Question

A diver shines a flashlight upward from beneath the water at a 35.2° angle to the vertical. at what angle does the light leave the water?

Answer 1

We can solve the problem by using Snell's law:
$n_i \sin \theta_i = n_r \sin \theta_r$
where 
$n_i$ is the refractive index of the first medium (in this case, air, so $n_i = 1.00$)
$n_r$ is the refractive index of the second medium (in this case, water, so $n_r = 1.33$)
$\theta _i$ is the angle of incidence of the light, with respect to the vertical, so $\theta_i = 35.2^{\circ}$
$\theta_r$ is the angle of refraction of the light inside the water, with respect to the vertical

Re-arranging the equation and using the data of the problem, we can find the the angle of refraction of the light inside the water:
$\theta_r = \arcsin ( \frac{n_i}{n_r} \sin \theta_i )=\arcsin ( \frac{1.00}{1.33} \sin 35.2^{\circ} )=25.7^{\circ}$