You place a point source of light at the bottom of a pool of water 1.00 m deep. The source emits rays upward in all directions. You notice that a circle of light is formed by the rays that are refracted into the air and that the rays outside of this circle are reflected back into the water. The index of refraction of water is 1.33. The radius R of the circle at the surface of the water is approximately
A) 0.75 m
B) 1.00 m
C) 1.13 m
D) 1.33 m
E) infinite

Question

You place a point source of light at the bottom of a pool of water 1.00 m deep. The source emits rays upward in all directions. You notice that a circle of light is formed by the rays that are refracted into the air and that the rays outside of this circle are reflected back into the water. The index of refraction of water is 1.33. The radius R of the circle at the surface of the water is approximately
A) 0.75 m 
B) 1.00 m 
C) 1.13 m 
D) 1.33 m 
E) infinite

Quizplus · Accepted Answer

The circle of light is formed by the rays that undergo total internal reflection at the water-air interface. At the point of total internal reflection, the angle of incidence is equal to the critical angle. We can use Snell's law at the water-air interface to find the critical angle:
sin(critical angle) = n_air / n_water = 1 / 1.33
critical angle = sin^-1(1/1.33) = 48.8°
Now we can use trigonometry to find the radius of the circle at the surface of the water. Let's call this radius r. The angle between the normal to the water surface and the ray that just undergoes total internal reflection is 90° - critical angle = 41.2°. Using basic trigonometry,
sin(41.2°) = r / 1.00 m
r = 1.13 m. Therefore, the correct choice is C.

You Place a Point Source of Light at the Bottom