Light of wavelength 655 nm is incident normally on a film of water 10 -4 cm thic

Question

Light of wavelength 655 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?

Please help me out with part c . I think the answer shouldn't have terms of pi.
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Explanation / Answer

a)

here by using the formula

lambda(water) = lambda / n

lambda(water) = 655 * 10^-9 / 1.33

lambda(water) = 492.48 * 10^-9 m

b)

N = 2 * t / lambda(water)

N = ( 2 * 10^-6 ) / (492.48 * 10^-9)

N = 4.061

c)

phase difference = - pie + 2 * pie * 2 * t / lambda(water)

= - pie + 2*pie * 4.061

= 7.122 * pie

then subtract 10pie then we get 2.878 * pie = 9.037 rad

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