Asked by Anonymous
A light source, S, is located 2.9 m below the surface of a swimming pool and 1.7 m from one edge of the pool. The pool is filled to the top with water. At what angle does the light reaching the edge of the pool leave the water?
Answers
Answered by
drwls
First you need to look up the index or refraction for air and water, the two materials in which the light will travel and refract.
Air has a refractive index of N2 = 1.0003, and water has a refractive index of N1 = 1.333. I use index 2 for air because it is the second medium the light passes through; water is the first.
The angle of incidence A1 for the light beam in the water is tan^-1 1.7/2.9 = tan^-1 0.5862 = 30.37 degrees.
The last step is to use Snell's law:
N1 sin A1 = N2 sin A2
Solve for the angle of refraction, A2
Sin A2 = 0.5057*(1.333/1.0003)= 0.6739
A2 = ?
Air has a refractive index of N2 = 1.0003, and water has a refractive index of N1 = 1.333. I use index 2 for air because it is the second medium the light passes through; water is the first.
The angle of incidence A1 for the light beam in the water is tan^-1 1.7/2.9 = tan^-1 0.5862 = 30.37 degrees.
The last step is to use Snell's law:
N1 sin A1 = N2 sin A2
Solve for the angle of refraction, A2
Sin A2 = 0.5057*(1.333/1.0003)= 0.6739
A2 = ?
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