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Young's Double slit Experiment

  1. Nov 6, 2007 #1
    i was going through the standard proof of Young's experiment using electric field concept .what the proof did is took two source [itex]S_{1}[/itex] and [itex]S_{2}[/itex] in the same line(asuuming such points do exist) and took a point [itex]P[/itex] in the same plane as those points and found out the field at the point .
    1.Was this done assuming hugyen's principle that is the points on the slit behave as secondary wavelets .otherwise i don't find reason why there are 'two' fields at the same place due to one source(the one which emits light actually)

    2. also doesn't the condition that fr each point A in slit 1 there exista a point B in slit 2 which lies on the vertical drawn from B guarantee that they are of same length geometry...t's like a bijective function from one set to other

    3.also why do we take point which lie in plane with the point P only to calculate the net field why not the contributions due to others
     
  2. jcsd
  3. Nov 6, 2007 #2
    another doubt
    4. in my book the author starts with the equation of waves originating from the slits [itex]S_{1}[/itex] and [itex]S_{2}[/itex] as [tex]E=E_{01}\sin(kx-\omega t[/tex] and so
    this means since the ditance between slits and source is large the waves are assumead to be plane waves
    clearly here the distance x is measured in the direction of propagation vetcor
    and i beleive the propagation vetcors of the the waves 'originating' from either slits is same (though i don't get why but still can't argue why they can't be the same) so why does the author tell the difference between the S1P and S2P as the path difference when actually it should be the difference between the length of perpendiculars drawn from S1 and S2 to the wavefront passing through P
     
  4. Nov 7, 2007 #3

    Claude Bile

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    Typically in the "ideal" two slit experiment, the slits are assume to be infinitely thin. This allows two-slit diffraction to be deconvolved from single-slit diffraction (slits of finite width will produce a diffraction pattern that is a convolution of the two). If one does this, then one can regard each slit as being a single point source (via Huygens' principle).

    To obtain the diffraction pattern, one simply needs to find the sum of the waves as a function of position.

    With regard to the propagation vectors - There is only a single incident propagation vector since the incident wave is a plane wave. The wave emerging from the aperture however will contain a whole spectrum of propagation vectors. Regarding the slits as point sources and calculating the path difference remains the easiest way to derive the diffraction pattern.

    Claude.
     
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