The index of refraction of a transparent liquid (similar to water but with a dif

Question

The index of refraction of a transparent liquid (similar to water but with a different index of refraction) is 1.28. 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 theta a = 40 degree with respect to the vertical. At what angle (with respect to the vertical) is the flashlight being held under transparent liquid? Answer in units of degree. 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? Answer in units of degree.

Explanation / Answer

Data: Angle of incidence, w = ? Angle of refraction, a = 40 deg Refractive index of water, nw = 1.28 Refractive index of air, na = 1 Solution: According to the Snell's law, nw * sin w = na * sin a 1.28 * sin w = 1 * sin 40             sin w = 0.5022                   w = 30.14 deg Ans: Angle of incidence in water, w = 30.14 deg

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