physics
posted by Steve on .
In searching the bottom of a
pool at night, a watchman
shines a narrow beam of light
from his flashlight, 1.3 m
above the water level, onto the
surface of the water at a point
2.7m from the edge of the
pool. Where does the spot of
light hit the bottom of the pool,
measured from the wall
beneath his foot, if the pool is
2.1m deep.

First use trig to determine the incidence angle of the beam hitting the water. That angle is
I = arctan 2.7/1.3 = arctan 2.077 = 64.3 degrees
The use Snell's law to calculate the angle of refraction, R, in the water. Assuming the refractive index of water is 1.33,
1.33 sin R = sin 64.3
sin R = 0.6775
R = 42.6 degrees
1.33 sin R
The beam goes an additional horizointal distance under the water, given by
X = 2.1 tan R
Add that to 2.7 m for the answer