Snell law and EM wave properties

Question

1.What is the speed of light through an unknown glass with an index of refraction of 1.6?

2.

A flashlight beam strikes the surface of a pane of glass (n = 1.02) at a 70° angle to the normal. What is the angle of refraction? (Assume air to have an index of refraction of 1.0003)

3.A diver shines a flashlight upward from beneath the water (n=1.33) at an angle of 21.3° to the vertical. At what angle does the light refract through the air above the surface of the water?

Answer 1

Hi Bryce!

Like the first one I answered, this problem set also uses Snell's Law:

n1 sin (theta1) = n2 sin (theta2)

where n is the index of refraction and theta is the angle, measured from the normal line (or vertical).

1. The light goes from air to glass, so use Snell's Law:

(1.0003)(sin 70 degrees) = (1.02)(sin theta2)

sin^-1[(1.0003)(sin 70 degrees)/1.02)] = theta2 = 67.1 degrees

2. We will use Snell's law again, but this time from water to air:

(1.33)(sin 21.3 degrees) = (1.0003)(sin theta2)

sin^-1[(1.33)(sin 21.3 degrees)/1.0003] = theta2 = 28.9 degrees