Question

What is the critical angle for light traveling from crown glass into air?

Answer 1

Let's write the Snell's Law:

$$n_isin\theta_i=n_rsin\theta_r,$$

here, $n_i=1.52$ is the index of refraction of the incident medium (or the crown glass), $n_r=1.0$ is the index of refraction of the refractive medium (or the air), $\theta_i$ is the angle of incidence, $\theta_r$ is the angle of refraction.

The critical angle is defined as the angle of incidence that provides an angle of refraction of $90^{\circ}.$Therefore, we can write:

$$n_isin\theta_c=n_rsin90^{\circ},$$ $$sin\theta_c=\dfrac{n_r}{n_i},$$ $$\theta_c=sin^{-1}(\dfrac{n_r}{n_i})=sin^{-1}(\dfrac{1.0}{1.52})=41.1^{\circ}.$$

Answer:

$\theta_c=41.1^{\circ}.$