The index of refraction of transparent liquid is 1.33. A flashlight held under t

Question

The index of refraction of transparent liquid is 1.33. 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 on angle of 44degree with respect to the vertical. At what angle theta_w. (with respect to the vertical) is the flashlight being held under water? The flashlight is slowly turned away from the vertical direction. At what angle will the beam no longer be visible above the surface of the pool?

Explanation / Answer

Since all other questions are solved I am here giving answer for problem 9.

9)

a) n1sin1 = n2sin2

1 = sin^-1[sin2*(n2/n1)]

1 = sin^-1[sin44*(1.00/1.33)] = 31.5 deg

b) Here 2=90

again,

1 = sin^-1[sin90*(1.00/1.33)] = 48.75 deg

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