# Interference between incident and reflected light rays in thin films

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?

• So you are asking why doesn't the wave reflected off the first surface interfere destructively with the incident wave? Commented Oct 10, 2019 at 16:22
• That's exactly right Commented Oct 10, 2019 at 16:28
• 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. Commented Oct 10, 2019 at 21:31
• if the comment were true then the object would never warm up as it absorbs light, but of course it does! Commented Oct 10, 2019 at 21:33