Chapter 27, Problem 18 Consult Interactive Solution 27.18 to review a model for

Question

Chapter 27, Problem 18 Consult Interactive Solution 27.18 to review a model for solving this problem. A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 622 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference? Number Units the tolerance is +/-2%

Explanation / Answer

For the film to look dark, the reflected waves must interfere destructively and/or transmitted waves interfere instructively.

u(oil) > u(water)

Let the thickness be t.

Path travelled by the ray reflected from oil = lambda/2

Path travelled by ray that entered oil and then got reflected by water = 2t(u(oil))

For destructive interference,

Path difference = 2t(u(oil)) - lambda/2 = (2n + 1)lambda/2 where n is a positive integer.

t = (n + 1)(lambda/2) / (u(oil))

putting different values of n, we get different values of t.

n = 1, t = lambda/2u

Similarly, for constructive interference of reflected waves,

2t(u(oil)) - lambda/2 = nlambda

t = (2n + 1)(lambda/4)/u = (2n + 1)lambda/4u

By destructive interference, we find ut

lambda/2u = t

ut = lambda/2 = 622/2 = 311 nm

Substitute this in constructive equation,

t = (2n + 1)lambda/4u

lambda = 4ut / (2n + 1)

lambda = 4(311) / (2n + 1)

Put n such that the wavelength lands in visible spectrum.

It comes out to be,

n = 1

lambda = 414.67 nm

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