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I have read that when a thin film becomes very thin (lambda>>film), it actually does not reflect any light since the incident wave and the reflected wave interfere destructively (and there is no light reflected off the bottom of the film). This would be because the reflected wave is phase shifted by half a wavelength when it hits the top of the film, which has a higher refractive index. It would therefore be perfectly out of phase with the incident ray.

But my question is, wouldn’t this destructive interference happen in the normal case too, even when there is a 2nd reflected wave from the bottom of the film?

Why are we in this case considering the interference of the two reflected waves only, and not the destructive interference with the incident wave?

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  • $\begingroup$ So you are asking why doesn't the wave reflected off the first surface interfere destructively with the incident wave? $\endgroup$
    – M. Enns
    Oct 10, 2019 at 16:22
  • $\begingroup$ That's exactly right $\endgroup$
    – jewelBeetle
    Oct 10, 2019 at 16:28
  • $\begingroup$ 2 photons never cancel each other out, that is a violation of conservation of energy. Classically many physicists use the theory of cancellation which is historical but inaccurate. Thin films work just fine for single photons as well (see also single photon double slit experiments). The modern thinking is to understand wave functions and their propagation, Feynman path integral theory is the more logical way to go. However the classical mathematical solution theory works, it coincides with path theory and the important property of light travelling in multiples of its wavelength. $\endgroup$
    – PhysicsDave
    Oct 10, 2019 at 21:31
  • $\begingroup$ if the comment were true then the object would never warm up as it absorbs light, but of course it does! $\endgroup$
    – PhysicsDave
    Oct 10, 2019 at 21:33

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The key difference is, with the interference of light reflecting off the front and back surfaces of the film the two interfering light waves are traveling in the same direction while in the case of interference between the incident light and the reflected light the waves are moving in opposite directions. You can have an interference pattern with light waves moving in opposite directions (a laser cavity comes to mind) but this wouldn't give destructive interference everywhere, rather you would have dark nodes and bright antinodes.

In practice, I can't imagine this being observable in a situation with a soap bubble since the phase, wavelength and amplitude of the incident light are not consistent like they are in a laser cavity.

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  • $\begingroup$ This makes sense @M. Enns, but then that would mean that the first part of my question is wrong? I.e. the reason we do not see any reflection of very thin films is not due to destructive interference between the incident and reflected ray. If there was interference, it would fluctuate in intensity and not lead to complete disappearance of the reflected light. Rather, it must be interference between the two reflected waves (as in the normal case) but because the film is very thin, the path length addition is negligible and the two reflected waves are practically out of phase. $\endgroup$
    – jewelBeetle
    Oct 10, 2019 at 20:07
  • $\begingroup$ 2 photons never cancel each other out, that is a violation of conservation of energy. Classically many physicists use the theory of cancellation which is historical but inaccurate. Thin films work just fine for single photons as well (see also single photon double slit experiments). The modern thinking is to understand wave functions and their propagation, Feynman path integral theory is the more logical way to go. However the classical mathematical solution theory works, it coincides with path theory and the important property of light travelling in multiples of its wavelength. $\endgroup$
    – PhysicsDave
    Oct 10, 2019 at 21:31

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