Fraunhofer diffraction simulation in Matlab
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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.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/170349/image.png)
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.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/170351/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/170353/image.png)
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回答(1 个)
Justin James Hyatt
2019-3-7
You are right, It is scaling the coordinates. There is an explanation here: https://ece661web.groups.et.byu.net/notes/complex_apertures.ppt
Let me break this down:
C*lambda*z/M/delta*1000
C=the original coordinates at the aperture
lambda = wavelength
z = distance to the image plane
M = the number of data points in the aperture plane (in each dimentsion)
delta = the sampling period
1000 = I don't know what this is for.
A good check if you've got the units right is the distance from the center to the first zero in the diffraction pattern should be 1.22*lamda*z/(Aperture Diameter)
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Lasse Thurmann
2019-5-21
The 1000 is most likely just a conversion from m to mm. But it sucks that it isn't stated explicitly.
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