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I have a doubt about how double slit experiment is made.

Let's think about the perforated wall, what are the requirement for it?

Can a photographic plate could be used as a wall ?

I see a problem here, as a photographic plate serve also as a detector, then the single photon experiment could end up with a single point in the wall that contains the slits and not in the plate located behind the slit's wall.

Of course the "one photon count", is knowing the emition before the slits (if this count were made after, it will be a detector making a measurement and would simply reflect the particle effect)

The wall with the slits could absorb the photon(removing electrons from the material) before something happen on the other side, if that happened, there would be no experiment, the photon won't pass through the slits, just because its energy was already used (supposing all energy is used and the photon doesn't split).

There must be some kind of specification about the material of the wall, need to be special, in the sense of needing a higher energy (than the test photon) to pickup electrons, then the light won't stop there, for that, I think the wall can't be made of a photosensible material, but I don't know, and here I am asking..

thanks for any answer

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I still feel that you are asking about two problems at once---the measurement problem and the behavior of the two slit system---and I don't think that this is a useful way to proceed. Understand each of them in isolation and then you'll find that you do not need to ask about the combined system. – dmckee Mar 4 '11 at 19:54

It doesn't matter whether the wall is photosensitive or not. If the photons' wave functions give some probability of hitting the wall and being absorbed, then some of the photons will hit the wall and be absorbed, whether they produce some chemical reaction in a photographic plate or just heat the wall.

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I see... then there is something missing in usual descriptions of this experiment, I couldn't find references to a photon hitting first wall in the "one photon test", thanks – HDE Mar 4 '11 at 13:41
@HDE No, nothing is really missing. If you actually do the calculations, you'll find that there is no difference at all between a wall that absorbs the photons that miss the slit and a wall that reflects those photons. It just does not matter what the properties of the wall are. I invite you to calculate the interference patterns for yourself if you doubt this. – spencer nelson Mar 4 '11 at 17:42

1. Of course wall sometimes absorbs the photon (light do not evade it in any way). To have interference, you need to have no information (in terms of excitation, creation of photographic grain of whatever) that could be used to say which path the photon 'has chosen'.

2. It is fine to use foil. See e.g. (so called Fresnel Zone Plates on film or polygraphic foil). Want to test it by yourself? Make an analog photography of two dark lines, and then use film as a diffractive agent. Or - in any DTP studio post a PostScript file e.g. from -> (don't be afraid it is in Polish; just download a few first ps/pdf files).

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+1, yes I will try to do it myself ( I don't understand Polish but thanks anyway) – HDE Mar 4 '11 at 13:45

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