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Class 10 · Science · Chapter 9

Light - Reflection and Refraction: Class 10 Important Questions with Answers

39 important questions for Class 10 Science Chapter 9, Light - Reflection and Refraction, the way they come in the board exam: MCQs, assertion–reason, case-based and short and long answer questions, with answers. Taken from our Class 10 Science notes for 2026-27.

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Jump to: Solving mirror numericals — 4 stepsCase-based questions20 MCQsAssertion–ReasonGive reasons

Solving mirror numericals — 4 steps

1. Write given values with signs➜2. Put in 1/v = 1/f − 1/u➜3. Find m = −v/u, h′ = m h➜4. State position, nature, size

Example 9.1 (NCERT) — convex rear-view mirror

R = 3.00 m; a bus is 5.00 m away. Find position, nature and size of image.

R = +3.00 m → f = +1.50 m; u = −5.00 m
1/v = 1/f − 1/u = 1/1.50 + 1/5.00 = (5.00 + 1.50)/7.50 = 6.50/7.50 → v = +1.15 m
m = −v/u = −1.15/(−5.00) = +0.23

Image 1.15 m behind the mirror; virtual, erect, 0.23 times the size of the bus.

Example 9.2 (NCERT) — concave mirror

Object 4.0 cm high at 25.0 cm from a concave mirror of focal length 15.0 cm. Where should the screen be? Nature and size?

h = +4.0 cm, u = −25.0 cm, f = −15.0 cm
1/v = 1/f − 1/u = −1/15 + 1/25 = (−5 + 3)/75 = −2/75 → v = −37.5 cm
h′ = −(v/u) h = −(−37.5/−25.0)(4.0) = −6.0 cm

Screen 37.5 cm in front of the mirror; image real, inverted, enlarged (6.0 cm).

NCERT p. 145 Q1: convex mirror, R = 32 cm → f = R/2 = +16 cm.
NCERT p. 145 Q2: concave mirror, real image 3 times enlarged, u = −10 cm. Real → m = −3. m = −v/u → −3 = −v/(−10) → v = −30 cm (30 cm in front of the mirror).

Board-type numericals (solved)

(a) CBSE 2025: object 10 cm in front of a concave mirror, f = 15 cm. → u = −10, f = −15; 1/v = −1/15 + 1/10 = (−2 + 3)/30 = 1/30 → v = +30 cm (behind the mirror); m = −v/u = +3 → virtual, erect, 3× enlarged.

(b) CBSE 2024: object 10 cm from a convex mirror, f = 15 cm. → u = −10, f = +15; 1/v = 1/15 + 1/10 = 5/30 → v = +6 cm behind; m = +0.6 → virtual, erect, diminished.

(c) CBSE 2024: a ray parallel to the axis, after reflection, crosses the axis 10 cm from the pole → f = 10 cm → R = 2f = 20 cm.

Unit check: keep everything in cm (or m) — do not mix. Write the final answer with sign and words (“30 cm behind the mirror, virtual and erect”); the sign alone does not get full marks.

Case-based questions (4 marks each)

Case 1 · The dentist's mirror. A dentist holds a small concave mirror of focal length 4 cm at 2 cm from a tooth to see an enlarged view.

(i) Why is a concave mirror used? (1) (ii) Find the image position and magnification. (2) (iii) Will the image form on a screen? Why? (1)

(i) Object within F gives an enlarged, erect image. (ii) u = −2, f = −4: 1/v = −1/4 + 1/2 = 1/4 → v = +4 cm (behind the mirror); m = −v/u = +2. (iii) No — it is virtual; the rays only appear to meet behind the mirror.
Case 2 · The projector. A school projector uses a convex lens of power +5 D to throw an image of a slide on a screen 1.0 m away from the lens.

(i) Find the focal length. (1) (ii) Where should the slide be placed? (2) (iii) Why is the slide put upside down? (1)

Full answer in the notes pack
Case 3 · The swimming pool. A child standing at the edge of a pool feels the pool is only 1 m deep, but it is actually deeper. Light from the bottom travels from water (n = 1.33) to air.

(i) Name the phenomenon. (1) (ii) Does light bend towards or away from the normal on leaving water? (1) (iii) Find the speed of light in water and explain why the pool looks shallower. (2)

(i) Refraction. (ii) Away from the normal (denser → rarer). (iii) v = 3 × 108/1.33 ≈ 2.26 × 108 m/s. The rays bend away from the normal at the surface; traced back, they seem to come from a point higher than the bottom, so the bottom appears raised.
Case 4 · The optician. An optician puts a +2.0 D lens and a −0.5 D lens together in the trial frame.

(i) Net power? (1) (ii) Nature and focal length of the combination. (1) (iii) An object is 1 m from this combination; where is the image? Is it real? (2)

Full answer in the notes pack

20 MCQs (1 mark each)

  1. Image in a plane mirror is
    (a) real, inverted (b) virtual, erect, laterally inverted (c) real, erect (d) virtual, inverted
  2. For a spherical mirror, R and f are related as
    (a) R = f (b) R = f/2 (c) R = 2f (d) R = 4f
  3. Focal length of a concave mirror by sign convention is
    (a) positive (b) negative (c) zero (d) infinite
  4. An object at C of a concave mirror forms an image
    (a) at F (b) at C, same size (c) behind (d) at infinity
  5. Shaving mirrors are
    (a) plane (b) convex (c) concave (d) cylindrical
  6. Image by a convex mirror is always
    (a) real (b) enlarged (c) virtual, erect, diminished (d) inverted
  7. If m = −2, the image is
    (a) virtual, enlarged (b) real, inverted, enlarged (c) real, diminished (d) virtual, diminished
  8. A ray through the centre of curvature of a concave mirror
    (a) goes through F (b) returns along the same path (c) becomes parallel (d) is absorbed
  9. Light bends towards the normal when going from
    (a) glass to air (b) water to air (c) air to glass (d) glass to water
  10. The emergent ray from a glass slab is
    (a) perpendicular (b) parallel to the incident ray (c) along the normal (d) reflected
  11. Refractive index of a medium is
    (a) v/c (b) c/v (c) c × v (d) c + v
  12. Light travels slowest in
    (a) water (b) glass (c) diamond (d) air
  13. A ray through the optical centre of a lens
    (a) bends towards F (b) goes undeviated (c) becomes parallel (d) reflects
  14. To get a same-size real image, the object is put at ___ of a convex lens
    (a) F (b) 2F (c) between F and O (d) infinity
  15. Magnifying glass is a
    (a) concave lens (b) convex lens with object inside F (c) convex mirror (d) plane mirror
  16. Power of a lens of focal length 50 cm (convex):
    (a) +0.5 D (b) +2 D (c) −2 D (d) +50 D
  17. SI unit of power of a lens:
    (a) metre (b) watt (c) dioptre (d) joule
  18. Lenses of +3 D and −1 D in contact give
    (a) +4 D (b) +2 D (c) −2 D (d) −3 D
  19. Lens formula is
    (a) 1/v + 1/u = 1/f (b) 1/v − 1/u = 1/f (c) v − u = f (d) m = −v/u
  20. A concave lens with f = −20 cm has power
    (a) +5 D (b) −5 D (c) −0.05 D (d) +0.2 D

Answer key

1 b
2 c
3 b
4 b
5 c
6 c
7 b
8 b
9 c
10 b
11 b
12 c
13 b
14 b
15 b
16 b
17 c
18 b
19 b
20 b

Why: 16 → 1/0.5 m = 2 D · 20 → 1/(−0.2 m) = −5 D · 12 → diamond has the largest n.

Assertion–Reason (10)

(a) Both A and R true, R explains A · (b) Both true, R does not explain A · (c) A true, R false · (d) A false, R true

  1. A: Convex mirrors are used as rear-view mirrors. R: They form erect images and have a wide field of view. → (a)
  2. A: A concave mirror can form a virtual image. R: The object is placed between P and F. → (a)
  3. A: The focal length of a convex mirror is negative. R: Its focus is behind the mirror. → (d)
  4. A: A pencil in water looks bent. R: Light changes speed when it goes from water to air. → (a)
  5. A: Kerosene is optically denser than water. R: Kerosene has a lower mass density than water. → (b)
  6. A: A ray falling normally on a glass slab does not bend. R: The angle of incidence is zero. → (a)
  7. A: A concave lens always forms a diminished image. R: A concave lens is a converging lens. → (c)
  8. A: Covering half a convex lens makes the image dimmer. R: Fewer rays pass through the lens. → (a)
  9. A: A lens of shorter focal length has more power. R: P = 1/f. → (a)
  10. A: The magnification of a real image is positive. R: Real images are inverted. → (d)

Give reasons (2 marks each)

1. Concave mirrors are used in car headlights.

With the bulb at the focus, reflected rays go out as a strong parallel beam that lights the road far ahead. 2

2. The bottom of a pool looks raised.

Light from the bottom bends away from the normal on entering air; the eye traces the rays back to a higher point. 2

3. A diamond sparkles more than glass.

Its refractive index (2.42) is very high, so light slows and bends a lot inside and is reflected many times before coming out. 2

4. A convex lens is called a converging lens.

It bends parallel rays inward so they meet at the principal focus. 2

5. A real image can be taken on a screen but a virtual one cannot.

For a real image the rays actually meet; for a virtual image they only appear to meet, so no light reaches a screen there. 2

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