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Diffraction of a extensive source

  1. Aug 8, 2010 #1
    1. The problem statement, all variables and given/known data

    There is an extensive source (since it isn't important it can be monochromatic). The waves goes through a condenser lens that makes them converge. So, what we have is the "image" reproduced at the other side of the lens, at the focal plane.
    So now: I understand that each dot of that image works again as a point source. Is that correct? (I assume that's correct.)

    Now supose we have a single slit at the focal plane.
    ---->How can I solve the diffraction?<---- (I'll use, for example, the kirchhoff integral in the Fraunhofer aproximation).

    2. Relevant equations

    Kirchhoff integral in fraunhoffer aproximation:

    [tex] E= \frac{-1}{2 \pi \lambda} [/tex] [tex] \frac{e^{i(wt-kr)}}{r}[/tex] [tex] \frac{1+ \cos( \Theta )}{2} [/tex] [tex] \int_{ }^{ } \int_{S}^{ } E_{incident}(r') [/tex][tex]e^{ik\frac{x}{r}x'}[/tex] [tex]e^{ik\frac{y}{r}y'} dx' dy' [/tex]

    [tex]E_{incident}(r') = [/tex] [tex]\frac{A}{z'} [/tex] [tex]e^{i(wt-kz') + ik [/tex] [tex]\frac{(x'^2+y'^2)}{z'} [/tex]

    3. The attempt at a solution

    Considering that I have a single source at each point of the slit,
    trying to solve the integral, I found that I cannot evaluate the waves in the slit, because the waves cannot be evaluated at the convergence point.

    So, what to do?

    I dont know why, but I think in consider the waves at the focal plane as a plane wave. Does that make any sense?
    Last edited: Aug 8, 2010
  2. jcsd
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