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A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.
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$33.9\textrm{ nm}$
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OpenStax College Physics for AP® Courses Solution, Chapter 27, Problem 78 (Problems & Exercises) (3:11)

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This is College Physics Answers with Shaun Dychko. Some white light is shining on an oil slick that has some thickness such that the additional path length that this second ray is traveling as it penetrates the oil and then reflects off the oil-water interface here this additional path length has to be less than the wavelength divided by 4 in order for this path length to be so insignificant that the phase shift due to reflection at this interface exactly cancels this ray here which does not experience a phase shift on reflection at the oil-water interface. So there's λ over 2 phase shift at this reflection because it's going from a medium of low index of refraction, which is just 1.0 for air and then the second medium is a high index of refraction of oil 1.40 and so the reflection here has a λ over 2 phase shift. And so that's enough to destructively interfere with the reflection here so long as this path length is insignificant and that's true we are told when the additional path length is a quarter of the wavelength So the thickness of this oil slick is what we want to find and the path length difference is going to be 2 times that thickness and we are assuming that the light is actually coming straight down although I have drawn it on an angle just to illustrate the two different reflections. So this Δx then is 2 times the thickness for a round trip and we can substitute 2t in place of Δx here then so that means 2 times the thickness is less than a quarter of the wavelength or divide by 2 on both sides and we get that the thickness then is less than one-eighth of the wavelength. Now I have this subscript n on the wavelength here because it's the wavelength that appears in the oil that matters here and that wavelength in the oil is the wavelength in a vacuum or air since they are pretty much the same divided by the index of refraction of the oil. So we substitute that in for λ n and so we have the maximum possible thickness we have to consider the minimum wavelength because this thickness has to be less than an eighth of every visible wavelength and so we have to consider the shortest wavelength that we need to cancel, which will be for blue and then having the thickness, you know, less than this particular wavelength... well, it will be less than every other wavelength that's greater than that too. Okay! So we have 380 nanometers—that's the shortest wavelength I guess that's violet actually not blue— and divide that by 8 times 1.40 and that is 33.9 nanometers is the maximum possible thickness for the oil slick in order to have destructive interference for every visible wavelength, in which case the oil slick will look black.