<p>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).</p>

CC BY-SA 4.0
2022-03-18T12:59:29.893Z
2022-03-18T12:59:29.893Z
Carl Witthoft
<p>I was reading <a href="https://astronomy.stackexchange.com/questions/48836/why-does-alignment-evaluation-image-from-jwst-look-like-this">this question</a> about the JWST's diffraction spikes, and I was rather surprised by the magnitude of the 4 sets of diffraction spikes.
<a href="https://i.sstatic.net/3QsgW.jpg" rel="noreferrer"><img src="https://i.sstatic.net/3QsgW.jpg" alt="JWST image diffraction spikes. 6 big spikes in a hexagonal pattern and two smaller spikes orthogonal" /></a>
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.<br />
<a href="https://i.sstatic.net/rIUME.png" rel="noreferrer"><img src="https://i.sstatic.net/rIUME.png" alt="enter image description here" /></a></p>
<p>I would have expected the vertical arm to have a much greater diffraction spike size than the honeycomb shape of the primary mirrors. <strong>Why is this not the case</strong>?</p>
<p>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.</p>

CC BY-SA 4.0
2022-03-18T01:37:10.187Z
I'm not sure how much of my answer [here](https://astronomy.stackexchange.com/a/48843/7982) 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.
uhoh
CC BY-SA 4.0
2022-03-18T03:40:33.300Z
@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!
Ingolifs
CC BY-SA 4.0
2022-03-18T09:16:31.903Z
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.
uhoh
CC BY-SA 4.0
2022-03-18T10:49:58.413Z
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.
pela
CC BY-SA 4.0
2022-03-18T23:45:11.563Z
You might be interested in looking at [WebbPSF](https://webbpsf.readthedocs.io/en/latest/) which is used to simulate the Point Spread Function for the JWST Exposure Time Calculator which makes uses of [POPPY](https://poppy-optics.readthedocs.io/en/latest/) (Physical Optics Propagation in PYthon) which simulates the Fraunhofer and Fresnel diffraction
astrosnapper
CC BY-SA 4.0
2022-03-19T02:50:58.560Z
@pela I still think there's a discrepancy in that the shadow of boom seems to have much less impact in the actual star image than in the FFT of the mirror 'selfie' .
Roger Wood
CC BY-SA 4.0
2022-03-19T12:31:59.987Z
@RogerWood As far as I understand how the selfie was taken, that wouldn't be a plane wave, so I wouldn't expect the diffraction to be similar.
pela
CC BY-SA 4.0
2022-03-19T17:29:27.860Z
@pela yes, I think the difference is that the diffraction around the booms does not occur at the surface of the mirror and the wavefronts do not interfere constructively along the length of the edges - as the they do coming from the edges on the surface of the mirror.
Roger Wood
<telescope><optics><hubble-telescope><james-webb-space-telescope>
General process of calculating what a telescope's diffraction spikes would look like?
Ingolifs
# General process of calculating what a telescope's diffraction spikes would look like?

(Asked by: username_0 on 2022-03-17)

I was reading [this question](https://astronomy.stackexchange.com/questions/48836/why-does-alignment-evaluation-image-from-jwst-look-like-this) 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](https://i.sstatic.net/3QsgW.jpg)](https://i.sstatic.net/3QsgW.jpg)
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](https://i.sstatic.net/rIUME.png)](https://i.sstatic.net/rIUME.png)

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.

**Tags:** telescope, optics, hubble-telescope, james-webb-space-telescope

> **Comments:**
> **username_1**: I'm not sure how much of my answer [here](https://astronomy.stackexchange.com/a/48843/7982) 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.  
> **username_0**: @username_1 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!  
> **username_1**: 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.  
> **username_2**: As @username_1 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.  
> **username_3**: You might be interested in looking at [WebbPSF](https://webbpsf.readthedocs.io/en/latest/) which is used to simulate the Point Spread Function for the JWST Exposure Time Calculator which makes uses of [POPPY](https://poppy-optics.readthedocs.io/en/latest/) (Physical Optics Propagation in PYthon) which simulates the Fraunhofer and Fresnel diffraction  
> **username_4**: @username_2 I still think there's a discrepancy in that the shadow of boom seems to have much less impact in the actual star image than in the FFT of the mirror 'selfie' .  
> **username_2**: @RogerWood As far as I understand how the selfie was taken, that wouldn't be a plane wave, so I wouldn't expect the diffraction to be similar.  
> **username_4**: @username_2 yes, I think the difference is that the diffraction around the booms does not occur at the surface of the mirror and the wavefronts do not interfere constructively along the length of the edges - as the they do coming from the edges on the surface of the mirror.  

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## 1 answers

### Answer by username_5 on 2022-03-18. Score: 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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