Question

A ray of light is incident normally on one of the faces of a solid prism of apex angle 30° and refractive index √2 . The angle of minimum deviation is ____________ . 

(A) 39° (B) 42° (C) 52° (D) 15°

Answer 1

The correct option is (D) 15°.

Explanation:

as Δ AQP is  30° – 60° – 90° triangle

Q₁ = 60° hence  θ₂ = 90° – 60° = 30°

At face AC,   i = θ₂ = 30°.

Let angle of deviation of the ray be r hence Shells law at face AC,

η = [(sin r) / (sin 30°)]

sin r = η × sin 30° = √2 × (1/2) = (1 / √2)

 r = 45°

Angle of deviation = δ = r – 30 = 45 – 30 = 15°.