Fraunhofer diffraction of circular aperture

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Lucrezia Cester
Lucrezia Cester 2020년 4월 9일
댓글: Image Analyst 2020년 4월 25일
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.

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Image Analyst
Image Analyst 2020년 4월 9일
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
  댓글 수: 7
Lucrezia Cester
Lucrezia Cester 2020년 4월 24일
Would you have a good book to recommend that talks about this specificalities?
Image Analyst
Image Analyst 2020년 4월 25일
I don't know of any off the top of my head.

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