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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?

my2cts
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user157588
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  • A phase difference of $\pi$ is required for destructive interference, not $\frac{\pi}{2}$. – Chris Nov 23 '17 at 04:13
  • 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? – FlatterMann Sep 24 '22 at 21:05

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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.

anna v
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  • 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. – FlatterMann Sep 24 '22 at 22:40
  • How does the referenced video show the quantum mechanical dependence of light? – Not_Einstein Sep 24 '22 at 23:34
  • 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. – flippiefanus Sep 25 '22 at 03:52
  • @Not_Einstein I do not think that the return to the source can be mathematically described with classical maxwell equations. – anna v Sep 25 '22 at 03:53
  • @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 https://physics.stackexchange.com/questions/248726/how-does-qed-explain-reflection/248741#248741) – anna v Sep 25 '22 at 04:00
  • Your microscopic description would not give interference, because if there is momentum transfer, the scattered particle would become entangled with the scatterer. For light, it is better to use a linear model in terms of reflection and transmission. – flippiefanus Sep 25 '22 at 04:09
  • @flippiefanus classically , as light, the scatter is with the whole lattice that makes up the film, the momentum arguments are the same for reflection. For momentum conservation the whole lattice should move with twice the momentum of the impinging and reflecting light, which implies energy loss from the beam that has to go somewhere, and I guess heat, unless the film vibrates, and then again it will go to heat – anna v Sep 25 '22 at 04:13
  • In other words, you are saying that all the light would be absorbed because of the momentum transfer caused by the reflection. Obviously this does not happen. – flippiefanus Sep 25 '22 at 04:32
  • @flippiefanus No, the ball scatters off the wall, it is not absorbed! But momentum conservation imposes the need of energy loss, and that energy has to be somewhere. The same with the light beam. – anna v Sep 25 '22 at 04:50
  • @flippiefanus in the simple ball scatter, ball in x derection hitting wall and bouncing back in x directioin, it looks elastic, but because it returns with the negative momentum, of the original direction, it means that the wall must take up twice the momentum, so momentum in the x direction is conserved. – anna v Sep 25 '22 at 04:56
  • In the context of the OP, you are saying that the reason that thin film appears dark is because the light is absorbed. Now you argue that it is because of momentum conservation. That is nonsense. Light is reflected off the interfaces with negligible energy loss, regardless of the fact that momentum is conserved. This is easily confirmed by experiment. – flippiefanus Sep 25 '22 at 08:43
  • l;ets tp this here, as we do not read the situation the same way. In the MIT video I link the energy goes back to the source, for example. In this case my opinion is that the energy goes finally to thermal energy, and I am trying to explain my POV with the ball example. For me if there is total destructive interference, black, the energy has to turn up as thermal energy. – anna v Sep 25 '22 at 11:53
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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.

flippiefanus
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