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This question has arisen out of a previous question regarding the various approximations in optical diffraction theory, and when they can each be applied.

I would like to know the steps that must be taken to numerically calculate the diffracted field observed after an arbitrary aperture geomerty, at small distances from the aperture. Specifically, for distances from the source much greater than the wavelength $z\gg \lambda$, but not far enough to satisfy the validity conditions for the Fresnel regime.

I was under the impression that the Rayleigh-Sommerfeld (RS) diffraction integral (Eq. (1) here) is basically always a good approach (as long as we are dealing with scalar diffraction, and that $z\gg\lambda$), and that for solving the above problem we should numerically integrate the RS integral?

However, it was mentioned in the comments in this question that "For near field conditions, one needs a beam propagation approach in terms of the angular spectrum". Is it true that I wouldn't be able to use the RS integral to solve such a problem, or rather is this equivalent to the RS integral, but instead working in frequency space (allowing FFT methods, etc)? Could someone elaborate on what this method is exactly, and outline the practical steps that would typically be done to solve such a problem.

Thank you

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To calculate the near field in a region where the field can include an evanescent field, one can use the angular spectrum approach. Here, I'll assume the optical field is a scalar field, as assumed for scalar diffraction theory (as opposed to vector diffraction theory).

Given an input field $f(x,y)$, which is a two-dimensional complex function, one first computes the angular spectrum, as the two-dimensional Fourier transform$^{\dagger}$ $$ F(a,b) = \int f(x,y)\exp(i2\pi(ax+by)) dx dy . $$ For propagation over a distance $z$, one needs to multiply the angular spectrum with a phase factor corresponding to the $z$-depending part of the plane wave. Let's call this the propagation phase factor: $$ \Phi(a,b,z) = \exp(-ik_z z) = \exp\left(-iz\sqrt{\frac{1}{\lambda^2}-a^2-b^2}\right) . $$ Note that if $a^2+b^2>1/\lambda^2$ then the square-root becomes imaginary. This situation represents the evanescent part of the part.

Finally, we reconstruct the field at $z$ by computing the inverse Fourier transform: $$ g(u,v) = \int F(a,b) \Phi(a,b,z) \exp(-i2\pi(ax+by)) da db . $$

This approach is rigorous and can be applied for any value of $z$.

Under the paraxial approximation one would expand the square root as a binomial expansion and thus recover the Fresnel integral (after interchanging the order of the integrations and evaluating the integral over $a$ and $b$.)


The historical approaches (Rayleigh-Sommerfeld, Kirchhoff, etc.) generally start from heuristic principles such as Huygens principle. One can show that the interchange of the order of integration here, without the paraxial approximation, gives a convolution kernel reminiscent of a spherical wave, which gives some qualitative link with the Huygens principle. However, the present way to do it gives a simpler approach based on linear systems theory that is completely rigorous. It produces the propagation phase factor directly from the $z$-dependent part of the plane wave, which is a solution of the Helmholtz equation

$\dagger$ Note my phase convention: phase increases with time and therefore it decreases with distance.

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  • $\begingroup$ Thanks for the description of what the "angular spectrum" is. Can you add additional explaination on where the Rayleigh-Sommerfeld integral fits into the story - since before reading your comment I would naively have taken that approach instead. Thank you! $\endgroup$
    – teeeeee
    Aug 31, 2020 at 10:25
  • $\begingroup$ These historical approaches (Rayleigh-Sommerfeld, Kirchhoff, etc.) generally start from heuristic principles such as Huygens principle. One can show that the interchange of the order of integration here, without the paraxial approximation, gives a convolution kernel reminiscent of a spherical wave, which gives some qualitative link with the Huygens principle. However, the present way to do it gives a simpler approach based on linear systems theory that is completely rigorous. $\endgroup$ Sep 3, 2020 at 10:48
  • $\begingroup$ So is this approach really a different thing to the Rayleigh-Sommerfeld integral? To obtain this angular spectrum propagation phase factor, do we have to go back to the Helmholz equation? Or can it be seen from RS integral? $\endgroup$
    – teeeeee
    Sep 8, 2020 at 8:49
  • $\begingroup$ In a roundabout way, you can probably get it from RS, but it comes out directly from the $z$-dependent part of the plane wave, which is a solution of Helmholtz. Personally I don't bother with RS. $\endgroup$ Sep 8, 2020 at 10:41
  • $\begingroup$ OK, I've added it in the answer. $\endgroup$ Sep 9, 2020 at 3:58

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