Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 25.17
A thin film of oil (n = 1.50) of thickness 0.40e-6 m is spread over a puddle of water (n = 1.33). The two visible wavelengths that will destructively interfere for reflection will have values of __________ nm and __________ nm in air. (Visible wavelengths in air are in the range of 400 nm to 700 nm.)Correct answer: 400 nm, 600 nm.
Light in air reflects off of the top of the oil film with a 180 degree phase shift ("fast off of slow"). Light in oil reflects off the top of the water with no phase shift ("slow off of fast"). The two reflections are out of phase.
For destructive interference of these two out of phase reflections, the path difference condition is:
delta(l) = m*lambda_oil,
where the path difference between the two reflected waves is twice the thickness of the oil film, or 2*t, and lambda_oil is related to the lambda_air wavelengths in air:
lambda_oil = lambda_air/n_oil,
as wavelengths are shorter in slower media, compared to the longest in air (or vacuum).
Substituting delta(l) = 2*t, and lambda_oil = lambda_air/n_oil into the destructive interference condition for out of phase reflections:
2*t = m*(lambda_air/n_oil),
2*t*n_oil/m = lambda_air,
with m = 0, 1, 2, 3, 4, ..., possible wavelengths in air that would destructively interfere would be infinite, 1200 nm, 600 nm, 400 nm, 300 nm, ..., respectively, of which only 400 nm and 600 nm are in the visible range in air.
Student responses
Section 31988
537 nm, 637 nm: 1 response
400 nm, 600 nm: 2 responses
429 nm, 687 nm: 1 response
(Blank/no work submitted: 9 responses)
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