Light of wavelength 475 nm is incident normally on a film of water
10-4 cm thick. The index of refraction of water is
1.33.
a. What is the wavelength of the light in the water?
b. How many wavelengths are contained in the distance 2t, where t is the thickness of the film? (Do not round your answer to a whole number.)
c. What is the phase difference between the wave reflected from the top of the air-water interface and the one reflected from the bottom of the water-air interface after it has traveled this distance? (in Rad)
light wave length = 475 nm
index of refraction of water is 1.33
a) wavelength of the light in the water = 475nm *(1/1.33) = 357.14 nm = 357.14*10-9 m
b) thickness of water film ,t = 10-4 cm = 10-6 m
distance travelled = 2t =2*10-6 m
no of wavelngths = 2*10-6 m / 357.14*10-9 m = 5.6 wavelengths
c) light is travelling from low index of refraction (1) of air to high index of refraction (1.33) of water, so the reflected wave is out of phase by pi radians.
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