A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at point A. The angle of incidence is 68 degrees. The depth of the lake is 3.0 m. On the flat lake-bottom is point B, directly below point A. (a) If refraction did not occur, how far away from point B would the laser beam strike the lakebottom? (b) Considering refraction, how far away from point B would the laser beam strike the lake-bottom?
a) Without refraction, the light would hit the bottom at a
distance of 3tan68 = 5.4569 m.
b). Here we need to use Snell's Law to find the angle of
refraction. If we call x2 the angle of refraction, the Snell's Law
tells us
1.00sin68 = 1.33sinx2.
Solving for x2 gives an angle of refraction of 45.79 degrees.
The distance from point A will be 3tan45.79 = 2.6276 m.
A narrow beam of light from a laser travels through air (n = 1.00) and strikes...
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