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Basic Double Slit Question

  1. Sep 24, 2008 #1
    So, after reading a few different explanations of the double slit experiment, there is one aspect I am confused about. The set up is that you shoot out electrons/photons/whatever at a wall with two small slits. The particles pass through one slit, the other, or both if you're not looking.

    But none of the explanations has mentioned the other possible outcomes. Walls must be made of atoms. When you fire an electron or photon at a wall of atoms, you're not always going to hit the hole. An electron might stick to a lose ion in the wall. A photon might be absorbed or it might bounce back. How are these accounted for in the amplitude distribution of the experiment? My thought was that the calculated amplitude distribution only describes the particles that successfully make it through the holes at all. This compounds with the fact that in many versions, the detector only detects a particle at a single position on the far wall, and there is no way to tell whether an individual particle even made it through a hole.

    I'm sure this is just a simplification in these "thought experiment" versions of the setup, but I just want to make sure I'm on the right track.
     
  2. jcsd
  3. Sep 25, 2008 #2
    Not completely sure if this the answer to the question you're asking. But the double slit experiment is only used to show wave interference, that coinciding waves with the same properties will constructively combine into a strong one, and waves whose crests do not match up will cancel each other out. The result of the slits is that one source of light is split into two sources which radiate light with the same properties (wavelength). When these two waves interact and interfere, and alternating pattern is formed on the surface.
     
  4. Sep 25, 2008 #3
    Right. :smile:
     
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