Answer:
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°.
