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I(x, y, D) = |U(x, y, D)|^2 =((4*|U0|^2*b^2)*(cos(k*x*a/2*L))^2)/(L^2*D^2)
where a=0.001,L=5e-7,D=1,U0=1, b=pi*r^2(area of the pinhole), r=0.0001, k=2*pi/L. x and y are the horizontal and vertical dimensions of the fringes on the viewing screen. i have attached an image of how the plots look like.
Thanks a lot!
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Accepted Answer
KSSV
on 11 Feb 2021
You have to proceed something like below:
x = -1:0.1:+1 ;
y = -1:0.1:+1 ;
[x,y] = meshgrid(x,y) ;
U0 = 5 ;
k = pi/4 ;
L = 2 ;
D = 2 ;
a = rand ;b = rand ;
U =sqrt(((4*U0^2*b^2)*(cos(k*x*a/2*L)).^2)/(L^2*D^2)) ;
imagesc(U)
colormap(gray)
You check the code and extend it to your case.
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