Question 5 of 24advanced🔧 ApplyNumerical5 marks

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Correct Answer

(i) Refractive index of prism material = 1.53

(ii) New angle of minimum deviation in water = 10°

Exercise: EXERCISES | Q: 9.6 | (Page 29)
For More Understanding

Explanation

This question tests understanding of the prism formula for minimum deviation. The formula n₂₁ = sin[(A + Dm)/2]/sin[A/2] relates refractive index, prism angle A, and minimum deviation angle Dm. For part (i), we substitute the given values directly. For part (ii), when the prism is immersed in water, the relative refractive index becomes n_prism/n_water, and we solve for the new Dm.

Solution Steps

  1. Step 1: Given data - Angle of minimum deviation Dm = 40°, Refracting angle of prism A = 60°

  2. Step 2: Using formula for refractive index: n₂₁ = sin[(A + Dm)/2]/sin[A/2]

  3. Step 3: Substituting values: n = sin[(60° + 40°)/2]/sin[60°/2] = sin(50°)/sin(30°)

  4. Step 4: Calculation: n = 0.766/0.5 = 1.532 ≈\approx 1.53

  5. Step 5: When prism is placed in water, relative refractive index = n_prism/n_water = 1.532/1.33 = 1.152

  6. Step 6: Using formula again: 1.152 = sin[(60° + Dm')/2]/sin(30°)

  7. Step 7: Solving: sin[(60° + Dm')/2] = 1.152×0.51.152 \times 0.5 = 0.576

  8. Step 8: Therefore, (60° + Dm')/2 = sin⁻¹(0.576) ≈\approx 35.1°

  9. Step 9: Hence, 60° + Dm' = 70.2°, giving Dm' ≈\approx 10°

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