N.B. This effect is of statistical nature, so no change will be measured.

Where do you have your misinformation from: the slit is not a cavity

even Wikipedia gives you the right information.

This is a standard exercise in wave optics or electomagnetism-but the way it's stated is incomprehensible.First of all, it's not a quantum problem, but a classical one. You can, of course, describe it in terms of photons, but that's not the typical case. The slits don't select frequencies, boundary conditions may do. And this is, also, a classical issue-the finite volume and the boundary conditions on the walls of the cavity leads to a discrete spectrum, not any quantum effect.In particular, if the space beyond the slits doesn't have other boundaries, electromagnetic waves are transverse and at large distances can be taken to be plane waves. Finally, If you do want to affect the diffraction pattern, you need a mechanism that affects the relative phase of the waves beyond the slits. A ``voltage'' can't, because it doesn't affect the phase, but the amplitude-that's the short answer. Any book on waveguides may provide a longer answer-but one can work it out for oneself: tedious, but straightforward.

Of course, if you consider the plane, containing the slits, to impose particular boundary conditions on the incident electromagnetic field, then you have to solve the wave equation with these boundary conditions to find the full transmitted field-and that  field, of course,  won't be, just,  due to the interference between the two slits, but much more complicated. By comparing with the simpler problem, you may, then, deduce the properties of the material of the plane, for example.

The nature of electromagnetic fields in relation to one another is superposition.  They occupy the same space without interacting.  The net field at any point is the simple linear sum of potentials from each source.  By this principle, one should conclude that a voltage would have no direct effect on photons.  Nor would other photons.  You can pile them up to infinite density and they have no electromagnetic effect (though if you literally did that, they would begin to have a gravitational effect).

Of course, EM fields do indeed interact with charged objects (or conductors, which necessarily must contain mobile charged objects).  So if there are charged objects present, the two fields may influence each other through the charged objects.  If the slits are constructed from material that does not change position or other properties in response to the voltage, then the condition for interaction between the voltage and photons is removed.  If the voltage is large enough to overcome the binding force of molecules or atoms, the slits will disappear destructively, but I presume that is not what you are talking about.

 ``does the diffraction pattern form because of the interaction between the incoming photons and the photons of the cavity?''

The answer is no. Much experimental work has been done during the last decennia studying interference of very weak beams (ideally, one particle/photon at a time entering the interferometer), impacts of the particle/photon on a screen being found to be particle-like. It was experimentally found that, when the experiment is repeated in this way, the interference pattern becomes gradually visible by accumulation of the results. Evidently, each single particle/photon ``knows'' that it should not impact at dark regions of the screen. Evidently, each single particle/photon ``knows'' that it should not impact at dark regions of the screen. No interaction between the particles/photons is necessary.

Remi: you are right BUT

He asks "Would application of voltage change the results of the double-slit experiment with PHOTONS?"

Remi

Please read the full question: he thinks it has to do with the slits acting as resonators?!

Harry, why don't you read the question, as you seem have done, correct it where necessary, and then give a reasonable answer?

* "...does the diffraction pattern form because of the interaction between the incoming photons and the photons of the cavity?"

A slit here is not a cavity. Rather, it is a narrow waveguide, with a high threshold  for incoming light (assuming the slit width less than the light's wavelength). Such a waveguide does not allow the light  to propagate along unattenuated. The light's intensity exponentially decays down the waveguide, and it can only squeeze through the slits if the screen is sufficiently thin. Farther, it would be totally wrong to think of the double-slit experiment in terms of interactions between different photons.  We can perform the experiment using only one photon at a time, and this single photon takes all possible paths (in our case - passes through both slits at once) and interferes with itself on the secondary screen. In particular, this interference determines the forbidden points on the screen, where the photon cannot land (dark fringes), and preferable points where it lands with highest probability (bright fringes). I do not see how and why the applied voltage could affect this process. As Remi Cornwall said quite correctly, the photons here are not affected by the external electromagnetic potential (in contrast with electron diffraction (Aharonov-Bohm effect) )     

Oscar

Then you explain to us what his idea is of a slit acting as a resonator and photons waiting for an with new photons coming?

Oscar

To disprove directly his main subidea: a photon interferes with itself. For this he can go to the answers in several similar questions here at the ResearchGate on double-slit and single photon interference. Secondly, where is the resonator? the wavelength is much smaller than the slit.