Question

A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted angle?

Answer 1

We can solve the problem by using Snell's law, which states 
$n_i \sin \theta_i = n_r \sin \theta_r$
where
$n_i$ is the refractive index of the first medium
$\theta_i$ is the angle of incidence
$n_r$ is the refractive index of the second medium
$\theta_r$ is the angle of refraction

In our problem, $n_i=1.00$ (refractive index of air), $\theta_i = 28.0^{\circ}$ and $n_r=1.63$ (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_r = \frac{1.00}{1.63} \sin 28.0^{\circ} = 0.288$
From which we find
$\theta_r = \arcsin (0.288)=16.7^{\circ}$