Posted by Steve on Sunday, September 7, 2008 at 1:24am.
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
physics - drwls, Sunday, September 7, 2008 at 5:38am
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
Answer This Question
More Related Questions
- Physics ~ Please help - The index of refraction of a transparent liquid (similar...
- Physics ~ missed this lecture pls help! - The index of refraction of a ...
- Physics - A light source, S, is located 2.9 m below the surface of a swimming ...
- Physics - A swimming pool filled with water to its edge, the depth of the water ...
- Physics - A very narrow beam of white light is incident at 56.10° onto the top ...
- Physics - A very narrow beam of white light is incident at 47.50° onto the top ...
- Physics - very narrow beam of white light is incident at 40.40° onto the top ...
- physics - A beam of light is emitted in a pool of water from a depth of 64.0 cm...
- Physics - A beam of light is emitted in a pool of water from a depth of 61.5 cm...
- physics - An Olympic pool has dimensions 50.0 m × 25.0 m × 2.68 m (length × ...