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Consider a metal plate, on which a light of appropriate wavelength is incident such that it creates an interference pattern. The wavelength is chosen such that it can cause detectable photo-electric emission.

If we observe only the part of the metal where the light interferes destructively, two things can happen:

  • Due to the destructive interference(wave nature of light), there is no energy reaching that part so no photo-emission.

  • There are two photons reaching the destructive interface, so the electrons in the atom can interact with either of them and cause photo-emission.

Which one of the two will happen(Or will something entirely different happen)? Has this experiment been done before?

udiboy1209
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1 Answers1

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First, note that the metal "plate" will need to be more like a strip -- it needs to be small enough to fit into one trough (or peak) in the interference pattern.

But this thin strip that interacts with light is just like any other photodetector (including your eye). So, when it is in a trough, no electrons are emitted.

Your intuitive picture of "two photons reaching the interface" is misleading you in this context. The amplitude of the photon wave function is (essentially) zero at the center of the trough. Thus, the rate of electron ejection is (essentially) zero.

Dave
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