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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?

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Light is neither a particle nor a wave. We use the particle and wave analogies to allow us to apply some level of intuition to the effects. The interference effects follow similar equations to waves in water so when we want to talk about the interference, we call light a wave. When we want to talk about effects such as your absorption of a quanta of light in the photoelectric effect, we talk about it as a particle. –  Jason A Oct 25 '13 at 1:14
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up vote 3 down vote accepted

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.

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You don't need to use a thin strip of metal, you can simply cover the metal plate with paper or something else and expose only the region of the trough. –  udiboy Oct 24 '13 at 19:08
secondly, are you saying that no photons will reach the trough at all, if we look at it from the particle view-point –  udiboy Oct 24 '13 at 19:09
The trough is a probability of going there trough, so if it is zero, the probability of a photon being there is zero. Classically as an electromagnetic wave, the energy in the trough is zero, thus no energy can be transmitted to that part of the surface, for it to be considered quantum mechanically as the originator of photoelectrons. –  anna v Nov 24 '13 at 5:17
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