Physics Asked by JobHunter69 on April 19, 2021
I was thinking that because the source plane is infinitely large would be analogous to having the source plane being infinitely close, which would imply a Fresnel region rather than Fraunhofer.
This is too big for a comment so I'm putting it in an answer.
From
https://en.wikipedia.org/wiki/Near_and_far_field
the near field is that part of the radiated field that is below distances shorter than the Fraunhofer distance[1] $d_f = 2D^2∕λ$ from the source of the diffracting edge or antenna of longitude or diameter D.
MY ANSWER: If the aperture size $D$ is taken to be infinite (if the aperture is assumed to be the film size) then the diffraction pattern would be from the Fresnel zone--as you say.
Did he also give an aperture size that you did not put in the question?
Answered by user45664 on April 19, 2021
For Fraunhofer diffraction the angles of the waves at the observation point should be small and the distance should be large with respect to the wavelength. Since the source plane is stated to be "infinitely large" the first condition cannot be met and the solution is not the Fourier transform. Less strictly speaking "infinitely large" probably is intended to mean "ignore edge effects" in which case the answer is the FT.
Answered by my2cts on April 19, 2021
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