Fraunhofer diffraction of circular aperture

23 visualizaciones (últimos 30 días)
Lucrezia Cester
Lucrezia Cester el 9 de Abr. de 2020
Comentada: Image Analyst el 25 de Abr. de 2020
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.

Respuesta aceptada

Image Analyst
Image Analyst el 9 de Abr. de 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 comentarios
Lucrezia Cester
Lucrezia Cester el 24 de Abr. de 2020
Would you have a good book to recommend that talks about this specificalities?
Image Analyst
Image Analyst el 25 de Abr. de 2020
I don't know of any off the top of my head.

Iniciar sesión para comentar.

Más respuestas (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by