Light of wavelength 418 nm is incident normally on a film of
water 1.0 μm 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 in the region where the two
reflected waves superpose?
The Attempt at a Solution
my attemp at a solution:
(a)lambda"water" = lambda/n = 445nm
(b)N = 2t/ lambda"water" = 4.5
(c) Light is going from a medium of smaller index of refraction
(air n = 1) to a medium with higher one (water n = 1.33) so the
reflected wave is out of phase by π respect to the transmitted
So the phase diff = π + 2π(2t/lambda"water") = -π +2π(4.5) = ???
My answers i have tried:
25.1 rad, 3.14rad, 31.4 rad, 1 rad, 4.72 rad, 0 rad 31.42rad 1 rad
Im really stumped...pls help!!!!!!!!!!!!!!!!!!!!!!!