water-filled convex container); clay's opacity is the actual disqualifier.
21 markNCERT Cl-10 Science, Exercise Q2
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object? (a) Between the principal focus and the centre of curvature (b) At the centre of curvature (c) Beyond the centre of curvature (d) Between the pole of the mirror and its principal focus.
Hint. Only one object position gives a concave mirror a virtual image at all.
Step 1 — Recall when a concave mirror gives a virtual image. A concave mirror only produces a virtual (and erect, enlarged) image when the object is placed between the pole and the principal focus.

Step 2 — Rule out the other options. At C, beyond C, or between F and C, a concave mirror always gives a real image, not a virtual one.

✦ Answer: (d) Between the pole of the mirror and its principal focus.
Where students slip. Picking 'between F and C' — that range gives a real, inverted, magnified image, not a virtual one; only the region between the pole and F gives a virtual image.
31 markNCERT Cl-10 Science, Exercise Q3
Where should an object be placed in front of a convex lens to get a real image of the size of the object? (a) At the principal focus of the lens (b) At twice the focal length (c) At infinity (d) Between the optical centre of the lens and its principal focus.
Hint. Same-size and real together pin down one exact position, not a range.
Step 1 — Recall the same-size condition. A convex lens gives a real image exactly the same size as the object only when the object is placed at twice the focal length (2F).

Step 2 — Rule out the others. At F, the image forms at infinity; between the optical centre and F, the image is virtual and magnified; at infinity, the image is a diminished point at F.

✦ Answer: (b) At twice the focal length.
Where students slip. Picking 'at the principal focus' — an object at F produces an image at infinity, not a same-size real image.
41 markNCERT Cl-10 Science, Exercise Q4
A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be (a) both concave. (b) both convex. (c) the mirror is concave and the lens is convex. (d) the mirror is convex, but the lens is concave.
Hint. Work out what a negative focal length means for a mirror, then separately for a lens.
Step 1 — Interpret the sign for the mirror. A negative focal length for a spherical mirror corresponds to a concave mirror.

Step 2 — Interpret the sign for the lens. A negative focal length for a lens corresponds to a concave (diverging) lens.

Step 3 — Combine. Since both have f = −15 cm, both must be concave.

✦ Answer: (a) both concave.
Where students slip. Assuming the same sign must mean opposite types for mirror vs. lens — it's true that the physical curvature looks different, but the sign convention for negative focal length points to 'concave' in both cases independently.
51 markNCERT Cl-10 Science, Exercise Q5
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be (a) only plane. (b) only concave. (c) only convex. (d) either plane or convex.
Hint. Check whether a concave mirror can keep giving an erect image at every possible distance.
Step 1 — Check the plane mirror. A plane mirror always gives an erect, same-size image, regardless of distance.

Step 2 — Check the convex mirror. A convex mirror always gives an erect, diminished image, regardless of distance too.

Step 3 — Check the concave mirror. A concave mirror gives an erect image only when the object is close (between the pole and focus) — at larger distances, the image becomes real and inverted.

✦ Answer: (d) either plane or convex.
Where students slip. Picking 'only convex' and forgetting that a plane mirror also always gives an erect image at any distance — the question's phrasing ('no matter how far') fits both plane and convex mirrors equally.
61 markNCERT Cl-10 Science, Exercise Q6
Which of the following lenses would you prefer to use while reading small letters found in a dictionary? (a) A convex lens of focal length 50 cm. (b) A concave lens of focal length 50 cm. (c) A convex lens of focal length 5 cm. (d) A concave lens of focal length 5 cm.
Hint. A magnifying glass needs to actually magnify, and needs a short focal length to do it well up close.
Step 1 — Rule out concave lenses. Concave lenses only ever diminish images, so they can't work as a magnifying glass.

Step 2 — Compare the two convex lenses. A convex lens used as a simple magnifier gives greater magnification the shorter its focal length is.

✦ Answer: (c) A convex lens of focal length 5 cm.
Where students slip. Picking the convex lens with the longer focal length (50 cm) — a longer focal length gives weaker magnification for a simple magnifying glass, the opposite of what's needed for reading small print.
73 marksNCERT Cl-10 Science, Exercise Q7
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Hint. Only one region in front of a concave mirror ever gives an erect image.
Step 1 — Recall the condition for an erect image in a concave mirror. A concave mirror gives an erect image only when the object is placed between the pole and the principal focus.

Step 2 — State the distance range. Since the focal length is 15 cm, the object must be placed at a distance less than 15 cm from the mirror (0 to 15 cm).

Step 3 — Describe the image. The image is virtual, erect, and larger than the object (magnified), forming behind the mirror.

Step 4 — Ray diagram (description). A ray parallel to the principal axis reflects through the focus F; a ray directed at the pole reflects symmetrically about the axis. Both reflected rays diverge and appear to meet behind the mirror when extended backward, forming the virtual, enlarged, erect image there.

✦ Answer: The object must be within 15 cm of the mirror (between the pole and focus). The image is virtual, erect and larger than the object.
Where students slip. Including distances beyond the focus (up to the centre of curvature) in the range — anywhere beyond F gives a real, inverted image, not an erect one; the erect-image zone is strictly between the pole and F.
83 marksNCERT Cl-10 Science, Exercise Q8
Name the type of mirror used in the following situations. (a) Headlights of a car. (b) Side/rear-view mirror of a vehicle. (c) Solar furnace. Support your answer with reason.
Hint. One property (parallel beams) is useful twice, in opposite directions; the other property (wide view) is useful once.
Step 1 — (a) Headlights. A concave mirror is used, with the bulb placed at its focus — this converts the diverging light from the bulb into a powerful, parallel beam, since rays originating at the focus reflect parallel to the axis.

Step 2 — (b) Side/rear-view mirror. A convex mirror is used, since it always gives an erect (if diminished) image and has a wider field of view, letting the driver see a larger area of traffic behind.

Step 3 — (c) Solar furnace. A (large) concave mirror is used, since it converges parallel incoming sunlight to a small focal area, concentrating enough energy there to generate very high temperatures.

✦ Answer: (a) Concave mirror (produces a parallel beam from a source at the focus). (b) Convex mirror (erect image, wide field of view). (c) Concave mirror (converges parallel sunlight to a small, very hot focal spot).
Where students slip. Using the same reasoning ('produces a parallel beam') for both the headlight and the solar furnace without noting they're opposite directions of the same idea — headlights send a source's light out as a parallel beam, while a solar furnace takes an incoming parallel beam and concentrates it to a point; both rely on concave mirrors, but for reversed reasons.
92 marksNCERT Cl-10 Science, Exercise Q9
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Hint. Think about whether every point of the object still has some path through the uncovered part of the lens.
Step 1 — State the outcome. Yes, the lens will still produce a complete image of the object.

Step 2 — Explain why. Every point on the object sends out rays in many directions, and enough of these rays still pass through the uncovered half of the lens to converge and form a full image — no part of the image is missing.

Step 3 — Note the actual effect of covering half the lens. Since fewer rays now reach the image point (only from the uncovered half), the image is complete but dimmer than before.

Step 4 — Experimental verification. Using a convex lens to focus the image of a candle flame onto a screen, then covering half the lens with black paper: the complete image of the flame is still seen on the screen, just fainter than with the whole lens uncovered.

✦ Answer: Yes — the image remains complete, only dimmer, since every object point still sends rays through the uncovered half of the lens, which are sufficient to form the full image at reduced brightness.
Where students slip. Predicting that half the image will be missing — covering half the lens reduces the amount of light forming the image (making it dimmer), it doesn't cut the image itself in half.
103 marksNCERT Cl-10 Science, Exercise Q10
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Hint. The object sits beyond 2F here — work out what that tells you about where the image should land, before you even calculate.
Step 1 — Assign values for the lens. u = −25 cm, f = +10 cm (converging lens), h = 5 cm.

Step 2 — Apply the lens formula. 1/v − 1/u = 1/f, so 1/v − 1/(−25) = 1/10, giving 1/v = 1/10 − 1/25 = (5−2)/50 = 3/50, so v = 50/3 ≈ 16.7 cm.

Step 3 — Find the magnification and image height. m = v/u = (50/3)/(−25) = −2/3. Height h′ = m × h = (−2/3)(5) = −10/3 ≈ −3.3 cm.

Step 4 — Interpret the signs. Since v is positive, the image is real, on the opposite side of the lens from the object; since m is negative, the image is inverted; since |m| < 1, it is diminished.

Step 5 — Ray diagram (description). A ray from the object parallel to the axis refracts through the far focus F; a ray through the optical centre goes straight through undeviated. Both rays meet at about 16.7 cm on the far side, forming a small, inverted, real image there — consistent with the object lying beyond 2F (20 cm).

✦ Answer: The image forms at v ≈ 16.7 cm (50/3 cm) on the far side of the lens, is about 3.3 cm tall, and is real, inverted and diminished.
Where students slip. Expecting a magnified image because the object is 'far away' — for a converging lens, an object beyond 2F actually gives a diminished (not magnified) real image between F and 2F on the other side.
113 marksNCERT Cl-10 Science, Exercise Q11
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Hint. A concave lens only ever forms a virtual image on the same side as the object — that fixes the sign of v before you even start.
Step 1 — Assign values for the concave lens. f = −15 cm (concave lens). Since a concave lens only forms virtual images, on the same side as the object, v = −10 cm.

Step 2 — Apply the lens formula. 1/v − 1/u = 1/f, so 1/(−10) − 1/u = 1/(−15), giving −1/u = −1/15 + 1/10 = (−2+3)/30 = 1/30, so u = −30 cm.

Step 3 — Interpret the result. The object is 30 cm in front of the lens — further from the lens than the (virtual) image, which is exactly what a diverging lens always does.

Step 4 — Ray diagram (description). A ray parallel to the axis refracts as if it came from the near-side focus; a ray through the optical centre passes straight through. Both diverge after the lens, and tracing them backward, they appear to meet at 10 cm on the same side as the object, giving a virtual, erect, diminished image there.

✦ Answer: The object is 30 cm from the lens.
Where students slip. Taking v as positive (+10 cm) — a concave lens never forms a real image, so its image distance must be assigned negative, on the same side as the object.
123 marksNCERT Cl-10 Science, Exercise Q12
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Hint. Convex mirror focal lengths are always positive — start from there.
Step 1 — Assign values for the convex mirror. u = −10 cm, f = +15 cm (convex mirror).

Step 2 — Apply the mirror formula. 1/v + 1/u = 1/f, so 1/v + 1/(−10) = 1/15, giving 1/v = 1/15 + 1/10 = (2+3)/30 = 5/30 = 1/6, so v = +6 cm.

Step 3 — Find the magnification. m = −v/u = −6/(−10) = +0.6.

Step 4 — Interpret the signs. Since v is positive, the image forms behind the mirror (virtual); since m is positive and less than 1, the image is erect and diminished.

✦ Answer: The image forms 6 cm behind the mirror, and is virtual, erect and diminished.
Where students slip. Assigning a negative focal length because 'mirrors usually have negative f' — that's only true for concave mirrors; a convex mirror's focal length is positive.
132 marksNCERT Cl-10 Science, Exercise Q13
The magnification produced by a plane mirror is +1. What does this mean?
Hint. Split the meaning of +1 into its size part and its sign part.
Step 1 — Interpret the magnitude. A magnification of magnitude 1 means the image is exactly the same size as the object.

Step 2 — Interpret the sign. Since the sign is positive, the image is virtual and erect.

✦ Answer: The image is the same size as the object (magnitude 1), and it is virtual and erect (positive sign) — exactly what a plane mirror always produces.
Where students slip. Explaining only the '1' (same size) and skipping the '+' sign's meaning — the question specifically has two parts to interpret, size and orientation/nature, both carried in that single number.
143 marksNCERT Cl-10 Science, Exercise Q14
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Hint. Get the focal length from the radius of curvature first, remembering the sign for a convex mirror.
Step 1 — Find the focal length. f = R/2 = 30/2 = 15 cm, positive since the mirror is convex: f = +15 cm.

Step 2 — Assign u and apply the mirror formula. u = −20 cm. 1/v + 1/(−20) = 1/15, so 1/v = 1/15 + 1/20 = (4+3)/60 = 7/60, giving v = 60/7 ≈ 8.6 cm.

Step 3 — Find the magnification and image size. m = −v/u = −(60/7)/(−20) = 3/7 ≈ 0.43. Height h′ = m × h = (3/7)(5) = 15/7 ≈ 2.1 cm.

Step 4 — Interpret the signs. Since v is positive, the image is virtual, behind the mirror; since m is positive and less than 1, it is erect and diminished.

✦ Answer: The image forms about 8.6 cm (60/7 cm) behind the mirror, is virtual and erect, and is about 2.1 cm (15/7 cm) tall.
Where students slip. Using R directly as the focal length instead of halving it — f = R/2 always, so skipping this step would double the focal length used in the rest of the calculation.
153 marksNCERT Cl-10 Science, Exercise Q15
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Hint. A screen can only catch a real image — check first whether this object position even produces one.
Step 1 — Assign values. u = −27 cm, f = −18 cm (concave mirror).

Step 2 — Apply the mirror formula. 1/v + 1/(−27) = 1/(−18), so 1/v = −1/18 + 1/27 = (−3+2)/54 = −1/54, giving v = −54 cm.

Step 3 — Interpret v. Since v is negative, the image is real, forming 54 cm in front of the mirror — this is where the screen should be placed to catch it.

Step 4 — Find the magnification and image size. m = −v/u = −(−54)/(−27) = −2. Height h′ = m × h = (−2)(7) = −14 cm.

Step 5 — Interpret the size and nature. Since m is negative, the image is inverted; since |m| = 2 > 1, it is magnified — consistent with the object lying between F (18 cm) and 2F (36 cm), which always gives an enlarged real image beyond 2F.

✦ Answer: The screen should be placed 54 cm from the mirror. The image is real, inverted, and 14 cm in size (magnified).
Where students slip. Forgetting to check that the object lies between F and 2F, which is exactly what predicts a magnified (not diminished) image beyond 2F — going straight to the formula without this sanity check makes it easy to misread the final answer's plausibility.
162 marksNCERT Cl-10 Science, Exercise Q16
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Hint. Negative power always signals the same lens type, regardless of the specific value.
Step 1 — Convert power to focal length. f = 1/P = 1/(−2.0) = −0.5 m = −50 cm.

Step 2 — Identify the lens type. Since both the focal length and power are negative, the lens must be concave (diverging).

✦ Answer: f = −50 cm; the lens is concave (diverging).
Where students slip. Reporting the focal length as positive 50 cm — the negative power given in the question must carry through to a negative focal length.
172 marksNCERT Cl-10 Science, Exercise Q17
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Hint. Positive power always signals the same lens type, regardless of the specific value.
Step 1 — Relate power and focal length. f = 1/P = 1/1.5 = 2/3 m ≈ +66.7 cm.

Step 2 — Identify the lens type. Since both the focal length and power are positive, the lens must be convex (converging).

✦ Answer: f ≈ +66.7 cm (2/3 m); the lens is converging (convex).
Where students slip. Concluding 'diverging' from the small-looking power value — the sign of the power (positive here), not its magnitude, is what determines converging vs. diverging.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Science textbook, Reprint 2026-27 (jesc109.pdf) — four in-text question sets (14 questions total, not 18 as some older manifests claim) plus one end-of-chapter Exercise (17 questions, correctly counted). Total internal reflection (critical angle, optical fibres, mirage) has been removed from the current chapter.. Questions are referenced from the NCERT textbook for identification.

All exercises in Light - Reflection and Refraction