NCERT Solutions for Class 10 Science Chapter 9: Light – Reflection and Refraction (NCERT 2026–27)

These Class 10 Science Chapter 9 solutions cover Light – Reflection and Refraction with every in-text “Questions” set and the end-of-chapter Exercises solved step by step. All mirror and lens numericals are worked out using the New Cartesian Sign Convention, the mirror formula 1⁄v + 1⁄u = 1⁄f, the lens formula and magnification — with answers verified and units shown, in line with the NCERT textbook for session 2026–27.

Class: 10 Subject: Science Chapter: 9 Topic: Light – Reflection and Refraction Type: Physics (Ray optics) Session: 2026–27
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Class 10 Science Chapter 9 – Overview

Chapter 9, Light – Reflection and Refraction, studies how light behaves when it bounces off polished surfaces and when it passes from one transparent medium into another, using the straight-line (ray) model of light. You revise the laws of reflection and apply them to spherical mirrors (concave and convex), learning the meaning of pole, centre of curvature, principal axis, principal focus, focal length and the relation R = 2f. The chapter then introduces refraction — the bending of light caused by a change in its speed — the laws of refraction including Snell’s law, and the idea of refractive index. It extends these ideas to spherical lenses (convex and concave) and finally to the power of a lens measured in dioptres. Image formation by mirrors and lenses is analysed with ray diagrams and solved numerically using the mirror formula, the lens formula and magnification.

Key Concepts & Formulae

Laws of reflection: (i) angle of incidence = angle of reflection; (ii) the incident ray, the normal and the reflected ray all lie in the same plane.

Spherical mirror terms: pole (P), centre of curvature (C), radius of curvature (R), principal axis, principal focus (F) and focal length (f).

Laws of refraction: (i) the incident ray, refracted ray and normal lie in the same plane; (ii) Snell’s law — the ratio sin i / sin r is constant for a given pair of media.

Refractive index: nm = (speed of light in air, c) ⁄ (speed of light in the medium, v).

Sign convention (New Cartesian): object is on the left; all distances are measured from the pole/optical centre; distances to the right are +, to the left are −; heights above the axis are +, below are −.

Mirror formula: 1⁄v + 1⁄u = 1⁄f   and   R = 2f

Magnification (mirror): m = h′⁄h = − v⁄u

Lens formula: 1⁄v − 1⁄u = 1⁄f

Magnification (lens): m = h′⁄h = v⁄u

Power of a lens: P = 1⁄f (in metres)   (SI unit: dioptre, D; 1 D = 1 m−1)

In-text “Questions” — Solutions

Questions (Page 135) — after Spherical Mirrors

1. Define the principal focus of a concave mirror.

ANSWERWhen a number of rays parallel to the principal axis fall on a concave mirror, after reflection they all meet at a single point on the principal axis. This point is called the principal focus (F) of the concave mirror. (For a concave mirror this focus is real.)

2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

ANSWER Given R = 20 cm. For a spherical mirror, f = R⁄2 = 20⁄2 = 10 cm.

3. Name a mirror that can give an erect and enlarged image of an object.

ANSWERA concave mirror — when the object is placed between the pole (P) and the principal focus (F), it forms a virtual, erect and enlarged image.

4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

ANSWERA convex mirror is preferred because it always forms an erect and diminished (virtual) image of the traffic behind, and being curved outwards it has a much wider field of view than a plane mirror. This lets the driver see a larger area of the road behind for safe driving.

Questions (Page 145) — after Mirror Formula and Magnification

1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

ANSWER Given R = +32 cm (convex mirror, R is positive). f = R⁄2 = 32⁄2 = +16 cm. The focal length is 16 cm (the positive sign confirms a convex mirror).

2. A 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?

ANSWER Object distance u = −10 cm. Image is real, so magnification m = −3 (real images have negative m). m = − v⁄u ⇒ −3 = − v⁄(−10) ⇒ −3 = v⁄10. ∴ v = −30 cm. The image is located 30 cm in front of the mirror (on the same side as the object); being real and inverted, it forms on a screen there.

Questions (Page 149) — after Refractive Index

1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

ANSWERIt bends towards the normal. Water is optically denser than air, so light slows down on entering water. A ray passing obliquely from a rarer medium (air) to a denser medium (water) always bends towards the normal.

2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m s−1.

ANSWER Refractive index n = c⁄v, so v = c⁄n. v = (3 × 108 m s−1) ⁄ 1.50 = 2 × 108 m s−1.

3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.

ANSWER Highest optical density: Diamond (refractive index = 2.42 — the largest in the table). Lowest optical density: Air (refractive index = 1.0003 — the smallest in the table). The larger the refractive index, the higher the optical density.

4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.

ANSWERRefractive indices: water = 1.33, kerosene = 1.44, turpentine = 1.47. The medium with the lowest refractive index has the highest speed of light. Hence light travels fastest in water (n = 1.33).

5. The refractive index of diamond is 2.42. What is the meaning of this statement?

ANSWERIt means that the ratio of the speed of light in vacuum (or air) to the speed of light in diamond is 2.42. In other words, light travels 2.42 times faster in vacuum than in diamond — so its speed in diamond is c⁄2.42 ≈ 1.24 × 108 m s−1.

Questions (Page 158) — after Power of a Lens

1. Define 1 dioptre of power of a lens.

ANSWER1 dioptre is the power of a lens whose focal length is 1 metre. That is, 1 D = 1 m−1.

2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

ANSWER For a convex lens, a real image equal in size to the object is formed when the object is at 2F; then the image is also at 2F. So image distance v = +50 cm and the object is at the same distance on the other side. m = v⁄u = −1 (real, inverted, same size) ⇒ u = −v = −50 cm. So the needle is placed 50 cm in front of the lens. Lens formula: 1⁄v − 1⁄u = 1⁄f ⇒ 1⁄50 − 1⁄(−50) = 1⁄50 + 1⁄50 = 2⁄50 = 1⁄25. ∴ f = +25 cm = +0.25 m. Power P = 1⁄f (in m) = 1⁄0.25 = +4 D.

3. Find the power of a concave lens of focal length 2 m.

ANSWER For a concave lens, f = −2 m. P = 1⁄f = 1⁄(−2) = −0.5 D. The negative sign confirms a concave (diverging) lens.

End-of-Chapter Exercises — Solutions

1. Which one of the following materials cannot be used to make a lens? (a) Water  (b) Glass  (c) Plastic  (d) Clay

ANSWER(d) Clay. A lens must be made of a transparent material that allows light to pass through and refract. Water, glass and plastic are transparent, but clay is opaque, so it cannot be used to make a lens.

2. 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.

ANSWER(d) Between the pole of the mirror and its principal focus. Only for this position does a concave mirror produce a virtual, erect and enlarged image; for all other positions it forms a real image.

3. 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.

ANSWER(b) At twice the focal length (at 2F1). When the object is at 2F1, a convex lens forms a real, inverted image of the same size at 2F2 on the other side.

4. 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.

ANSWER(a) both concave. By the sign convention, a concave mirror has a negative focal length and a concave lens also has a negative focal length, so f = −15 cm fits a concave mirror and a concave lens.

5. 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.

ANSWER(d) either plane or convex. A plane mirror always gives an erect image, and a convex mirror also always gives an erect (and diminished) image for every object position; a concave mirror would give an inverted image for distant objects.

6. 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.

ANSWER(c) A convex lens of focal length 5 cm. A convex lens acts as a magnifying glass, and the shorter the focal length the greater the magnification. So a convex lens of 5 cm focal length gives a larger magnified image than one of 50 cm. (Concave lenses only diminish.)

7. 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.

ANSWER Range of object distance: the object should be placed between the pole and the principal focus, i.e. at a distance between 0 and 15 cm (less than the focal length) from the mirror. Nature of the image: virtual and erect. Size: the image is larger (enlarged) than the object. Ray diagram (in words): Place the object AB between P and F. From the top of the object draw one ray parallel to the principal axis — after reflection it passes through F. Draw a second ray directed towards the centre of curvature C — it reflects back along the same line. After reflection these two rays diverge, so they do not meet in front of the mirror; produced backwards (behind the mirror) they appear to meet, forming a virtual, erect, enlarged image behind the mirror.

8. 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.

ANSWER (a) Concave mirror. When the bulb is placed at the focus of a concave mirror, the reflected light comes out as a powerful parallel beam, giving a strong forward beam. (b) Convex mirror. It always forms an erect, diminished image and has a wide field of view, so the driver can see a large area of traffic behind the vehicle. (c) Concave mirror. A large concave mirror concentrates parallel rays of sunlight at its focus, producing intense heat for the solar furnace.

9. 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.

ANSWER Yes, the lens will still produce a complete image of the object. Each part of a lens forms a full image, because rays from every point of the object can pass through any portion of the lens and still converge to form the image. Experiment: Cover the upper half (or lower half) of a convex lens with black paper and form the image of a distant object on a screen. A complete image is still obtained, but it is fainter (less bright) because the covered half blocks some of the light, so fewer rays reach the screen and the intensity of the image decreases.

10. 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.

ANSWER Given: h = +5 cm, u = −25 cm, f = +10 cm (converging/convex lens). Lens formula: 1⁄v − 1⁄u = 1⁄f ⇒ 1⁄v = 1⁄f + 1⁄u = 1⁄10 + 1⁄(−25). 1⁄v = 1⁄10 − 1⁄25 = (5 − 2)⁄50 = 3⁄50. ∴ v = 50⁄3 = +16.7 cm (image on the opposite side of the lens). Magnification m = v⁄u = (16.7)⁄(−25) = −0.67. h′ = m × h = (−0.67)(5) = −3.3 cm. Result: the image is formed at 16.7 cm on the other side of the lens; it is real, inverted and diminished (about 3.3 cm tall). Ray diagram (in words): object beyond 2F1; one ray parallel to the axis refracts through F2, another through the optical centre passes straight; they meet between F2 and 2F2, forming a small inverted real image.

11. 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.

ANSWER Given: f = −15 cm (concave lens). A concave lens forms a virtual, erect image on the same side as the object, so v = −10 cm. Lens formula: 1⁄v − 1⁄u = 1⁄f ⇒ 1⁄u = 1⁄v − 1⁄f = 1⁄(−10) − 1⁄(−15). 1⁄u = −1⁄10 + 1⁄15 = (−3 + 2)⁄30 = −1⁄30. ∴ u = −30 cm. The object is placed 30 cm in front of the lens. Ray diagram (in words): a ray parallel to the axis appears, after refraction, to diverge from F1 on the same side; a ray through the optical centre goes straight. Their backward extensions meet at 10 cm, giving a virtual, erect, diminished image between the object and the lens.

12. 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.

ANSWER Given: u = −10 cm, f = +15 cm (convex mirror, f positive). Mirror formula: 1⁄v + 1⁄u = 1⁄f ⇒ 1⁄v = 1⁄f − 1⁄u = 1⁄15 − 1⁄(−10). 1⁄v = 1⁄15 + 1⁄10 = (2 + 3)⁄30 = 5⁄30 = 1⁄6. ∴ v = +6 cm (behind the mirror). m = − v⁄u = −(6)⁄(−10) = +0.6. Result: the image is formed 6 cm behind the mirror; it is virtual, erect and diminished (0.6 times the object size).

13. The magnification produced by a plane mirror is +1. What does this mean?

ANSWER m = +1 means the image is the same size as the object (since |m| = 1, h′ = h). The positive sign shows the image is virtual and erect. So a plane mirror forms a virtual, erect image of exactly the same size as the object — which agrees with everyday experience.

14. 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.

ANSWER Given: h = +5.0 cm, u = −20 cm, R = +30 cm. Focal length f = R⁄2 = +15 cm (convex mirror). Mirror formula: 1⁄v = 1⁄f − 1⁄u = 1⁄15 − 1⁄(−20) = 1⁄15 + 1⁄20. 1⁄v = (4 + 3)⁄60 = 7⁄60 ⇒ v = 60⁄7 = +8.6 cm (behind the mirror). m = − v⁄u = −(8.6)⁄(−20) = +0.43. h′ = m × h = (0.43)(5.0) = +2.2 cm. Result: the image is formed about 8.6 cm behind the mirror; it is virtual, erect and diminished, about 2.2 cm tall.

15. 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.

ANSWER Given: h = +7.0 cm, u = −27 cm, f = −18 cm (concave mirror). Mirror formula: 1⁄v = 1⁄f − 1⁄u = 1⁄(−18) − 1⁄(−27) = −1⁄18 + 1⁄27. 1⁄v = (−3 + 2)⁄54 = −1⁄54 ⇒ v = −54 cm. So the screen should be placed 54 cm in front of the mirror (same side as the object) to get a sharp image. m = − v⁄u = −(−54)⁄(−27) = −2. h′ = m × h = (−2)(7.0) = −14 cm. Result: the image is real, inverted and enlarged, 14 cm tall, formed 54 cm in front of the mirror.

16. Find the focal length of a lens of power −2.0 D. What type of lens is this?

ANSWER P = 1⁄f ⇒ f = 1⁄P = 1⁄(−2.0) = −0.5 m = −50 cm. The negative power and focal length show that this is a concave (diverging) lens.

17. 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?

ANSWER f = 1⁄P = 1⁄(+1.5) = +0.667 m = +66.7 cm (about +0.67 m). The positive power and focal length show that this is a converging (convex) lens.

Extra Practice Questions

Short Answer Type Questions

Q1. State the New Cartesian Sign Convention for the object distance and image distance in a spherical mirror.

ANSWERThe object is always placed to the left of the mirror, and all distances are measured from the pole. Distances measured to the right of the pole (along +x) are positive and those to the left (along −x) are negative. Hence the object distance u is always negative, while a real image (in front) has negative v and a virtual image (behind) has positive v.

Q2. Why does a ray of light bend when it passes from air into glass?

ANSWERLight travels at different speeds in different media. Glass is optically denser than air, so light slows down on entering glass. This change in speed at the air–glass interface causes the ray to change direction and bend towards the normal — the phenomenon of refraction.

Q3. A convex lens has focal length 20 cm. Find its power.

ANSWERf = +20 cm = +0.20 m. P = 1⁄f = 1⁄0.20 = +5 D. The positive sign confirms a converging (convex) lens.

Q4. Why is the emergent ray parallel to the incident ray when light passes through a rectangular glass slab?

ANSWERRefraction occurs at both parallel faces of the slab. At the air–glass face the ray bends towards the normal, and at the glass–air face it bends away from the normal by an equal and opposite amount. Because the two faces are parallel, these bendings cancel out, so the emergent ray is parallel to the incident ray — only laterally shifted sideways.

Q5. The refractive index of glass with respect to air is 1.5. Find the speed of light in glass (c = 3 × 108 m s−1).

ANSWERv = c⁄n = (3 × 108)⁄1.5 = 2 × 108 m s−1.

Long Answer Type Questions

Q1. With the help of a table, describe the nature, position and size of the image formed by a concave mirror for different positions of the object.

ANSWER For a concave mirror, the image depends on where the object lies in relation to P, F and C:
Position of objectPosition of imageSizeNature
At infinityAt the focus FHighly diminished, point-sizedReal and inverted
Beyond CBetween F and CDiminishedReal and inverted
At CAt CSame sizeReal and inverted
Between C and FBeyond CEnlargedReal and inverted
At FAt infinityHighly enlargedReal and inverted
Between P and FBehind the mirrorEnlargedVirtual and erect

Q2. An object 4.0 cm tall is placed 30 cm in front of a concave mirror of focal length 20 cm. Find the position, size and nature of the image.

ANSWER Given: h = +4.0 cm, u = −30 cm, f = −20 cm. 1⁄v = 1⁄f − 1⁄u = −1⁄20 + 1⁄30 = (−3 + 2)⁄60 = −1⁄60. ∴ v = −60 cm (60 cm in front of the mirror). m = − v⁄u = −(−60)⁄(−30) = −2 ⇒ h′ = (−2)(4.0) = −8.0 cm. Result: a real, inverted, enlarged image 8.0 cm tall is formed 60 cm in front of the mirror (object lies between C and F).

Q3. Explain how a convex lens forms images for different object positions, and state one practical use for each main case.

ANSWER When the object is at infinity, the image is highly diminished, real and inverted at F2 — used in a camera/telescope objective to capture distant scenes. Beyond 2F1, the image is real, inverted and diminished (between F2 and 2F2) — as in a camera. At 2F1, the image is real, inverted and the same size at 2F2. Between F1 and 2F1, the image is real, inverted and enlarged beyond 2F2 — used in a projector. At F1, the image is at infinity. When the object is between F1 and the optical centre, the image is virtual, erect and enlarged on the same side — this is how a convex lens acts as a magnifying glass for reading small print. Thus a single convex lens serves cameras, projectors and magnifiers depending on object position.

MCQs & Assertion–Reason

1. The focal length of a spherical mirror of radius of curvature R is:

(a) 2R    (b) R    (c) R⁄2    (d) R⁄4

2. The mirror used as a shaving/make-up mirror to get an enlarged erect image is:

(a) plane    (b) convex    (c) concave    (d) cylindrical

3. A ray of light passing from a rarer to a denser medium bends:

(a) away from the normal    (b) towards the normal    (c) along the normal    (d) does not bend

4. The SI unit of power of a lens is:

(a) metre    (b) dioptre    (c) candela    (d) lux

5. The magnification produced by a plane mirror is:

(a) +1    (b) −1    (c) greater than 1    (d) less than 1

6. For a real image formed by a concave mirror, the magnification m is:

(a) positive    (b) negative    (c) zero    (d) always +1

7. A convex lens of focal length 0.5 m has a power of:

(a) +0.5 D    (b) +1 D    (c) +2 D    (d) +5 D

8. The refractive index of water is 1.33. The speed of light in water is about:

(a) 3.0 × 108 m s−1    (b) 2.25 × 108 m s−1    (c) 1.33 × 108 m s−1    (d) 4.0 × 108 m s−1

9. A concave lens always forms an image that is:

(a) real and inverted    (b) virtual, erect and enlarged    (c) virtual, erect and diminished    (d) real and the same size

10. An object at the centre of curvature C of a concave mirror forms an image that is:

(a) at F, diminished    (b) at C, same size, real and inverted    (c) behind the mirror, virtual    (d) at infinity

Answer key: 1-(c), 2-(c), 3-(b), 4-(b), 5-(a), 6-(b), 7-(c), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A convex mirror is used as a rear-view mirror in vehicles.

Reason: A convex mirror always forms an erect, diminished image and has a wide field of view.

A-R 2. Assertion: A ray of light bends towards the normal when it enters water from air.

Reason: Light travels faster in water than in air.

A-R 3. Assertion: The power of a concave lens is negative.

Reason: The focal length of a concave lens is negative by the sign convention.

A-R 4. Assertion: A concave mirror can form a real, inverted and enlarged image.

Reason: This happens when the object is placed between the principal focus F and the centre of curvature C.

A-R 5. Assertion: When one half of a convex lens is covered with black paper, only half the image is formed.

Reason: Each part of a lens forms a complete image of the object.

Answer key: 1-(A), 2-(C), 3-(A), 4-(A), 5-(D).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Forgetting to apply the sign convention — object distance u is always negative; check the sign of f for the mirror/lens type before substituting.
  • Mixing up the two formulae: mirror uses 1⁄v + 1⁄u = 1⁄f with m = −v⁄u; lens uses 1⁄v − 1⁄u = 1⁄f with m = v⁄u.
  • Confusing real (negative m, inverted) with virtual (positive m, erect) images.
  • Using f when the question gives R — remember f = R⁄2.
  • Writing power without converting focal length to metres first.

How to score full marks in this chapter

Always start a numerical by writing the given data with correct signs, then state the formula, substitute, and finish with the value plus its unit and the nature of the image. Use the magnification sign to state whether the image is real/inverted or virtual/erect. For theory questions, link every answer to the underlying reason — speed change for refraction, R = 2f for mirrors, and field of view for the convex rear-view mirror. Practise the ray-diagram descriptions so you can explain image formation even without drawing.

Frequently Asked Questions

What is Class 10 Science Chapter 9 about?

Chapter 9, Light – Reflection and Refraction, explains the laws of reflection and refraction, image formation by concave and convex mirrors and lenses, the New Cartesian Sign Convention, the mirror and lens formulae, magnification, refractive index and the power of a lens.

What are the main formulae in Chapter 9?

The mirror formula 1/v + 1/u = 1/f with R = 2f, the lens formula 1/v − 1/u = 1/f, magnification m = −v/u for mirrors and m = v/u for lenses, and power P = 1/f (in metres), measured in dioptres.

Why is a convex mirror used as a rear-view mirror?

Because a convex mirror always forms an erect, diminished (virtual) image and gives a much wider field of view than a plane mirror, letting the driver see a large area of traffic behind the vehicle for safe driving.

Are these Class 10 Science Chapter 9 solutions free?

Yes. All solutions are free and follow the official NCERT Science textbook for session 2026–27, with every in-text question and exercise solved step by step.

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