Fraunhofer diffraction simulation in Matlab

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.

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.

Answer 1

[Justin James Hyatt]

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)

[Lasse Thurmann]

The 1000 is most likely just a conversion from m to mm. But it sucks that it isn't stated explicitly.