Question

A light ray traveling through medium 1, index of refraction n1n1n_1 = 1.75, reaches the interface between medium 1 and medium 2, index of refraction n2n2 = 1.24. Part A At what minimum angle with respect to the normal must the ray be incident on the interface in order to be totally internally reflected?

Answer 1

Answer:

θ₁ = 35.32°

Explanation:

given,

refractive index of medium 1 = n₁ = 1.75

refractive index of medium 2 = n₂ = 1.24

condition to describe the refracted angle

 $\theta_{refracted} + \theta_{reflected}=90^0$

 $\theta_2 = 90^0-\theta_1$...(1)

Using Snell's Law

   n₁ sin θ₁ = n₂ sin θ₂

θ₁ , θ₂ is the angle of incidence and refractive index

n₁. n₂ is the refractive index medium 1 and medium 2

   1.75 x  sin θ₁ = 1.24 x sin θ₂

From equation (1)

   1.75 x  sin θ₁ = 1.24 x sin (90-θ₁)

   1.75 sin θ₁ = 1.24 cos θ₁

      tan θ₁ = 0.708

         θ₁ = 35.32°

Hence, angle of incidence is equal to θ₁ = 35.32°