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Double slit experiment

  1. Jan 28, 2010 #1
    I'm reading the double slit experiment (Cohen-Tannoudji chapter 1 - complement D) and it is said:
    "We are going to consider Young's double slit experiment again to demonstrate how complementary and uncertainty relations are intimately related. Assume that the plate P, in which the slits are pierced, is mounted so that it can move vertically in the same plane. Consider a photon, its momentum changes when it crosses P. Conservation of momentum implies that the plate P absorbs the difference, but the momentum thus transferred to P depends on the path of the photon." Then the text demonstrates that the position of P is only known to within [tex]\Delta x>\frac{\lambda d}{a}[/tex] where a is the separation of the slits and d is the distance between P and the screen where the photon strikes.
    Then the text says: "But [tex]\frac{\lambda d}{a}[/tex] is precisely the fringe separation we expect to find on the screen. If the vertical position of the slits is defined only to within an uncertainty greater than the fringe separation, it is impossible to observe the interference pattern."
    I don't understand why I can't see the interference pattern... please help!!
     
  2. jcsd
  3. Jan 28, 2010 #2
    If the fringe separation on the interference pattern would be 'd' if the experiment was performed while P is at a constant position, then those fringes would be smeared into one another if P were allowed to have random fluctuations of size 'd'.

    I don't have that book so I haven't seen the diagram, but I'm assuming they are talking about the fact that if P is loose, then it will be randomly knocked about by the photons?

    Torquil
     
  4. Jan 29, 2010 #3
    Oh... it was so simple :smile:
    Thank you!
     
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