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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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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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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