The index of refraction of a transparent liquid is 1.31. A flashlight held under

Question

The index of refraction of a transparent liquid is 1.31. A flashlight held under the transparent liquid shines out of the transparent liquid in a swimming pool. This beam of light exiting the surface of the transparent liquid makes an angle of a= 23o with respect to the vertical. At what angle (with respect to the vertical) is the flashlight being held under the transparent liquid?

The flashlight is slowly turned away from the vertical direction. At what angle will the beam no longer be visible about the surface of the pool?

Please include steps and explanations.

Explanation / Answer

snells law:

N1sin1 = N2sin2

1.31sin1=1sin(23)

1= 17.35 degrees

1.31sin1=1sin(90) <------for no beam to be visible, there must be total internal reflection where no beam is refracted out of the liquid

1=49.76 degrees

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