Question

What is the critical angle for the interface between water and diamond?

Answer 1

Answer:

33 degrees

Explanation:

The critical angle for a light ray moving from a denser medium with refractive index $n_1$ to a second medium with refractive index $n_2$ is given by

$\theta_c = sin^{-1} (\frac{n_2}{n_1})$

In this case, the critical angle occurs when light moves from diamond to water. The index of refraction of the two materials are:

$n_d = 2.42$ for diamond

$n_w = 1.33$ for water

So the critical angle is

$\theta_c = sin^{-1} (\frac{1.33}{2.42})=sin^{-1}(0.550)=33.3^{\circ}$