A thin film having an index of refraction of 1.50 is surrounded by air. It is il

Question

A thin film having an index of refraction of 1.50 is surrounded by air. It is illuminated normally by white light. Analysis of the reflected light shows that the wavelengths 360, 450, and 600 nm are the only missing wavelengths in the visible portion of the spectrum. That is, for those wavelengths, there is destructive interference.

(a) What is the thickness of the film?
nm

(b) What wavelengths in the visible portion of the spectrum are brightest in the reflected interference pattern? (Enter your answers from smallest to largest.)

(c) If this film were resting on glass that has an index of refraction of 1.60, what wavelengths in the visible spectrum would be missing from the reflected light? (Enter your answers from smallest to largest.)
nm
nm
nm

Explanation / Answer

a)

here by using the formula

t = m * lambda / 2 * n

t = 4 * 450 * 10^-9 / 2*1.5 = 600 nm

b)

here thw formula

lambda = 2 * 1.5 * 600 * 10^-9 / ( m + 0.5 )

lambda = 1800 * 10^9 / ( m + 0.5 )

so by putting the values of m from 0 to 5 we get

m = 0

lambda = 3600 nm

m = 1

lambda = 1200 nm

m = 2

lambda = 720 nm

m = 3

lambda = 514 nm

m = 4

lambda = 400 nm

m = 5

lambda = 327 nm

so the visible spectrum are 72 nm , 514nm and 400nm

c)

here also the values are same as above and the answer is

the missing wavelengths in the visible sepctrum are 720nm , 514 nm and 400nm

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