Question

A light beam in glass (n = 1.5) reaches an air-glass interface, at an angle of 60 degrees from the surface. What is the angle of the refracted light beam from the normal in air?Note: sin(30) = 0.5, sin(45) = 0.71, sin(60) = 0.87 (all angles are in degrees)

Answer 1

Answer:

θ₂ = 35.26°

Explanation:

given,

refractive index of air, n₁ = 1

refractive index of glass, n₂ = 1.5

angle of incidence, θ₁ = 60°

angle of refracted light, θ₂ = ?

using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

1 x sin 60° = 1.5 sin θ₂

sin θ₂ = 0.577

θ₂ = sin⁻¹(0.577)

θ₂ = 35.26°

Hence, the refracted light is equal to  θ₂ = 35.26°