A red light is submerged 1.9 m beneath the surface of a liquid with an index of
ID: 1898493 • Letter: A
Question
A red light is submerged 1.9 m beneath the surface of a liquid with an index of refraction 1.33. What is the radius of the circle from which light escapes from the liquid into the air above the surface? Answer: A beam of light is emitted 7.79cm beneath the surface of a liquid and strikes the surface 7.02cm from the point directly above the source. If total internal reflection occurs, calculate the index of refraction of the liquid. Answer: A beam of light is emitted in a pool of water from a depth of 74.4cm. Where must it strike the air water interface, relative to the spot directly above it, in order that the light does not exit the water? Answer: A beam of light in air strikes a piece of glas (n = 1.53) and is partially reflected and partially refracted. Find the angle of incidence if the angle of reflection is twice the angle of refraction. Answer:Explanation / Answer
(4) c = arcsin(1/1.33) = 0.8509 radians; the radius you want is such that
r/1.9 m = tan(0.8509)
r = 2.167 m. The color of the light is irrelevant given the index of refraction.
(5) tan c = 7.02/7.79
c = 0.733453 radians, and since 1/n = sin c , n = 1.494
(6) r/74.4 = tan c = tan(arcsin(1/n)) where n is the index of refraction for water. So:
r = 74.4 tan(arcsin(1/n)) and if we let n = 1.333 which is the Wikipedia value for water's index of refraction, we get r = 84.4 cm
(7) nairsin air = nglasssin glass and substituting the actual values for index of refraction,
sin air = 1.53 sin glass
Since the first angle is twice the second, we have
sin 2glass = 1.53 sin glass
2 sin glass cos glass = 1.53 sin glass
cos glass = 1.53/2
glass = 40.09 degrees
air = incidence = 80.18 degrees
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