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## Homework Statement

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?

## Homework Equations

## 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

wave right?

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!!!!!!!!!!!!!!!!!!!!!!!