Dodgy step in the Far field approximation

  1. Dec 16, 2012 #1
    The Fresnel diffraction integral is:

    [itex] A(x_0 , y_0 ) = \frac{i e^{-ikz}}{λz} \int \int dx dy A( x , y ) e^{\frac{-ik}{2z} [(x - x_0)^2 + (y - y_0)^2]} [/itex]

    When we want to obtain the Fraunhofer diffraction integral from here, we need to somehow convert it to:

    [itex] A(x_0 , y_0 ) = \frac{i e^{-ikz}}{λz} \int \int dx dy A( x , y ) e^{\frac{+ik}{z} [x x_0 + y y_0]} [/itex]

    So I thought we should do it as follows:

    [itex] \frac{-ik}{2z} [(x - x_0)^2 + (y - y_0)^2] = \frac{-ik}{2z} [x^2 + x_0^2 + y^2 + y_0^2 - 2x x_0 - 2y y_0 ] [/itex]

    And then it seems that we should neglect: [itex] x^2 + x_0^2 + y^2 + y_0^2 [/itex] since they're all much smaller than z.
    Then we get the correct solution.

    But I don't see why we could do that, and leave out the [itex] - 2x x_0 - 2y y_0 [/itex]. After all they are of the same order... Please help!
     
  2. jcsd
  3. Dec 16, 2012 #2
    There might be an assumption that the aperture is small compared to the image space (x0,y0). Considering this is a far-field approximation, that tends to make sense.
     
  4. Dec 17, 2012 #3
    Thanks,

    It does, but then we couldn't neglect [itex] x_0^2 + y_0^2 [/itex]
     
  5. Dec 17, 2012 #4

    mfb

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    Those terms do not depend on the integration variables, it is possible to pull them out of the integral. They give a prefactor, which might be irrelevant, or accounted for in some other way.
     
  6. Dec 17, 2012 #5
    They're just a part of a phase! Got it. Thanks :)
     
  7. Dec 17, 2012 #6
    Hold on, but wouldn't that mean that Fraunhofer approximation works best away from the optical axis - where we're allowed to say: [itex] x_0 , y_0 >> x , y [/itex] ? (I don't think that's the case)
     
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