Fraunhofer diffraction of circular aperture

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Hello,
I am trying to see how a speckle pattern changes with distance after it scatters from a rough surface.
I am using Fraunhofer in its Fourier form, but I do not understand where distance comes into play. For example, the code below
Lambda=.633; %Wavelength of laser (micron)
D=50; %Diameter of aperture (micron)
Z_Meters=.1; %Screen distance in meters
Z=Z_Meters*10^6; %Screen distance in microns
MeshSpacing=1; %Sampling across aperture (micron)
MeshSize=200; %Size of Screen (micron)
% Calculate and show the Amplitude across the aperture.
[XGrid,YGrid]=meshgrid((-MeshSize/2:MeshSpacing:MeshSize/2),(-MeshSize/2:MeshSpacing:MeshSize/2));
R=sqrt(XGrid.^2+YGrid.^2);
A=R<=D/2; %The Amplitude across aperture (plane wave has constant amplitude, phase)
figure;imshow(A);title('Amplitude Across Aperture')
% Do the the 2D fft and show the result.
U=fftshift(fft2(A));
I=abs(U).^2;
figure;imshow(I,[]);title('Intensity at Screen')
I am only taking the FT of the circular aperture, how would my output change if instead of 0.1 meters from the screen, I was at 100 meters? I do not see how I can include the distance anywhere. Does someone know the answer?
Cheers.

Accepted Answer

Image Analyst
Image Analyst on 9 Apr 2020
The Fraunhofer diffraction pattern is the Fourier Transform of the aperture times the illumination pattern at the aperture, which is what you get at a distance of infinity from the aperture. The Fresnel pattern is what you get at non-infinite distances. I don't know the formula for the Fresnel equation off the top of my head (despite having a Ph.D. in optics), but you can look it up, like maybe here in WIkipedia
  7 Comments
Image Analyst
Image Analyst on 25 Apr 2020
I don't know of any off the top of my head.

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