Question

A diver shines a flashlight upward from beneath the water (n = 1.33) at a 32.7° angle to the vertical. At what angle does the light leave the water?

Answer 1

45.93°

Step-by-step explanation:

The angle of incidence is given as 32.7°

The refractive index of the water that is
`n_1=1.33`

Refractive index of the air that is
`n_2=1` (because the refractive index of air is 1 )

We have to find the angle at which the light leave the water means angle of refraction

So according to snell's law
`n_1sini=n_2sinr`

`1.33sin32.7=1* sinr`

`sinr=0.7185`

r =45.93°

So the light leave the water at an angle of 45.93°