Question

A ray of light is incident normally on one of the faces of a prism of apex 30° and n = √2 What is the angle of deviation of the ray ? 

(A) 45° 

(B) 30° 

(C) 15° 

(D) 60°

Answer 1

The correct option is (C) 15°.

Explanation:

ΔAQP is 30 – 60 – 90 triangle 

∴ θ₁ = 60 hence θ₂ = 90 – 60 = 30° 

At face AC, I = θ₂ = 30° 

Let angle of deviation of the ray ber hence shells law at face AC. 

η = [(sin r) / (sin 30)] 

sin r = n ∙ sin30 = √2 × (1/2) = (1 / √2) 

∴ r = sin–1(1 / √2) = 45° 

∴ angle of deviation = δ = r – 30° = 45 – 30 = 15°