At what angle will it emerge into the air?

Question

A swimmer shines a beam of light in water up toward the surface. It strikes the air-water interface at 30°. At what angle will it emerge into the air?
_____ ° (to the surface)


** Refraction and Reflection Unit - Physics

Answer 1

Use Snells law here

n1*Sin(theta1) = n2*Sin(theta2)

You do not state where the 30 degree angle is measured from (ie from the normal or from the interface)

We will assume that the 30 degree angle is wrt the normal and that n for water = 1.33

1.33*Sin(30) = 1*Sin(theta2)

Solving gives theta2 = 41.6 degrees to the normal. Thus it emerges at an angle of 48.4 degrees to the surface.

If we assume the other case, that it struck at an angle of 30 to the surface, then the angle wrt the normal is 60 degrees

1.33*Sin(60) = 1*Sin(theta2)

This gives sin(theta2) > 1 which is not possible, meaning that the light is totally internally reflected.

Answer 2

I concur. 41.68 degrees refracted angle.

Snell's Law

sin(theta1) / sin(theta2) = Refractive Index = 1/1.33 = 0.75188

Sin(theta1) = sin(60) = 0.500

Sin(theta2) = 0.500/0.75188 = 0.665

Arcsin(0.665) = 41.68 degrees

Answer 3

n(air) *sin(air) = n(water) sin (water)
n(air)=1 n(water ) =1.333

so sin (air ) =1.333*sin30°=1.333*0.5=0.665= 41°.8