Question

Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface with another material at a critical angle of incidence. Find the second material's index of refraction n n when the required critical angle is 75.1 ∘ .

Answer 1

Answer:

1.30

Explanation:

To calculate the critical angle we have ti use the formula:

$sin\theta_c=\frac{n_2}{n_1}$

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

$n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30$

hope this helps!!