Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Insights Author
    2015 Award

    Staff: Mentor

    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)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Dodgy step in the Far field approximation
  1. Far Field Radiation (Replies: 1)

Loading...