A diver shines a flashlight upward from beneath the water (n=1.33) at a 36.2° angle to the vertical. At what angle does the light leave the water? Express your answer using three significant figures?

The flashlight leaves the water at an angle of 51.77°.

Explanation:

if n1 = 1.33 is the refractive index of water and ∅1 is the angle at which the flashlight shine beneath the water, and n2 = 1.0 is the refractive index of air and ∅2 is the angle the flashlight leaves the water.

Then, according to Snell's law:

n1*sin (∅1) = n2*sin (∅2)

sin (∅2) = n1*sin (∅1) / n2

= (1.33) * sin (36.2) / (1.0)

= 0.7855055*379

∅2 = 51.77°

Therefore, the flashlight leaves the water at an angle of 51.77°.