Fraunhofer diffraction simulation in Matlab

riharenda009 · Feb 14, 2018

Hello, below is the screenshot from the book INTRODUCTION TO MODERN DIGITAL HOLOGRAPHY With MATLAB (TING-CHUNG POON, JUNG-PING LIU). This book is free to download. There is a script for Fraunhofer diffraction pattern and equation 1.40 is Fraunhofer diffraction formula in terms of Fourier transform.

$Book page showing MATLAB code for Fraunhofer diffraction of a circular aperture (Table 1.3), with equation 1.40 for the Fraunhofer diffraction pattern below the code.$

I don't understand commands C=C*lambda*z/M/delta*1000 and R=R*lambda*z/M/delta*1000;. They are probably scaling coordinates in image plane, but I don't know how. Can anybody explain it? Thank you very much. Here are the results of the simulation. Coordinates are in milimeters.

3D surface plot in MATLAB of an intensity pattern: a sharp central peak reaching about 0.9 on a 0 to 1 vertical axis, surrounded by faint concentric ring ripples on a flat dark blue base, axes labelled 0, 5, 10, 15.

$Grayscale plot titled "Modulus of the Fraunhofer diffraction pattern": a bright central spot surrounded by concentric rings that fade outward, with fainter speckled rings beyond.$

Dr Transport · Feb 14, 2018

if you comment out the C and R at the bottom and re plot the data, you'll find that the x-y coordinates vary between 1 and 512, they get the image back into reasonable coordinates for an image plot.

Fraunhofer diffraction simulation in Matlab

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