Question

A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 29.7◦ and the angle of refraction is 16.3◦ , find the critical angle for light traveling from the liquid back into the air. answer in units of ◦

Answer 1

When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
$\theta_c = \arcsin ( \frac{n_r}{n_i} )$ (2)
where $n_r$ is the refractive index of the second medium and $n_i$ is the refractive index of the first medium.

We can find the ratio $n_r / n_i$ by using Snell's law:
$n_i \sin \theta_i = n_r \sin \theta_r$ (1)
where
$\theta_i$ is the angle of incidence
$\theta_r$ is the angle of refraction

By using the data of the problem and re-arranging (1), we find
$\frac{n_r}{n_i} = \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566$

and if we use eq.(2) we can now find the value of the critical angle:
$\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}$