Dodgy step in the Far field approximation

  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. 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. Thanks,

    It does, but then we couldn't neglect [itex] x_0^2 + y_0^2 [/itex]
     
  5. mfb

    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. They're just a part of a phase! Got it. Thanks :)
     
  7. 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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