ICSE – Grade 10 – Physics – Ch 04 – QA

1. A ray of light travels from air into glass. Speed of light in air = 3 × 10⁸ m/s, in glass = 2 × 10⁸ m/s. Find refractive index of glass.
   Solution:
   Refractive index (n) = Speed in air / Speed in glass = (3 × 10⁸) / (2 × 10⁸) = 1.5

2. Calculate the speed of light in a diamond. Refractive index of diamond = 2.42; speed in air = 3 × 10⁸ m/s.
   Solution:
   Speed in diamond = Speed in air / Refractive index = (3 × 10⁸) / 2.42 ≈ 1.24 × 10⁸ m/s

3. A light ray passes from air (n = 1.00) into water (n = 1.33) with angle of incidence = 30°. Find angle of refraction.
   Solution:
   Using Snell’s Law: n₁ sin i = n₂ sin r
    1.00 × sin(30°) = 1.33 × sin r
    0.5 = 1.33 × sin r → sin r = 0.5 / 1.33 ≈ 0.3759
    Angle r = sin⁻¹(0.3759) ≈ 22.09°

4. Real depth of a pool is 2.5 m. Refractive index of water is 1.33. Calculate the apparent depth.
   Solution:
   Apparent depth = Real depth / Refractive index = 2.5 / 1.33 ≈ 1.88 m

5. A coin appears raised by 4 cm when in water. Refractive index of water = 1.33. Calculate real depth.
   Solution:
   Real depth = Apparent depth × Refractive index = 4 × 1.33 = 5.32 cm

6. A beam travels from glass to air. Critical angle = 42°. What is refractive index of glass?
   Solution:
   sin C = 1/n → sin 42° = 1/n → 0.6691 = 1/n → n = 1 / 0.6691 = 1.495

7. A light ray enters a glass slab at 45° and refracts at 28°. Find refractive index.
   Solution:
   n = sin i / sin r = sin 45° / sin 28° = 0.7071 / 0.4695 ≈ 1.506

8. In a glass prism, angle of prism = 60° and angle of minimum deviation = 38°. Find refractive index.
   Solution:
   n = sin[(A + D) / 2] / sin(A / 2) = sin(49°) / sin(30°) = 0.7547 / 0.5 = 1.509

9. A ray travels from a medium of refractive index 1.5 into air. Calculate critical angle.
   Solution:
   sin C = 1 / n = 1 / 1.5 = 0.6667 → C = sin⁻¹(0.6667) = 41.8°

10. Light passes from air to a medium and bends towards the normal. What can be inferred about the speed in the medium if n = 1.6?
    Solution:
    Speed in medium = c / n = (3 × 10⁸) / 1.6 = 1.875 × 10⁸ m/s

11. A light ray enters a prism with angle of incidence = 40° and emerges at 60°. If angle of prism = 60°, calculate angle of deviation.
    Solution:
    Angle of deviation = i + e – A = 40° + 60° – 60° = 40°

12. Refractive index of water = 1.33. Find the speed of light in water.
    Solution:
    Speed = 3 × 10⁸ / 1.33 ≈ 2.26 × 10⁸ m/s

13. A ray is incident at 90° to a glass slab. What will be the angle of refraction?
    Solution:
    Normal incidence implies angle = 0°. Thus, angle of refraction = 0°

14. Refractive index = 1.5. Find sin of critical angle.
    Solution:
    sin C = 1/n = 1/1.5 = 0.6667

15. Apparent depth = 15 cm. Refractive index = 1.5. Find real depth.
    Solution:
    Real depth = Apparent depth × Refractive index = 15 × 1.5 = 22.5 cm

16. A light ray travels from water (n = 1.33) to glass (n = 1.5) with angle of incidence = 30°. Find angle of refraction.
    Solution:
    1.33 × sin 30° = 1.5 × sin r
     0.665 = 1.5 × sin r → sin r = 0.4433 → r = 26.3°

17. A ray of light from glass to water strikes at angle 60°. Critical angle is 48°. Will Total Internal Reflection occur?
    Solution:
    Yes, since 60° > 48°. Yes, Total Internal Reflection occurs.

18. In a prism, angle of prism = 60°. Minimum deviation = 30°. Calculate refractive index.
    Solution:
    n = sin[(A + D)/2] / sin(A/2) = sin(45°)/sin(30°) = 0.7071/0.5 = 1.414

19. A prism deviates light by 40°. Angle of incidence = 50°, prism angle = 60°. Find angle of emergence.
    Solution:
    Deviation = i + e – A → 40 = 50 + e – 60 → e = 50°

Answer: 50°

20. A ray in water bends away from the normal on entering air. Why?
    Solution:
    Because speed of light increases and refractive index decreases. Light travels faster in air than in water.

21. A light ray in air strikes a prism at 45° and emerges at 65°. If the prism angle is 60°, calculate the deviation.
    Solution:
    Deviation = i + e – A = 45 + 65 – 60 = 50°

22. A coin appears 20 cm deep in a tank. If refractive index of water is 1.33, find actual depth.
    Solution:
    Real depth = 20 × 1.33 = 26.6 cm

23. A medium has speed of light = 2.25 × 10⁸ m/s. What is its refractive index?
    Solution:
    n = 3 × 10⁸ / 2.25 × 10⁸ = 1.33

24. Calculate the speed of light in a medium of refractive index 1.8.
    Solution:
    Speed = 3 × 10⁸ / 1.8 = 1.67 × 10⁸ m/s

25. A prism has a refractive index of 1.5 and angle of prism is 60°. Find minimum angle of deviation.
    Solution:
    Using n = sin[(A + D)/2] / sin(A/2), solve for D:
     1.5 = sin[(60 + D)/2] / sin 30 → sin[(60 + D)/2] = 0.75 → (60 + D)/2 = 48.6° → D = 37.2°

26. If angle of incidence in water is 50° and angle of refraction in air is 70°, find refractive index of water.
    Solution:
    n = sin i / sin r = sin 50° / sin 70° = 0.766 / 0.940 = 0.815

27. A diver sees the sun at an apparent angle of 30° from the vertical. Refractive index of water is 1.33. Find the real angle.
    Solution:
    sin i = n × sin r = 1.33 × sin 30° = 1.33 × 0.5 = 0.665
     i = sin⁻¹(0.665) = 41.8°

28. A glass block is 6 cm thick. Lateral shift observed is 2 cm. Find angle of refraction if angle of incidence is 60°.
    Solution:
    Using formula for lateral shift: Shift = t sin(i – r) / cos r
     Given: shift = 2, t = 6, i = 60°
     Trial method gives r ≈ 35° (approximate by substitution)

29. Light takes 1.5 ns to pass through a medium of 45 cm thickness. Find speed of light in medium.
    Solution:
    Speed = distance / time = 0.45 / (1.5 × 10⁻⁹) = 3 × 10⁸ m/s

30. A ray travels from medium A to B with i = 50°, r = 30°. Find n₂/n₁.
    Solution:
    n₂/n₁ = sin i / sin r = sin 50° / sin 30° = 0.766 / 0.5 = 1.532

31. Refractive index of a material is 1.6. What is the angle of incidence for which angle of refraction is 30°?
    Solution:
    sin i = n × sin r = 1.6 × sin 30° = 1.6 × 0.5 = 0.8 → i = sin⁻¹(0.8) = 53.1°

32. Speed of light in a plastic block is 2 × 10⁸ m/s. What is its refractive index?
    Solution:
    n = 3 × 10⁸ / 2 × 10⁸ = 1.5

33. Find angle of incidence if angle of refraction is 45° and refractive index = 1.41.
    Solution:
    sin i = n × sin r = 1.41 × sin 45° = 1.41 × 0.707 = 0.997 → i = sin⁻¹(0.997) = 85.6°

34. Light passes from glass (n = 1.5) into air. What is the maximum angle of incidence for which no refraction occurs (i.e., Total Internal Reflection)?
    Solution:
    Critical angle C = sin⁻¹(1 / 1.5) = sin⁻¹(0.6667) = 41.8°

35. A pencil appears bent in water. If real depth is 12 cm and apparent depth is 9 cm, find refractive index.
    Solution:
    n = Real / Apparent = 12 / 9 = 1.33

36. Time taken by light to cross a 60 cm thick slab of glass is 3 × 10⁻⁹ s. Find speed of light in glass.
    Solution:
    Speed = 0.6 / (3 × 10⁻⁹) = 2 × 10⁸ m/s

37. A ray is incident at 50° in a medium and refracts at 32° in another. Calculate the refractive index.
    Solution:
    n = sin 50° / sin 32° = 0.766 / 0.530 = 1.445

38. A slab of thickness 4 cm shows a lateral shift of 1 cm. Angle of incidence = 45°. Estimate angle of refraction.
    Solution:
    Using shift formula: Shift = t sin(i – r) / cos r, estimate r by substitution. Approximate value: 29°

39. Light enters a medium with refractive index 1.52. What is speed of light in that medium?
    Solution:
    Speed = 3 × 10⁸ / 1.52 ≈ 1.97 × 10⁸ m/s

40. A ray bends towards the normal. What does this tell about the two media?
    Solution:
    It travels from a rarer to a denser medium. Medium 2 has higher refractive index.

41. Apparent depth of a stick in oil is 7 cm. If refractive index is 1.4, find real depth.
    Solution:
    Real depth = Apparent × Refractive index = 7 × 1.4 = 9.8 cm

42. A slab of width 5 cm causes a lateral shift of 1.2 cm. Angle of incidence is 60°. Estimate angle of refraction.
    Solution:
    Estimate using shift = t sin(i – r)/cos r. Substituting and trial: r ≈ 34°

43. A diver sees a fish at 1.2 m. Actual depth is 1.6 m. Find refractive index.
    Solution:
    n = Real / Apparent = 1.6 / 1.2 = 1.33

44. In which medium does light travel faster: n = 1.45 or n = 1.65?
    Solution:
    Lower n means higher speed. n = 1.45

45. A laser beam in water bends at 25° in air. If incidence angle is 18°, find refractive index of water.
    Solution:
    n = sin r / sin i = sin 25° / sin 18° = 0.4226 / 0.3090 = 1.367

46. Light bends away from the normal when entering from medium A to B. Which is denser?
    Solution:
    Since it bends away, it enters a rarer medium. Medium A is denser.

47. What happens to speed and wavelength of light when it enters a denser medium?
    Solution:
    Both decrease. Speed and wavelength decrease.

48. Refractive index of diamond = 2.42. What is its critical angle?
    Solution:
    sin C = 1/2.42 = 0.4132 → C = sin⁻¹(0.4132) = 24.4°

49. Light in air has wavelength 600 nm. Find its wavelength in glass (n = 1.5).
    Solution:
    Wavelength = 600 / 1.5 = 400 nm

50. If angle of refraction = 90°, what is the angle of incidence called?
    Solution:
    Critical angle