NASA has just released a telescope alignment evaluation image from JWST for the star HD 84406 2MASS J17554042+6551277. It looks like this:

JWST Alignment Image

Higher resolution at Wikipedia

To my untrained eye, the star looks like it would look when viewed from a mirror that has a small smudge at the center. There are multiple lines at 45 degree originating from the star (and also couple of fainter ones at left-center and bottom-center of image).

Is this because of the brightness of the star (it is "only" 258 about 2000 light years away)? Or are there some image post-processing steps that have not been applied to the image? Would the star look the same, if, for example, Hubble Space Telescope photographed it?

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    $\begingroup$ I read somewhere that the JWST telescope alignment evaluation image was intentionally overexposed (by an extremely large amount) so as to allow the ground team to see if the expected diffraction pattern emerged. It did, and that appearance was deemed to indicate successful fine alignment. Unfortunately, I cannot find that article anymore. $\endgroup$ Mar 17 at 20:10
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    $\begingroup$ This looks absolutely gorgeous. It's also the first good news I've read in months. Thanks. $\endgroup$ Mar 18 at 17:12
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    $\begingroup$ Note that the star in this image is not HD84406. HD84406 is a star in the constellation Ursa Major of apparent magnitude V=+6.94. But the star in the image is actually 2MASS J17554042+6551277 (Tycho-2 4212-1079-1). It is a star in the constellation Draco of apparent magnitude V=+10.95. Best regards. $\endgroup$
    – Albert
    Mar 21 at 9:55
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    $\begingroup$ The source explaining that this is an image of the star 2MASS J17554042+6551277 and not the star HD 84406 is: nasa.gov/press-release/… The distance to 2MASS J17554042+6551277 is 1988 light-years, not 258 light-years. Best regards. $\endgroup$
    – Albert
    Mar 21 at 14:41
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    $\begingroup$ @Albert Thanks for that information. I have updated the question $\endgroup$
    – RedBaron
    Mar 22 at 10:13

2 Answers 2


A quick check by pasting the image into PowerPoint and rotating a line shows that the spikes have threefold symmetry; they're at -30°, 30° and 90°.

This is exactly what you would see from diffraction by the "spider web" of the dark edges that separate the 18 hexagonal subunits of the primary.

But it's also exactly what you would see from a single giant hexagonal aperture.

The devil is in the details, since the pattern will change depending on how wide of a range of wavelengths is being passed, which will tend to smear out some aspects of the power spectrum.

The secondary mirror is supported by a spider with elements at 60°, 90 and 120°. The three diffraction spikes they will produce will be perpendicular to them, but also spaced every 30° degrees rather than every 60°.

I took the Fourier transform of the monochrome image illustrating JWST's clear aperture from @pela's answer and we can instantly see similarities.

The horizontal spike at 0° is the diffraction pattern of the vertical element of the spike, and the light/dark banding in it (characteristic of slit diffraction) is nicely reproduced.

The other two that should appear at at +/- 30° are hidden under the sixfold star pattern of the mirror's "hexagonal theme".

How much of the six-pointed star's power is from the "spider web" of the internal gaps between elements versus the mirror's external jagged edge versus just a big giant hexagonal hole? It's difficult to say without a more careful analysis with a full model.

The spotted pattern within the arms of the stars in some images below will smear out once it is averaged over wavelength (smearing the power spectrum by scaling radially)

JWST ugly simulation

Here's the files processed

and here's the script:

import numpy as np
import matplotlib.pyplot as plt
from scipy.ndimage import gaussian_filter

fnames = 'rIUME.png', 'modified_1.png', 'modified_2.png', 'big_hex.png'
imgs = [(plt.imread(fname)[:, :527, :3].sum(axis=2) / 3. > 0.5).astype(float)
        for fname in fnames]

for img in imgs:
    img[:60] = 0. # blank out text

# img = gaussian_filter(img, sigma=1, mode='mirror', order=0) doesn't change conclusion

imgs = [img - img.mean() for img in imgs] # reduces zero frequency strength

# s0, s1 = img.shape
# w = np.hanning(s0)[:, None] * np.hanning(s1) # windowing not necessary in this case

fts = [np.fft.fftshift(np.fft.fft2(img)) for img in imgs]

powers = [np.abs(ft)**2 for ft in fts]
log_powers = [np.log10(p/p.max()) for p in powers] # log power

fig, axes = plt.subplots(len(log_powers), 2)
for img, lp, row in zip(imgs, log_powers, axes):
    row[0].imshow(img, cmap='gray')
    row[1].imshow(lp, vmin=-6, cmap='afmhot')
    for ax in row:
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    $\begingroup$ This is a lovely presentation; extra points for including the image links and source code! $\endgroup$ Mar 18 at 15:33

Any telescope will have diffraction of the light due to the edges between mirror and non-mirror. This sets a fundamental limit to the telescope's resolution, given by the size of its primary mirror. Earth-based telescopes will usually not have this problem, since the atmospheric seeing will dominate, but James Webb is in space, and so is "diffraction-limited".

The diffraction pattern depends on the shape of the mirror, as well as on anything that is in front of the mirror, i.e. the secondary mirror and the arms holding it (called "struts" or "spider"). In general, an edge will result in spikes perpendicular to that edge. A round mirror, such as Hubble's, results in concentric rings, but its secondary mirror is mounted on a plus-shaped spider, and hence point sources observed with Hubble will have plus-shaped spikes, as seen below:

HSTpattern NGC 6397 globular cluster. Credit: NASA/ESA/H. Richer.

James Webb's mirror is made up of 18 hexagonal segments. The edges of the outermost mirrors hence all follow three different directions, which are aligned at 60° to each other. This is the reason for the six brightest spikes.

Additionally, Webb's secondary mirror is mounted on three arms. The two lower arms are actually aligned with the hexagonal pattern (on purpose, I presume), so the diffraction caused by those fall on top of four of the six spikes. But the upper boom is vertical in the images, and thus gives the fainter, horizontal spikes.

You can see the alignment here on this "selfie" (created using a specialized pupil imaging lens inside of the NIRCam instrument that was designed to take images of the primary mirror segments instead of images of space):

JWSTselfie Credit: NASA.

Below you see the resulting diffraction pattern from various spiders:


Diffraction is seen for point sources, which will typically mean stars, because they are bright. The brighter a source is, the brighter the spikes will be. If you exposed for long enough, you'd also see a diffraction pattern from the fainter point sources. Almost all the other sources in this image are galaxies. In principle you also see diffraction from extended sources, but as John Doty comments the pattern will be convolved with the surface brightness of the source, smearing out the spikes.

The bright galaxy seen ~edge-on at ~10 o'clock in the JWST image is a very distant one, with a redshift of 0.285 (source: Grant Tremblay's tweet), placing it at a distance of almost 3.8 billion lightyears.

  • $\begingroup$ You don't get diffraction spikes for non-point sources right? $\endgroup$
    – ProfRob
    Mar 17 at 22:44
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    $\begingroup$ @ProfRob You get the diffraction pattern, but it's convolved with the source's intensity distribution. If that intensity distribution isn't point-like, the pattern is spread out, not so spiky, and thus more difficult to see. $\endgroup$
    – John Doty
    Mar 17 at 23:19
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    $\begingroup$ I only have one minor nit-pick about "any telescope". IIRC, diffraction spikes only happen on mirrored telescopes with spider veins holding a secondary mirror; i.e. not refractors. $\endgroup$
    – coblr
    Mar 18 at 18:15
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    $\begingroup$ @coblr In pela's answer he only stated any telescope would have diffraction effects, not that it would have diffraction spikes. Although a refractor won't have diffraction spikes, the image will still show diffraction effects. For a simple circular aperture, the image will have a series of unevenly spaced rings. $\endgroup$ Mar 18 at 20:18
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    $\begingroup$ @Stephen C. Steel, thanks for pointing that out and pela for the edits. $\endgroup$
    – coblr
    Mar 22 at 0:49

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