Question

(a) A ray of light is incident at angle θ on a right-angled prism at point X. At point Y, it emerges along the prism surface. Calculate the refractive index of the prism in terms of the incident angle.

(b) Show that for an equilateral prism kept in air, minimum deviation occurs when the angle of incidence i = sin^-1 (n/2), where n is the refractive index of the material of the prism.

Answer 1

(a)

At the 1st surface, using Snell's law 

sin θ = n sin r₁

sin r₁ = sin θ/n

r₂ = A - r₁ = 90 - r₁

At the second interface, 

sin r₂ = sin 90/n

sin r₂ = 1/n

sin (90 - r₁) = 1/n

cos r₁ = 1/n

Squaring both sides 

cos²r₁ = 1/n² 

1 - sin²r₁ = 1/n² 

1 - (sin²θ/n²) = 1/n² 

Solving, n = √(1+sin²θ)

(b) For an equilateral prism A = 60°

Using Snell's law at the first surface, 

sin i = n sin r 

At minimum deviation r = A/2 = 60/2 = 30°

sin i = n sin(30) 

sin i = n(1/2) 

i = sin^-1 (n/2)