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CBSE · Class 10 · Science

Light — Reflection and Refraction

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The radius of curvature of a concave mirror is 24 cm. Its focal length is:
  1. −24 cm
  2. 12 cm
  3. −13 cm
  4. −12 cm
2.
A concave lens has a focal length of −200 cm. Its power is:
  1. −0.5 D
  2. 0.5 D
  3. −0.01 D
  4. −200 D
3.
Light travels at 3 × 108 m/s in vacuum and 2.4 × 108 m/s in a medium. The refractive index of the medium is:
  1. 1.75
  2. 2.4
  3. 3
  4. 1.25
4.
An object 2 cm tall forms a real, inverted image 10 cm tall. The magnification is:
  1. −5
  2. 5
  3. −0.2
  4. −10
5.
The angle of incidence equals the angle of reflection. This is:
  1. a law of reflection
  2. the lens formula
  3. the mirror formula
  4. Snell's law
6.
An object is placed 56 cm in front of a concave mirror of focal length 24 cm. The image forms at:
  1. −21 cm
  2. −42 cm
  3. 42 cm
  4. −66 cm
7.
An object is placed 42 cm from a convex lens of focal length 24 cm. The image forms at:
  1. 24 cm
  2. −56 cm
  3. 80 cm
  4. 56 cm
8.
Assertion (A): A convex lens of focal length 25 cm has a power of −4 D.
Reason (R): Power of a lens = 1 ÷ focal length in metres, and a convex lens has a positive focal length.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
An object is 60 cm in front of a concave mirror of focal length 10 cm. Find the position and nature of the image.
10.
An object is 18 cm in front of a convex mirror of focal length 36 cm. Find the position and nature of the image.
11.
An object is 42 cm from a convex lens of focal length 36 cm. Find the image distance and the magnification.
12.
An object is 10 cm from a concave lens of focal length 40 cm. Find the position and nature of the image.
13.
An object 6 cm tall is placed 10 cm from a convex lens of focal length 5 cm. Find the image distance and the height of the image.
14.
Two thin lenses of power +10 D and −2.5 D are placed in contact. Find the combined power and focal length. (Give the answer to 2 decimal places where needed.)
15.
The refractive index of medium A is 1.25 and of medium B is 2. Find the refractive index of B with respect to A. (Give the answer to 2 decimal places where needed.)
16.
Where should an object be placed in front of a convex lens of focal length 9 cm to get a real image the same size as the object?
Section C — Word problems4 × 4 = 16 marks
17.
A 8 cm tall candle stands 48 cm in front of a concave mirror of focal length 40 cm. Where should a screen be placed to get a sharp image? Describe the image and find its height.
18.
A car's rear-view convex mirror has a focal length of 12 cm. A scooter is 36 cm behind it. Find the position of the image and its magnification.
19.
A projector lens forms a sharp image on a screen 90 cm away from a slide placed 45 cm from it. Find the focal length of the lens and the magnification.
20.
An optician prescribes a lens of power +2.5 D. Find its focal length, and say whether it is a converging or a diverging lens. (Give the answer to 2 decimal places where needed.)
AmbifulWorksheets
Practice that never repeats
CBSE · Class 10 · Science

Answer key

Light — Reflection and Refraction · AW-10S-LIG-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) −12 cmf = R/2, negative for a concave mirror
2(a) −0.5 DP = 1 ÷ f (in m) = 100 ÷ −200
3(d) 1.25n = c ÷ v = 3 ÷ 2.4
4(a) −5m = image height ÷ object height = −10 ÷ 2; negative for a real, inverted image
5(a) a law of reflection—
6(b) −42 cm1/v = 1/f − 1/u = 1/(−24) − 1/(−56), so v = −42 cm
7(d) 56 cm1/v = 1/f + 1/u = 1/24 + 1/(−42), so v = 56 cm
8(d) A is false, but R is true.A is false: P = 100 ÷ 25 = +4 D, not −4 D. R is true.
9v = −12 cm (in front of the mirror); real, inverted and diminished1/v = 1/f − 1/u gives v = −12 cm; m = −v/u = −0.2
10v = 12 cm (behind the mirror); virtual, erect and diminished1/v = 1/f − 1/u gives v = +12 cm; m = 0.67
11v = 252 cm; m = −61/v = 1/f + 1/u gives v = 252 cm; m = v/u = 252 ÷ (−42)
12v = −8 cm; virtual, erect and diminished1/v = 1/f + 1/u gives v = −8 cm (same side as the object); m = 0.8
13v = 10 cm; height −6 cm (inverted)v = 10 cm; m = v/u = −1; h′ = m × 6
14+7.5 D; 13.33 cmP = P1 + P2 = +7.5 D; f = 100 ÷ P cm
151.6n_BA = n_B ÷ n_A = 2 ÷ 1.25
1618 cm from the lens (u = −18 cm)m = v/u = −1, so v = −u; 1/v − 1/u = 1/9 gives u = −18 cm, v = 18 cm
17240 cm in front of the mirror; real, inverted and magnified; height 40 cmv = −240 cm; m = −5; h′ = m × 8
189 cm behind the mirror; m = 0.25 (virtual, erect, diminished)1/v = 1/f − 1/u = 1/12 − 1/(−36); m = −v/u
1930 cm; m = −21/f = 1/v − 1/u = 1/90 − 1/(−45); m = v/u
20+40 cm; converging (convex)f = 100 ÷ P cm; positive power means a convex lens