I was reading this question about the JWST's diffraction spikes, and I was rather surprised by the magnitude of the 4 sets of diffraction spikes. JWST image diffraction spikes. 6 big spikes in a hexagonal pattern and two smaller spikes orthogonal The large hexagonal spike pattern I believe is formed from the honeycomb shape of the primary mirrors, while the small horizontal spikes are formed from the one supporting truss that isn't aligned with the hexagonal axes.
enter image description here

I would have expected the vertical arm to have a much greater diffraction spike size than the honeycomb shape of the primary mirrors. Why is this not the case?

Furthermore, how would I go about calculating the diffraction spike pattern for an arbitrary 2d secondary mirror support structure shape observing a point source? My intuition tells me that this will be some kind of integral transform. There at least ought to be a way of approximating it without having to calculate the full wave equation propagating through the lenses.

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    $\begingroup$ I'm not sure how much of my answer here also applies here as well. "I would have expected the vertical arm to have a much greater diffraction spike size than the honeycomb shape..." "greater" can have two components; strength or intensity, and length or extent. The wider the feature, the tighter its diffraction pattern. So very thin features will diffract out much farther, but those spikes will be weaker in total power and much weaker in power per pixel. $\endgroup$
    – uhoh
    Mar 18, 2022 at 1:37
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    $\begingroup$ @uhoh I read that question before you posted your answer, and yes, it does basically answer mine too. Should've known it was a fourier transform! $\endgroup$
    – Ingolifs
    Mar 18, 2022 at 3:40
  • $\begingroup$ Technically what we see is the abs(FT)^2 = power since the photodetectors don't measure the electric field directly. That also means all the cool information stored in the phase of the FT is lost, which is one thing that makes back-calculating wavefront errors from point spread functions so challenging sometimes. But I think your question here can and should have a more complete and informative answer than that quick python script. $\endgroup$
    – uhoh
    Mar 18, 2022 at 9:16
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    $\begingroup$ As @uhoh shows, the diffraction is dominated by the edges from mirror to non-mirror, while the mirror-to-mirror edges can almost be neglected. If you count all edges going in a given direction and compare their total length to that of the mid-boom, you'll see that mirror edges dominate. $\endgroup$
    – pela
    Mar 18, 2022 at 10:49
  • $\begingroup$ You might be interested in looking at WebbPSF which is used to simulate the Point Spread Function for the JWST Exposure Time Calculator which makes uses of POPPY (Physical Optics Propagation in PYthon) which simulates the Fraunhofer and Fresnel diffraction $\endgroup$ Mar 18, 2022 at 23:45

1 Answer 1


Diffraction is easily (in the sense that your college professor will give it as a homework problem :-) ) calculated given the structure of any aperture. If you start with any optics textbook and read about Fresnell and Fraunhofer zones, you'll get the basic idea. For complicated structures, the solution is basically a superposition of the diffraction pattern from each aperture (for example, a single-slit pattern applied repeatedly for a row of identical slit apertures).


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