In case of interference, we know, energy is neither destroyed, nor created; but only redistributed. But in the case of an extremely thin film, due to a reflection and hence a phase difference of $\pi$, the film always appears dark due to destructive interference. So, where does the energy go?
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$\begingroup$ A phase difference of $\pi$ is required for destructive interference, not $\frac{\pi}{2}$. $\endgroup$– Chris ♦Commented Nov 23, 2017 at 4:13
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$\begingroup$ Where does the idea come from that an extremely thin film is always dark? There is an atomic layer of water on basically everything, including your windows, which are clearly not dark. Just curious where this obviously false phenomenological statement originates? Can you give a textbook reference? $\endgroup$– FlatterMannCommented Sep 24, 2022 at 21:05
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2 Answers
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The logical answer is : into an increase in the motion of the atoms on which individual photons scattered off the thin film, i.e. heat.
This is a fascinating similar phenomenon with monochromatic laser light showing destructive interference. It is instructive to look, as it shows the quantum mechanical dependence of light.
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$\begingroup$ Can you give me an actual physical example of a thin film that is dark? If that was the case, then deep black would be the cheapest paint of all... we would just have to make a really thin layer of paint of any color. That is obviously not so. I think the question is based on an unphysical understanding of the properties of thin films. My guess is that it implicitly assumes that the dielectric constant of the material of the thin film goes towards infinity. If it doesn't, then the thin film will surely not look dark. It will become ever more transparent the thinner it gets. $\endgroup$ Commented Sep 24, 2022 at 22:40
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$\begingroup$ How does the referenced video show the quantum mechanical dependence of light? $\endgroup$ Commented Sep 24, 2022 at 23:34
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$\begingroup$ This answer is not correct. The thin film does not need to be absorbing. So it would not have the ability to convert the energy into heat. $\endgroup$ Commented Sep 25, 2022 at 3:52
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$\begingroup$ @Not_Einstein I do not think that the return to the source can be mathematically described with classical maxwell equations. $\endgroup$– anna vCommented Sep 25, 2022 at 3:53
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$\begingroup$ @flippiefanus If there is a scatter, and reflection is a scatter, there is always momentum transfer and because of this some energy lost (ball scatter on the wall) . That energy ends up in heat is all I am saying. ( relevant my answer here physics.stackexchange.com/questions/248726/…) $\endgroup$– anna vCommented Sep 25, 2022 at 4:00
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Whenever you have destructive interference somewhere, you always produce constructive interference somewhere else. The reason is that all processes are unitary. Therefore, the device or setup that you use to produce the destructive interference can be modeled as a four port system with two input ports and two output ports.
For the example of the thin film, the two input ports are represented by the two sides of thin film (light incident from opposite sides of the thin film) and the two output ports are also the two sides (light propagating away from the thin film in opposite directions). Light from either direction can either be reflected or transmitted. The thin film has two interfaces each of which introduces reflection and/or transmission.
If light is incident on the thin film from only one side and the thin film introduces a relative $\pi$ phase shift for the two reflections, so that the combined reflection is zero due to destructive interference, then the transmission must (thanks to the requirements of unitarity) be constructive so that all the light passes through the thin film. This is the basic idea of an anti-reflection coating.
answered Sep 25, 2022 at 3:51