Class 10 Science NCERT Solutions Chapter 9: Light — Reflection and Refraction | Boundless Maths
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Chapter 9: Light — Reflection
and Refraction

Class 10 Science NCERT Solutions Chapter 9 (CBSE 2026-27) — every in-text question, every embedded Activity question, and all 17 back-of-chapter Exercises on this one page, solved with full working.

Light — Reflection and Refraction opens the physics portion of Class 10 Science and moves from the reflection of light by concave and convex spherical mirrors — the New Cartesian sign convention, the mirror formula, and magnification — to refraction of light at a boundary between two media, refractive index, and finally refraction by convex and concave spherical lenses, the lens formula, and the power of a lens used in eyeglass prescriptions. This is one of the most numerical-heavy chapters in Class 10 Science, and these NCERT solutions work through every Activity-based conceptual question and the full 17-question Exercise set with complete step-by-step working for every calculation, using the sign convention throughout.

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In-text & Exercise Coverage
Overview

What Chapter 9 Is Really About

Light — Reflection and Refraction is the first physics chapter of Class 10 Science, and it builds two parallel toolkits — one for mirrors, one for lenses — that share the same underlying logic. For spherical mirrors, it works through the pole, centre of curvature, principal focus and focal length, how concave and convex mirrors form different images depending on where the object sits, and how the New Cartesian sign convention turns all of this into one reliable formula (1/v + 1/u = 1/f) and a magnification rule. It then turns to refraction — why a pencil looks bent in water, why a pool looks shallower than it is — and refractive index, before applying the exact same logic to spherical lenses: convex and concave, their own sign convention, the lens formula (1/v − 1/u = 1/f), and finally the power of a lens, the number opticians actually write on a prescription. Every question woven through the chapter is solved here, section by section, in the order the textbook presents it: the four "QUESTIONS" boxes spread across the chapter, the questions embedded directly inside Activities, and the full 17-question Exercise set at the end, with complete numerical working shown for every calculation.

🪞

Spherical Mirrors

Concave and convex mirrors, ray diagrams, the mirror formula, magnification, and the New Cartesian sign convention.

🌊

Refraction & Refractive Index

Why light bends at a boundary between media, Snell's law, and how refractive index relates to the speed of light.

🔍

Spherical Lenses & Power

Convex and concave lenses, the lens formula, magnification, and the power of a lens in dioptres.

Section A

In-text Questions — Spherical Mirrors (Page 142)

4 Questions
9.2 · Q1Define the principal focus of a concave mirror.▾

The principal focus of a concave mirror is the point on its principal axis where a beam of light rays travelling parallel to the principal axis converges (meets) after reflection from the mirror.

9.2 · Q2The radius of curvature of a spherical mirror is 20 cm. What is its focal length?▾
f = R/2 = 20/2 = 10 cm
9.2 · Q3Name a mirror that can give an erect and enlarged image of an object.▾

A concave mirror — when the object is placed between the pole (P) and the principal focus (F).

9.2 · Q4Why do we prefer a convex mirror as a rear-view mirror in vehicles?▾

A convex mirror always gives an erect (though diminished) image, and it has a wider field of view than a plane mirror of the same size, because of its outward curvature. This lets the driver see a much larger area of traffic behind the vehicle, aiding safe driving.

Section B

In-text Questions — Mirror Formula (Page 144–145)

2 Questions
9.2.4 · Q1Find the focal length of a convex mirror whose radius of curvature is 32 cm.▾
f = R/2 = 32/2 = +16 cm (positive, since it's a convex mirror)
9.2.4 · Q2A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?▾

Since the image is real, magnification m = −3 (negative sign for real image), and u = −10 cm.

m = −v/u
−3 = −v/(−10)
v = −30 cm

The image is located 30 cm in front of the mirror (on the same side as the object), since it is a real image.

Section C

In-text Questions — Refraction (Page 149)

5 Questions
9.3.2 · Q1A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?▾

The ray bends towards the normal. Water is optically denser than air (has a higher refractive index), and when light travels from a rarer medium (air) to a denser medium (water), it slows down and bends towards the normal.

9.3.2 · Q2Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass?▾
n = c/v ⟹ v = c/n = (3×108)/1.50 = 2×108 m/s
9.3.2 · Q3Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.▾

Highest optical density: Diamond (refractive index 2.42). Lowest optical density: Air (refractive index 1.0003).

9.3.2 · Q4You are given kerosene, turpentine and water. In which of these does the light travel fastest?▾

Refractive indices: kerosene = 1.44, turpentine oil = 1.47, water = 1.33. Since speed of light (v = c/n) is inversely related to refractive index, light travels fastest in the medium with the lowest refractive index — water (n = 1.33).

9.3.2 · Q5The refractive index of diamond is 2.42. What is the meaning of this statement?▾

This means the speed of light in air (or vacuum) is 2.42 times the speed of light in diamond — light travels 2.42 times faster in air/vacuum than it does inside diamond.

Section D

In-text Questions — Power of a Lens (Page 158)

3 Questions
9.3.8 · Q1Define 1 dioptre of power of a lens.▾

One dioptre (1 D) is the power of a lens whose focal length is 1 metre. 1 D = 1 m−1.

9.3.8 · Q2A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it, equal to the size of the object. Where is the needle placed? Also find the power of the lens.▾

Since the image is real, inverted, and the same size as the object, m = −1 — this happens specifically when the object is at 2F1 (twice the focal length), giving v = 2f = 50 cm.

f = 50/2 = 25 cm = 0.25 m
Object distance u = −2f = −50 cm (needle placed 50 cm in front of the lens, at 2F1)
Power P = 1/f = 1/0.25 = +4 D
9.3.8 · Q3Find the power of a concave lens of focal length 2 m.▾

Since it's a concave lens, f = −2 m (by sign convention).

P = 1/f = 1/(−2) = −0.5 D
Section E

Embedded Activity Questions

6 Questions
Activity 9.1Viewing your face in a shining spoon's curved surfaces — compare the images from the concave (inner) and convex (outer) sides.▾

Looking into the concave (inward-curving) side of the spoon at close range shows a magnified, erect image of your face. Moving the spoon farther away, the image grows larger up to a point, then becomes inverted and diminished beyond the focal point. Looking into the convex (outward-curving) side always shows a diminished, erect image, regardless of distance. This is because the inward-curving surface behaves like a concave mirror (which can give both magnified virtual images up close and real, inverted images from farther away), while the outward-curving surface behaves like a convex mirror (always virtual, erect, diminished).

Activity 9.5Observing a pencil's image in a convex mirror — is it erect or inverted, diminished or enlarged, and what happens as the pencil moves away?▾

The image of the pencil in a convex mirror is always erect and diminished, regardless of the pencil's distance from the mirror. As the pencil moves farther away, the image becomes progressively smaller and moves closer to the focus (behind the mirror).

Activity 9.6Comparing full-length images of a distant tree in plane, concave, and convex mirrors — which mirror succeeds, and why?▾

A plane mirror or concave mirror generally cannot show the full-length image of a tall/large nearby object within a small mirror. A convex mirror, however, can show the full-length image of even a large object, because its outward curvature gives it a much wider field of view, letting it fit the whole object into its diminished image — which is why a small convex mirror (like the one at Agra Fort) can capture the full image of a large structure like the Taj Mahal.

Page 145–146Does a pencil appear displaced to the same extent with kerosene or turpentine as with water? Will letters rise to the same height with a plastic slab as with glass?▾

No — the extent of the displacement (or apparent raising) effect differs depending on the specific pair of media involved, because each medium has a different refractive index, and the amount of bending of light at the interface depends on this refractive index.

Activity 9.9Does a line under a glass slab appear bent at the edges when the slab is placed at an angle to it? What about when the slab is normal to the line?▾

When the slab is placed at an angle (obliquely) to the line, the part of the line under the slab appears bent/displaced at the edges, due to refraction (bending towards the normal entering glass, away from the normal exiting it). When the slab is placed normal (perpendicular) to the line, the line does not appear bent — light travelling along the normal (zero angle of incidence) passes straight through without bending, even though it still changes speed.

Page 147What happens when a light ray is incident normally on the interface of two media?▾

It passes straight through into the second medium without any bending or change in direction — the ray continues along the same straight-line path, although its speed changes as it crosses into the new medium.

Section F

End-of-Chapter Exercises (Q1–Q17)

17 Questions
Q1Which one of the following materials cannot be used to make a lens? (a) Water (b) Glass (c) Plastic (d) Clay▾

Answer: (d) Clay — clay is opaque, not transparent, so light cannot pass through it; a lens must be made of a transparent material.

Q2The image formed by a concave mirror is virtual, erect and larger than the object. Where should the object be? (a) Between F and C (b) At C (c) Beyond C (d) Between P and F▾

Answer: (d) Between the pole of the mirror and its principal focus.

Q3Where should an object be placed in front of a convex lens to get a real image the size of the object? (a) At principal focus (b) At twice focal length (c) At infinity (d) Between optical centre and principal focus▾

Answer: (b) At twice the focal length (2F1).

Q4A spherical mirror and a thin spherical lens each have focal length −15 cm. The mirror and lens are likely to be (a) both concave (b) both convex (c) mirror concave, lens convex (d) mirror convex, lens concave▾

Answer: (a) both concave — a negative focal length by sign convention corresponds to a concave mirror and a concave lens alike.

Q5No matter how far you stand from a mirror, your image appears erect. The mirror is likely (a) only plane (b) only concave (c) only convex (d) either plane or convex▾

Answer: (d) either plane or convex — both always give erect images regardless of object distance; a concave mirror only gives an erect image when the object is between the pole and focus.

Q6Which lens would you prefer for reading small letters in a dictionary? (a) Convex, f=50cm (b) Concave, f=50cm (c) Convex, f=5cm (d) Concave, f=5cm▾

Answer: (c) A convex lens of focal length 5 cm — a convex (magnifying) lens with a short focal length gives greater magnification when the object is placed within the focal length.

Q7We wish to obtain an erect image using a concave mirror of focal length 15 cm. Range of object distance? Nature of image, and is it larger or smaller? Ray diagram.▾

For a concave mirror to produce an erect image, the object must be placed between the pole (P) and the focus (F) — object distance between 0 and 15 cm in front of the mirror.

Nature of image: virtual, erect, and enlarged (larger than the object).

Ray diagram:

Ray diagram: concave mirror, object between P and F Principal axis P F C Object Image Solid = real ray path · dashed = extended backward to locate the image

Both reflected rays diverge in front of the mirror; extending them backward (dashed) locates the virtual, erect, enlarged image behind the mirror.

Q8Name the type of mirror used in (a) headlights of a car (b) side/rear-view mirror (c) solar furnace. Support your answer with reason.▾

(a) Headlights: Concave mirror — placing the bulb at the principal focus produces a powerful, parallel beam of light for illuminating the road.

(b) Side/rear-view mirror: Convex mirror — always gives an erect, diminished image with a wider field of view, letting the driver see more of the traffic behind.

(c) Solar furnace: Concave mirror (large) — converges parallel rays of sunlight to a sharp focus, producing intense heat at that point.

Q9One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Explain.▾

Yes, the lens will still form a complete image, though dimmer than before. Every unobstructed part of the lens still receives light rays from every point of the object and refracts them towards the image — covering half the lens only blocks half the rays reaching each image point, reducing the image's brightness/intensity, but the remaining half of the lens can still converge rays from every point of the object to form the complete image.

Q10An object 5 cm in length is held 25 cm from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image.▾

Given: h = +5 cm, u = −25 cm, f = +10 cm (convex lens)

1/v − 1/u = 1/f
1/v = 1/f + 1/u = 1/10 + 1/(−25) = 5/50 − 2/50 = 3/50
v = +16.67 cm ≈ +16.7 cm
m = v/u = 16.67/(−25) = −0.667
h′ = m × h = −0.667 × 5 = −3.33 cm ≈ −3.3 cm

The image is real, inverted, diminished, formed about 16.7 cm on the other side of the lens, with a height of about 3.3 cm.

Ray diagram:

Ray diagram: convex lens, object beyond 2F1 Principal axis O F1 2F1 F2 2F2 Object Image Teal ray: parallel to axis, refracts through F2 Brown ray: through the optical centre O, undeviated

Both real rays converge on the far side of the lens, giving a real image that can be captured on a screen.

Q11A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed? Draw the ray diagram.▾

Given: f = −15 cm, v = −10 cm (concave lens, virtual image on same side as object)

1/v − 1/u = 1/f
1/u = 1/v − 1/f = 1/(−10) − 1/(−15) = −3/30 + 2/30 = −1/30
u = −30 cm

The object is placed 30 cm from the lens.

Ray diagram:

Ray diagram: concave lens Principal axis O F1 F2 Object Image Teal ray: parallel to axis, appears to diverge from F1 Brown ray: through the optical centre O, undeviated

A concave lens always diverges rays — the image is located by extending the real, diverging rays backward until they meet, on the object's own side of the lens.

Q12An 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.▾

Given: u = −10 cm, f = +15 cm (convex mirror)

1/v + 1/u = 1/f
1/v = 1/f − 1/u = 1/15 − 1/(−10) = 2/30 + 3/30 = 5/30 = 1/6
v = +6 cm

Since v is positive, the image is virtual, formed 6 cm behind the mirror.

m = −v/u = −(6)/(−10) = +0.6

Since m is positive and less than 1: the image is virtual, erect, and diminished.

Q13The magnification produced by a plane mirror is +1. What does this mean?▾

The positive sign means the image is virtual and erect, and the magnitude of 1 means the image is exactly the same size as the object — consistent with the known properties of a plane mirror's image (always virtual, erect, same size).

Q14An object 5.0 cm in length is placed 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.▾

Given: h = +5.0 cm, u = −20 cm, R = +30 cm, so f = R/2 = +15 cm

1/v + 1/u = 1/f
1/v = 1/f − 1/u = 1/15 − 1/(−20) = 4/60 + 3/60 = 7/60
v = 60/7 ≈ +8.57 cm
m = −v/u = −(8.57)/(−20) ≈ +0.43
h′ = m × h ≈ 0.43 × 5.0 ≈ +2.14 cm

The image is formed about 8.6 cm behind the mirror, is virtual, erect, diminished, with a height of about 2.1 cm.

Q15An 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 should a screen be placed to get a sharp image? Find its size and nature.▾

Given: h = +7.0 cm, u = −27 cm, f = −18 cm (concave mirror)

1/v + 1/u = 1/f
1/v = 1/f − 1/u = 1/(−18) − 1/(−27) = −3/54 + 2/54 = −1/54
v = −54 cm

The screen should be placed 54 cm in front of the mirror.

m = −v/u = −(−54)/(−27) = −2
h′ = m × h = −2 × 7.0 = −14.0 cm

The image is real, inverted, and enlarged (twice the object size), with a height of 14.0 cm.

Q16Find the focal length of a lens of power −2.0 D. What type of lens is this?▾
f = 1/P = 1/(−2.0) = −0.5 m = −50 cm

Since the focal length is negative, this is a concave (diverging) lens.

Q17A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is it diverging or converging?▾
f = 1/P = 1/(+1.5) ≈ +0.67 m (≈ +66.7 cm)

Since the power (and focal length) is positive, this is a convex (converging) lens.

Common Questions

Frequently Asked Questions

The mirror formula, 1/v + 1/u = 1/f, relates the image distance (v), object distance (u), and focal length (f) of a spherical mirror. Under the New Cartesian sign convention, all distances are measured from the pole, distances measured in the direction of incident light (to the right) are positive, and distances measured against it (to the left) are negative — so a concave mirror has negative focal length and a convex mirror has positive focal length.
The power of a lens is the reciprocal of its focal length (in metres), measured in dioptres (D). A convex (converging) lens has positive power, and a concave (diverging) lens has negative power. One dioptre is the power of a lens with a focal length of exactly 1 metre.
Yes — numerical problems using the mirror formula and lens formula, ray diagrams for image formation, and questions on refractive index and power of a lens are among the most heavily tested topics in the physics portion of Class 10 Science, appearing almost every year in the CBSE board exam.
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