White light travels in air and strikes the surface of a placid (calm) lake at 25
ID: 1635862 • Letter: W
Question
White light travels in air and strikes the surface of a placid (calm) lake at 25.0 degree relative to normal to the surface. The index of refraction of water is n_water = 1.33. The index of refraction of air is n_air = 1.00. The lake is 20.0 m deep. (A) What is the angle (theta) relative to normal the light makes as it is refracted in the water column? (B) Now consider the range of visible wavelengths (400-700 ran). 400 nm blue light has an index of refraction of 1.339 and red light has an index of refraction of 1.331. If the angle of incident which light is still 25.0 degree relative to normal to the surface what is the expected distance between blue and red colors at the bottom of the lake due to dispersion? (C) If incoming light is reflected off the bottom of the lake and returns upward toward the sky, what is the critical angle (theta_c) needed so that none of the light breaks the surface of the water to the air above? Would a light ray (described above incoming from the sun at 25.0 degree) break the surface?Explanation / Answer
4. given angle of incidence, i = 25 deg
refractive index of water, n = 1.33
so from snells law, let angle of refraction = r
then sin(i)/sin(r) = n
a. r = arcsin(sin(i)/n) = arcfsin(sin(25)/1.33) = 18.52 degrees
b. let angle of refraction for blue light be Rb [ nb = 1.339]
angle of refraction of red light be Rr [ nr = 1.331]
from snells law
sin(i)/sin(Rb) = nb
Rb = 18.3983 deg
sin(i)/sin(Rr) = nr
Rr = 18.512 deg
so distance between the two at bottom of the lake
x = d*[tan(Rr) - tan(Rb)]
where d is depth of the lake
x = 20[tan(Rr) - tan(Rb)] = 0.04419 m = 4.410 cm
c. let critial angle be ic
then sin(ic)/sin(90) = 1/1.333
ic = 48.60 deg
but r = 188.52 degrees
so this light ray would break the syurface
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