Summary

Image plane of a flat-top beam under the effect of the first 21 Zernike polynomials. The beam starts in the so-called Fourier plane, which is imaged onto the image plane by an ideal lens.

A lens of focal length f is a Fourier-transform device between the place at distance f before it and the plane at distance f after it. These planes are called "Fourier" and "Image" plane respectively.

(*https://mathematica.stackexchange.com/questions/33574/whats-the-\
correct-way-to-shift-zero-frequency-to-the-center-of-a-fourier-transf*)
fftshift[dat_?ArrayQ, k : (_Integer?Positive | All) : All] := 
 Module[{dims = Dimensions[dat]}, 
  RotateRight[dat, 
   If[k === All, Quotient[dims, 2], 
    Quotient[dims[[:w:k]], 2] UnitVector[Length[dims], k]]]]

ZernikePoly = 
 Table[Table[
     If[m < 0, 
      Sqrt[(2 (n + 1))/(\[Pi] (1 + KroneckerDelta[0, m]))]
        ZernikeR[n, -m, \[Rho]] Sin[m \[Theta]], 
      Sqrt[(2 (n + 1))/(\[Pi] (1 + KroneckerDelta[0, m]))]
        ZernikeR[n, m, \[Rho]] Cos[m \[Theta]]], {m, -n, n, 2}], {n, 
     0, 5}] /. {\[Rho] -> Sqrt[x^2 + y^2], \[Theta] -> ArcTan[x, y]} //
   FullSimplify

(*The flat-top beam size (HeavisideTheta) has radius 1*)
(*The Fourier plane is taken to be much larger than the aperture size*)
(*The Fourier transform is the result of the imaging setup*)
(*Each Zernike polynomial is simulated with a coefficient of 2*)

Column[Map[
  ArrayPlot[
    Abs[fftshift[
      Fourier[Table[
        HeavisideTheta[1 - Sqrt[x^2 + y^2]]/Sqrt[\[Pi]]
          E^(-I 2 #), {y, -20, 20, 0.051}, {x, -20, 20, 0.051}]]]]^2, 
    PlotRange -> {{393 - 60, 393 + 60}, {393 - 60, 393 + 60}}, 
    PlotTheme -> "Scientific"] & , ZernikePoly, {2}], Center]
Export["ZernikeAiryImage.jpg", %]

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