Question

The critical angle for a substance is measured at 53.7 degrees. light enters from air at 45.0 degrees. at what angle it will continue?

Answer 1

When light travels from a medium with greater refractive index $n_1$ to a medium with smaller refractive index $n_2$, there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
In this problem, we know that the critical angle is$\theta_c = 53.7^{\circ}$, so we can find the ratio between the refractive indices of the two mediums:
$\frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81$
and since the second medium is air (n=1.00), the refractive index of the first medium is
$n_1= \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23$

In the second part of the problem, we have light entering from air ($n_i = 1.00$) at angle of incidence of $\theta_i = 45.0 ^{\circ}$, into the second medium with $n_r = 1.23$. By using Snell's law, we can find the angle of refraction of the light inside the medium:
$n_i \sin \theta_i = n_r \sin \theta_r$
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574$
$\theta_r = \arcsin(0.574)=35.1^{\circ}$