Question

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?

Answer 1

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

Step-by-step 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°.