In firing a single photon at the center divider of a double slit, does it ALWAYS go through the slits? If we think of a single photon approaching the slits as a wave function, and we fire the photons at the midpoint of the two slits, one at a time, then I would think the probability function is highest at this midpoint.  That would mean that, depending on the separation of the two slits, some of the photons would impact the wall separating the slits.  If I fire a photon at a thin piece of material, why would it not at least occasionally 'hit' the material, (with a possible photoelectric effect).  How is it that ALL the photons avoid this 'wall' and pass through one slit or the other?
 A: Short answer: When the photon hits the walls or the web of the double slit, obviously an interaction takes place. The photon is absorbed and finally EM radiation is re-emitted.
Long answer tl;dr: What happens at the boundary regions between the slit walls? The photon passes through unhindered? No, it also has an interaction. It gets polarized, the electric and magnetic field components are aligned. Clearly to prove by the fact, that, if one aligns two polarizers by 90° to each other, no more photons pass through. But if you place a third polarizer between them at 45°, then light passes through. And this is only possible if the field components of the photon are rotated by the polarizers.
The question whether photons are reflected at the center bar is understandable, but more interesting is what happens between the edge region and the photon. Instead of reflection, transmission occurs. And here the interaction has a direct influence on the intensity distribution on the observation screen. Instead of intensity distribution one could also speak of diffraction pattern. But not of interference pattern, because photons (almost) never interfere. The edges diffract the photon in such a way that it hits certain parts of the screen with higher probability and others rather not. It would be to investigate in how far phononic and other lattice oscillations produce the selective deflection of photons (as well as electrons). But of course it is easier to calculate with the wave property and interference of the particles in the double-slit experiment.
A: Don't try to analyze a diffraction problem using particles: you'll just confuse the issues.
Thinking of waves, well, what happens when waves impinge on opaque material? You see this every day. Some of the energy is reflected, some is absorbed. There's nothing particularly profound beyond this that happens in the double slit experiment. For narrow slits, you may need to account for polarization, but classical electrodynamics is fine for that, no quanta needed.
A: You are correct. To the left of the "wall with slits" we typically approximate that we are illuminating the wall/slits with a plane wave. If the wall is much bigger than the diameter of the beam then what matters is the fraction $\eta$ of the slit area compared to the total beam area. It is only this fraction $\eta$ of photons that make it through the slits and allow us to perform an interference experiment. If you put a big power meter before and after the slits you will see a power reduction by an amount $\eta$. If you're performing an experiment where you try to send photon or electron waves through the slit if you have a flux of 1 photon or electron per second but $\eta$ is 0.01 then it will take on average 100 s for a single point to be detected on your detection screen.
