<p>Also see answers to</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/14351/7982">Could mirrors be replaced with CCDs?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/32521/7982">Is Digital Adaptive Optics Possible?</a></li>
</ul>
<p>The problem is that light imaging detectors convert amplitude to power during the detection process. Phase is lost.</p>
<p>If you had maps of both the magnitude of the electric field and its relative phase, and had this at each wavelength of light, you could do exactly what you propose and correct in software.</p>
<p>In fact, you could then throw away the telescope and image the sky without it.</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/32014/7982">How would a Fast Fourier Transform Telescope work without a mirror, dish or lenses?</a></li>
</ul>
<p>That's exactly how a radio telescope array works! Dozens of antennas detect waves from a patch of sky, they're all brought into a computer and for each wavelength the source shape is reconstructed from the <em>amplitude and phase</em> information available in the electronic signals.</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/13930/7982">Would Adaptive Optics be Useful in Radio Astronomy?</a></li>
</ul>
<p>But CCD detectors or photographic emulsion plates or our eyes convert the photons to other forms of energy, then average it out over time.</p>
<p>Phase is lost, so reconstruction is no longer possible.</p>

CC BY-SA 4.0
2021-05-15T22:15:01.933Z
So why don't optical telescopes use other detection methods that _do_ preserve phase?
Vikki
CC BY-SA 4.0
2021-05-15T22:25:08.960Z
@Vikki-formerlySean they will some day! Visible light, say 600 nm has a frequency of 500,000,000,000,000 Hz (500,000 GHz) which is not manageable with normal electronics. In order to look at many wavelengths at the same time rather than an incredibly sharp 1 GHz wide optical bandpass, you need to digitize at this absurdly high rate and process. Basically, if the detector preserved the phase, you'd then have to either *record* the phase in real time or *process* it. If you had an extremely narrow bandpass filter (0.001 nm) and could cut the spectrum down to GHz, you could down-convert it to
uhoh
CC BY-SA 4.0
2021-05-15T22:26:24.977Z
@Vikki-formerlySean an electronic signal in GHz range and then process normally. That's roughly the kind of thing I'd asked about in [Has optical interferometry been done at radio frequency using heterodyning with a laser in a nonlinear material?](https://astronomy.stackexchange.com/q/36633/7982)
uhoh
CC BY-SA 4.0
2021-05-15T22:33:49.580Z
@Vikki-formerlySean there are other options but I'm not fully informed of the background so can't really write about it. You could build a one-pixel optical telescope and use a 0.000001 nm filter and re-expand the wavefront, interfere with a laser in quadrature, and digitize two interferograms at kHz rates or something like that, but I'm not sure that that would be at all useful. Since [astronomical seeing](https://en.wikipedia.org/wiki/Astronomical_seeing) is somewhat achromatic we can do it at kHz rates without worrying too much about spectrum, but for this application it's a problem.
uhoh
CC BY-SA 4.0
2021-05-16T02:23:15.840Z
@Vikki-formerlySean There's work on this, but it's hard: https://www.scientificamerican.com/article/quantum-astronomy-could-create-telescopes-hundreds-of-kilometers-wide/
John Doty
CC BY-SA 4.0
2021-05-16T04:07:58.633Z
At least as I understand it, adaptive optics in telescopes don't work by processing the image after capture.  Instead, they measure the distortion from the atmosphere, and use computers & actuators to slightly change the shape of the mirror to compensate. So it would seem that you could in principle use the same technology to change the shape of an "almost perfect" mirror to be closer to perfect.
jamesqf
CC BY-SA 4.0
2021-05-16T07:46:44.187Z
@jamesqf certainly "...and then correct for the distortion *using computers?*" provides substantial latitude in how you might use those computers. It's true you could spend 10x the money for a deformable mirror, a few hundred (or more) actuators or you might be able to design a correcting element like they did for Hubble using computers, but I took it to mean apply the computers to the images obtained by conventional means, as did (at)zabop's answer. I think you can certainly post your suggestion as a new answer, since the use of computers is involved...
uhoh
CC BY-SA 4.0
2021-05-16T15:42:06.723Z
Phase information doesn't help at all with normal light. You's need a coherent light source and pixels smaller than the wavelength for this. What is needed is directional sensitivity in each pixel, like it is used in commercial light field cameras.
asdfex
CC BY-SA 4.0
2021-05-16T18:20:48.640Z
@asdfex there is no such thing as completely coherent nor completely incoherent light. It's a continuum; lasers are a lot more coherent than light bulbs, but light bulbs have non-zero coherence and lasers are not perfectly coherent. Then you have to address longitudinal and transverse coherence separately. ***There is no such thing as "normal light".*** There's just photons. As long as you read through what I wrote carefully and note the filter bandwidths I specified in each case, you absolutely can heterodyne filtered light with a laser and down-convert.
uhoh
CC BY-SA 4.0
2021-05-16T18:21:56.407Z
It's no different than heterodyning "normal radio signals from a radio or TV transmitter with a stable local oscillator do down-convert them to an IF,  a detection or demodulation circuit and then to audio. cf. [Radio over Fiber](https://en.wikipedia.org/wiki/Radio_over_fiber)
uhoh
CC BY-SA 4.0
2021-05-16T18:26:29.650Z
What you write about radio telescopes is fine - but it doesn't apply to optical telescopes, at least not on the length scale that is relevant here. Phase helps if your spatial precision is much better than the wavelength, but this is not the case with any current imaging technology.
asdfex
CC BY-SA 4.0
2021-05-15T21:21:45.027Z
2021-05-15T22:28:17.373Z
2021-05-15T22:28:17.373Z
uhoh
<p>The image which is recorded by CCDs are the <a href="https://en.wikipedia.org/wiki/Convolution" rel="noreferrer">convolution</a> of the true image and a <a href="https://en.wikipedia.org/wiki/Point_spread_function" rel="noreferrer">point spread function</a>. If the PSF is not a nice function, such as this one, for example <a href="https://en.wikipedia.org/wiki/Point_spread_function#/media/File:Hubble_PSF_with_flawed_optics.jpg" rel="noreferrer">[source]</a>:</p>
<p><a href="https://i.sstatic.net/kfoNM.png" rel="noreferrer"><img src="https://i.sstatic.net/kfoNM.png" alt="enter image description here" /></a></p>
<p>then (I think) it is hard to deconvolve the detected image from the PSF to get the true image. For example, you don't necesserily know the exact PSF if the telescope is only &quot;roughly&quot; the right shape, so you don't know what to deconvolve.</p>

CC BY-SA 4.0
2021-05-16T16:49:51.650Z
it should be possible to determine the PSF, if you know what the result should look like. A laser pointer shined from an airplane should suffice as a calibration pattern?
John Dvorak
CC BY-SA 4.0
2021-05-16T19:58:35.820Z
I think you are probably right.
zabop
CC BY-SA 4.0
2021-05-18T08:59:18.153Z
Deconvolution is extremely error-prone operation, any noise will corrupt resulting data exponentially. It can be somewhat done in space, where there are no air disturbances and when mirror temperature is kept stable; on Earth this is not viable solution. Sure deconvolution is used in astronomy image processing to remove diffraction and other image distortions, which cannot be avoided by making better mirrors :)
Arvo
CC BY-SA 4.0
2021-05-18T10:25:46.303Z
This image shows the problem very well - one of the main measures taken from these images is the amount of light received from the star (it's brightness).  As can be seen in the image, that light has spread over a wide area and so it has to be summed up to get the total.  However, you can also see it is now covering other objects and separating out which light came from which object is hard and introduces uncertainties.  In a perfect world with perfect mirrors and no atmosphere, each object would be contained in a single pixel and so easily measurable.
Jonathan Twite
CC BY-SA 4.0
2021-05-20T10:47:46.643Z
@JonathanTwite In the perfect world you describe there would still be point spread functions, because the diffraction limit of a finite-sized mirror. So each object would *not* be "contained in a single pixel" (unless you made your pixels absurdly large, in which case you'd have many different objects in each pixel...).
Peter Erwin
CC BY-SA 4.0
2021-05-20T11:35:44.513Z
@PeterErwin in a perfect world, you'd have infinitely large mirrors... that nevertheless don't collapse under their own gravity or get deformed by tidal effects and solar wind.
John Dvorak
CC BY-SA 4.0
2021-05-15T13:39:01.837Z
2021-05-15T13:39:01.837Z
zabop
<p>The short answer is that blurring <em>destroys information</em>, and no amount of correction after the fact can bring it back.</p>
<p>A slightly less short answer:  As others have mentioned, the blur caused by imperfections in the mirror is a convolution of the true image with the &quot;point spread function&quot; created by the imperfections.</p>
<p>Convolution of the image is mathematically equivalent to multiplying the Fourier transform of the image with the Fourier transform of the point spread function.  Wherever the transform of that point-spread function is zero or close to it, the image information at the corresponding frequencies is destroyed.</p>
<p>The point-spread function created by random imperfections looks like a Gaussian distribution.  The Gaussian distribution has the interesting property that its Fourier transform is also a Gaussian, centered on 0-frequency with bandwidth inversely proportional to the width of the point-spread function.</p>
<p>In other words, the point-spread function is a low-pass filter. It destroys the high-frequency information in the image, where all the fine detail lives.  The wider the point spread, the worse it is.</p>

CC BY-SA 4.0
2021-05-17T00:29:29.817Z
2021-05-17T00:29:29.817Z
Matt Timmermans
<p>They do corrections in computers.  But it's not a perfect thing.  Just how you can't tell the difference between 1 + 2 + 1 and 1.3 + 1.4 + 1.3, the equations we solve don't have just one solution.  They have many.  So we have to do guesses in the algorithm.  Given that science is exploring that which is just outside of our knowledge, its useful to get actual raw data rather than having to manipulate it first.</p>
<p>I think my favorite example of this is the <a href="https://www.jpl.nasa.gov/edu/news/2019/4/19/how-scientists-captured-the-first-image-of-a-black-hole/" rel="nofollow noreferrer">famous black hole picture</a> from 2019.  It was, of course, a computer reconstruction of the data, using software from a team headed by <a href="https://www.bbc.com/news/science-environment-47891902" rel="nofollow noreferrer">Katie Bouman</a>.  She has a wonderful <a href="https://www.youtube.com/watch?v=BIvezCVcsYs" rel="nofollow noreferrer">TED talk</a> on the technology where she points out the challenges with biasing algorithms to detect what we expect to see.  One of the tools they relied on to gain confidence that the algorithm wasn't just showing them what they expected to see was to train the algorithm on pictures of her friends and faculty.  If the resulting image processed after training on these pictures was basically the same as the one from generated from processing after training on pictures of known stellar objects, they were reasonably confident it wasn't bias.</p>
<p>But its always easier to just have better raw data.</p>

CC BY-SA 4.0
2021-05-16T18:27:25.840Z
Yes, that's done with radio astronomy; just as I described in [my answer](https://astronomy.stackexchange.com/a/43817/7982). But the question was about correction optical mirrors and at 500,000 GHz, so the analogy to radio astronomy doesn't work.
uhoh
CC BY-SA 4.0
2021-05-16T18:32:28.010Z
@uhoh The equations may be different, but if computerized sharpening could prevent the need for quality measurements, then we would have certainly replaced our telescopes with computers.  It's still a game of imperfect information.
Cort Ammon
CC BY-SA 4.0
2021-05-16T18:54:36.230Z
We don't have analog to digital converters that work at 500,000 GHz nor the rest of the electronics, so even you wanted to try to run the world's largest supercomputer to deal with the rate, light frequencies are just too high to treat like radio waves given the current state of the art in electronics. It may be possible some day in the not-too-distant future, but not yet.
uhoh
CC BY-SA 4.0
2021-05-17T16:01:36.110Z
1+3+1 = 5. 1.3+1.4+1.3 = 4. I think I can tell the difference.
WaterMolecule
CC BY-SA 4.0
2021-05-17T16:07:48.280Z
@WaterMolecule ... I need to stop posting math when I'm sleepy!  Fixed!
Cort Ammon
CC BY-SA 4.0
2021-05-20T05:06:35.643Z
The maths example makes no sense to me. Yes, they both add up to 4, but so does -2, 8, -2. How does this look illustrate the point you’re trying to make? Or are you trying to demonstrate the effect of rounding? If so, 1.4 rounds to 1, not 2.
Chappo Hasn't Forgotten
CC BY-SA 4.0
2021-05-20T06:38:30.823Z
@ChappoHasn'tForgottenMonica If all you have is 4, you cannot determine which of those two cases was the true un-blurred data.  You can get somewhere with a convolution to invert a point spread function, but eventually it gets to a point where there are simply more than one possibility for what the image could be.  This is especially true in the presence of noise.
Cort Ammon
CC BY-SA 4.0
2021-05-20T10:52:58.893Z
Minor nitpick: there were actually four different teams (one with Katie Bouman) working independently on the reconstruction of the M87 black hole image, testing different methods. The final result was the combination of all four teams' work.
https://physicstoday.scitation.org/do/10.1063/PT.6.1.20190411a/full/
Peter Erwin
CC BY-SA 4.0
2021-05-16T06:02:30.163Z
2021-05-17T16:07:55.530Z
2021-05-17T16:07:55.530Z
Cort Ammon
<p>In situations where one is using a camera to look at a brightly glowing or brightly illuminated object at relatively short range, and one can observe the object for a long time without it changing, the total amount of information present in the incoming photons will be much greater than the amount of useful information one would want in a final image.  Using more photons will make it possible to reduce the level of noise in a picture, but halving the amount of noise will require roughly quadrupling the number of photons to be collected.  When photons are plentiful, it's possible to produce images with far less noise than when photons are scarce, but reducing noise levels requires having a lot of photons.</p>
<p>It's possible to construct a digital filter to undo the effects of precisely modeled optical defects or limitations that would cause images to be blurry.  Unfortunately, no matter how perfectly such a filter is constructed, using one to sharpen blurry images tends to amplify noise as much as it would amplify useful signal.  The more photons one can exploit to reduce noise, the less of a problem this will be.  Unfortunately, in astronomical photography, it's necessary to make due with a relatively limited number of photons.  While it might theoretically be possible to use a suitably constructed digital filter to sharpen images when performing astronomical photography, the purpose of using large mirrors is to collect as many photons as possible, and much of the reason for doing that is to minimize noise.  If use of a filter would double the amount of noise, and the increased noise wasn't acceptable, one would need to increase the mirror area by a factor of four to collect enough photons to compensate.  Conversely, if one would be willing to tolerate a factor-of-two increase in the amount of noise present in an image, that would imply the mirror was four times as big as needed to meet the noise targets.</p>
<p>In short, when photons are scarce, trying to use digital sharpening filters to correct blurriness caused by imperfect optics may be possible, but achieving a certain image quality with imperfect optics would require increasing the size of the optics so much that it's less practical than simply using higher quality optics to start with.</p>

CC BY-SA 4.0
2021-05-17T10:46:52.607Z
I believe this answer is beside the point. The optics determine the image. But sharpening a blurry image will never and under no circumstances reveal more details which are already lost in the optics (it may make better visible small differences otherwise overlooked - but not bring back those which are not there). One may try to rescue some part with applying the inverse PSF, but that only goes so far
planetmaker
CC BY-SA 4.0
2021-05-17T14:43:27.460Z
@planetmaker: If one had absolutely precise (infinite resolution) measurements of the optical defect, and absolutely precise measurements of the image produced with the defective optics, and absolutely precise knowledge of the distances to the object being observed, then one could reconstruct the original image perfectly.  The difficulty is that small uncertainties in one's measurements of the defect, the image, or--for nearby objects--the distances, often translate into larger uncertainties in the reconstructed image.  Astronomical distances are large enough that uncertainties there...
supercat
CC BY-SA 4.0
2021-05-17T14:48:19.163Z
you describe the knowledge of the PSF in your comment which then can be used to deconvolve the obtained image with. Your answer only talks about sharpening the obtained image and dealing with noise - both of which will do no good to improve the data quality.
planetmaker
CC BY-SA 4.0
2021-05-17T14:49:14.367Z
...don't really matter, but uncertainties in the observed images are significant even before one tries to correct for optical defects.  Sorry my answer failed to properly describe the type of filter I had in mind.  I'll edit it and let you know when I'm done.
supercat
CC BY-SA 4.0
2021-05-17T15:01:51.650Z
@planetmaker: Okay, tell me if that's better.  My intended point was that there are times when such filtering may allow one to extract more visual information from a scene than could be resolved directly with one's optics, but the number of photons one would need to grab to achieve a given level of detail would vastly exceed the number of photons one would have to grab if using better optics.  Consider, for example, using a camera whose sensor resolution is four times the highest special frequency of interest in an object which is "at infinity", but the focus...
supercat
CC BY-SA 4.0
2021-05-17T15:12:12.017Z
...is a bit short of infinity, thus effectively applying a Gaussian filter.  If one took enough pictures while the object was stationary, but with variations in lighting of +/- a few pixel values, gamma correcting and averaging them would yield an image which contained information at the frequency of interest.  If the area of interest was a checkerboard whose spacial frequency is 1/4 the captured resolution, the averaged captured image would have a checkerboard of gray on very very slightly lighter gray.  The amount of information one would need to capture would be much larger than the...
supercat
CC BY-SA 4.0
2021-05-17T15:15:35.367Z
...useful amount of information that could be extracted, possibly by multiple orders of magnitude.  Perhaps not a problem if one can capture an essentially unlimited amount of optical information, but not so good if every photon is precious.
supercat
CC BY-SA 4.0
2021-05-16T19:34:30.020Z
2021-05-17T14:52:26.650Z
2021-05-17T14:52:26.650Z
supercat
<blockquote>
<p>correct for the distortion</p>
</blockquote>
<p>An imperfect mirror does not produce a distorted image - it produces a blurry image. With light-field sensors and phase imaging, one could possibly correct for the blur, but it is much more challenging problem than normal lens distortion correction.</p>
<p><a href="https://en.wikipedia.org/wiki/Distortion_(optics)" rel="noreferrer">Distortion</a> refers to a systematic change in how shapes are projected in an image. It results from a lens or mirror with good, accurate geometry that just does not produce a rectilinear projection.</p>
<p>Random imperfections in a mirror do not cause distortion. <strong>Every point in the surface of a mirror contributes to every pixel in the result image.</strong> If a single part of the mirror is at slightly wrong angle, it does not cause a distortion in one point of the image. Instead, it projects the same image at a slightly different alignment on the same sensor. <sup>(1)</sup></p>
<p>In the case of a starfield, this would cause ghost images of very dim stars to appear next to the real stars. Repeat this for a thousand imperfections, and the result is just blurry dots. <a href="https://en.wikipedia.org/wiki/Deconvolution" rel="noreferrer">Deconvolution</a> is a process that can be used to remove blurriness, but noise and other uncertainties limit its effectiveness.</p>
<p><sub>(1) This may be a bit unintuitive if you think about <a href="https://en.wikipedia.org/wiki/Distorting_mirror" rel="noreferrer">funhouse mirrors</a> where the image is distorted. Those work differently because they act the part of a planar mirror, where indeed each part of image is reflected by a single part of a mirror. But planar mirrors cannot form an image by themselves, instead the lens in your eye is the critical component of the image accuracy.</sub></p>

CC BY-SA 4.0
2021-05-16T19:14:02.437Z
Every spot on a funhouse mirror actually reflects a pupil-sized region of the actual image, which is larger than a single point, but generally small relative to the size of the mirror.  In many mirror-based cameras, however, the effective pupil size is often essentially the same as the diameter of the biggest mirror.
supercat
CC BY-SA 4.0
2021-05-17T04:07:27.567Z
@supercat Thanks for the explanation, I had been wondering if there is a more exact way to describe the difference!
jpa
CC BY-SA 4.0
2021-05-16T07:39:22.117Z
2021-05-16T07:47:58.610Z
2021-05-16T07:47:58.610Z
jpa
<p>Modern telescopes go to great lengths to have perfectly shaped parabolic mirrors. My question is, why go to the trouble of having a perfect mirror? Why not take a mirror roughly the right shape, and then correct for the distortion using computers?</p>

CC BY-SA 4.0
2021-05-17T18:57:47.637Z
What if you "correct" wrong? Ask any professional photographer about whether it's better to have good source shot to work with or correct a blurry, poor-lit, or unfocused image in post-production.
MonkeyZeus
<observational-astronomy><telescope>
Why do small mirror imperfections matter with modern computers
DanielOfTaebl
# Why do small mirror imperfections matter with modern computers

(Asked by: username_0 on 2021-05-15)

Modern telescopes go to great lengths to have perfectly shaped parabolic mirrors. My question is, why go to the trouble of having a perfect mirror? Why not take a mirror roughly the right shape, and then correct for the distortion using computers?

**Tags:** observational-astronomy, telescope
> **Comments:**
> **username_1**: What if you "correct" wrong? Ask any professional photographer about whether it's better to have good source shot to work with or correct a blurry, poor-lit, or unfocused image in post-production.  

---

## 6 answers

### Answer by username_18 on 2021-05-16. Score: 45 | Accepted answer

> correct for the distortion

An imperfect mirror does not produce a distorted image - it produces a blurry image. With light-field sensors and phase imaging, one could possibly correct for the blur, but it is much more challenging problem than normal lens distortion correction.

[Distortion](https://en.wikipedia.org/wiki/Distortion_(optics)) refers to a systematic change in how shapes are projected in an image. It results from a lens or mirror with good, accurate geometry that just does not produce a rectilinear projection.

Random imperfections in a mirror do not cause distortion. **Every point in the surface of a mirror contributes to every pixel in the result image.** If a single part of the mirror is at slightly wrong angle, it does not cause a distortion in one point of the image. Instead, it projects the same image at a slightly different alignment on the same sensor. (1)

In the case of a starfield, this would cause ghost images of very dim stars to appear next to the real stars. Repeat this for a thousand imperfections, and the result is just blurry dots. [Deconvolution](https://en.wikipedia.org/wiki/Deconvolution) is a process that can be used to remove blurriness, but noise and other uncertainties limit its effectiveness.

(1) This may be a bit unintuitive if you think about [funhouse mirrors](https://en.wikipedia.org/wiki/Distorting_mirror) where the image is distorted. Those work differently because they act the part of a planar mirror, where indeed each part of image is reflected by a single part of a mirror. But planar mirrors cannot form an image by themselves, instead the lens in your eye is the critical component of the image accuracy.

> **Comments:**
> **username_16**: Every spot on a funhouse mirror actually reflects a pupil-sized region of the actual image, which is larger than a single point, but generally small relative to the size of the mirror.  In many mirror-based cameras, however, the effective pupil size is often essentially the same as the diameter of the biggest mirror.  
> **username_18**: @supercat Thanks for the explanation, I had been wondering if there is a more exact way to describe the difference!  

---

### Answer by username_2 on 2021-05-15. Score: 26

Also see answers to

* [Could mirrors be replaced with CCDs?](https://astronomy.stackexchange.com/q/14351/7982)
* [Is Digital Adaptive Optics Possible?](https://astronomy.stackexchange.com/q/32521/7982)

The problem is that light imaging detectors convert amplitude to power during the detection process. Phase is lost.

If you had maps of both the magnitude of the electric field and its relative phase, and had this at each wavelength of light, you could do exactly what you propose and correct in software.

In fact, you could then throw away the telescope and image the sky without it.

* [How would a Fast Fourier Transform Telescope work without a mirror, dish or lenses?](https://astronomy.stackexchange.com/q/32014/7982)

That's exactly how a radio telescope array works! Dozens of antennas detect waves from a patch of sky, they're all brought into a computer and for each wavelength the source shape is reconstructed from the *amplitude and phase* information available in the electronic signals.

* [Would Adaptive Optics be Useful in Radio Astronomy?](https://astronomy.stackexchange.com/q/13930/7982)

But CCD detectors or photographic emulsion plates or our eyes convert the photons to other forms of energy, then average it out over time.

Phase is lost, so reconstruction is no longer possible.

> **Comments:**
> **username_3**: So why don't optical telescopes use other detection methods that _do_ preserve phase?  
> **username_2**: @Vikki-formerlySean they will some day! Visible light, say 600 nm has a frequency of 500,000,000,000,000 Hz (500,000 GHz) which is not manageable with normal electronics. In order to look at many wavelengths at the same time rather than an incredibly sharp 1 GHz wide optical bandpass, you need to digitize at this absurdly high rate and process. Basically, if the detector preserved the phase, you'd then have to either *record* the phase in real time or *process* it. If you had an extremely narrow bandpass filter (0.001 nm) and could cut the spectrum down to GHz, you could down-convert it to  
> **username_2**: @Vikki-formerlySean an electronic signal in GHz range and then process normally. That's roughly the kind of thing I'd asked about in [Has optical interferometry been done at radio frequency using heterodyning with a laser in a nonlinear material?](https://astronomy.stackexchange.com/q/36633/7982)  
> **username_2**: @Vikki-formerlySean there are other options but I'm not fully informed of the background so can't really write about it. You could build a one-pixel optical telescope and use a 0.000001 nm filter and re-expand the wavefront, interfere with a laser in quadrature, and digitize two interferograms at kHz rates or something like that, but I'm not sure that that would be at all useful. Since [astronomical seeing](https://en.wikipedia.org/wiki/Astronomical_seeing) is somewhat achromatic we can do it at kHz rates without worrying too much about spectrum, but for this application it's a problem.  
> **username_4**: @Vikki-formerlySean There's work on this, but it's hard: https://www.scientificamerican.com/article/quantum-astronomy-could-create-telescopes-hundreds-of-kilometers-wide/  
> **username_5**: At least as I understand it, adaptive optics in telescopes don't work by processing the image after capture.  Instead, they measure the distortion from the atmosphere, and use computers & actuators to slightly change the shape of the mirror to compensate. So it would seem that you could in principle use the same technology to change the shape of an "almost perfect" mirror to be closer to perfect.  
> **username_2**: @jamesqf certainly "...and then correct for the distortion *using computers?*" provides substantial latitude in how you might use those computers. It's true you could spend 10x the money for a deformable mirror, a few hundred (or more) actuators or you might be able to design a correcting element like they did for Hubble using computers, but I took it to mean apply the computers to the images obtained by conventional means, as did (at)zabop's answer. I think you can certainly post your suggestion as a new answer, since the use of computers is involved...  
> **username_6**: Phase information doesn't help at all with normal light. You's need a coherent light source and pixels smaller than the wavelength for this. What is needed is directional sensitivity in each pixel, like it is used in commercial light field cameras.  
> **username_2**: @asdfex there is no such thing as completely coherent nor completely incoherent light. It's a continuum; lasers are a lot more coherent than light bulbs, but light bulbs have non-zero coherence and lasers are not perfectly coherent. Then you have to address longitudinal and transverse coherence separately. ***There is no such thing as "normal light".*** There's just photons. As long as you read through what I wrote carefully and note the filter bandwidths I specified in each case, you absolutely can heterodyne filtered light with a laser and down-convert.  
> **username_2**: It's no different than heterodyning "normal radio signals from a radio or TV transmitter with a stable local oscillator do down-convert them to an IF,  a detection or demodulation circuit and then to audio. cf. [Radio over Fiber](https://en.wikipedia.org/wiki/Radio_over_fiber)  
> **username_6**: What you write about radio telescopes is fine - but it doesn't apply to optical telescopes, at least not on the length scale that is relevant here. Phase helps if your spatial precision is much better than the wavelength, but this is not the case with any current imaging technology.  

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### Answer by username_7 on 2021-05-15. Score: 15

The image which is recorded by CCDs are the [convolution](https://en.wikipedia.org/wiki/Convolution) of the true image and a [point spread function](https://en.wikipedia.org/wiki/Point_spread_function). If the PSF is not a nice function, such as this one, for example [[source]](https://en.wikipedia.org/wiki/Point_spread_function#/media/File:Hubble_PSF_with_flawed_optics.jpg):

[![enter image description here](https://i.sstatic.net/kfoNM.png)](https://i.sstatic.net/kfoNM.png)

then (I think) it is hard to deconvolve the detected image from the PSF to get the true image. For example, you don't necesserily know the exact PSF if the telescope is only "roughly" the right shape, so you don't know what to deconvolve.

> **Comments:**
> **username_8**: it should be possible to determine the PSF, if you know what the result should look like. A laser pointer shined from an airplane should suffice as a calibration pattern?  
> **username_7**: I think you are probably right.  
> **username_9**: Deconvolution is extremely error-prone operation, any noise will corrupt resulting data exponentially. It can be somewhat done in space, where there are no air disturbances and when mirror temperature is kept stable; on Earth this is not viable solution. Sure deconvolution is used in astronomy image processing to remove diffraction and other image distortions, which cannot be avoided by making better mirrors :)  
> **username_10**: This image shows the problem very well - one of the main measures taken from these images is the amount of light received from the star (it's brightness).  As can be seen in the image, that light has spread over a wide area and so it has to be summed up to get the total.  However, you can also see it is now covering other objects and separating out which light came from which object is hard and introduces uncertainties.  In a perfect world with perfect mirrors and no atmosphere, each object would be contained in a single pixel and so easily measurable.  
> **username_11**: @JonathanTwite In the perfect world you describe there would still be point spread functions, because the diffraction limit of a finite-sized mirror. So each object would *not* be "contained in a single pixel" (unless you made your pixels absurdly large, in which case you'd have many different objects in each pixel...).  
> **username_8**: @PeterErwin in a perfect world, you'd have infinitely large mirrors... that nevertheless don't collapse under their own gravity or get deformed by tidal effects and solar wind.  

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### Answer by username_12 on 2021-05-17. Score: 7

The short answer is that blurring *destroys information*, and no amount of correction after the fact can bring it back.

A slightly less short answer: As others have mentioned, the blur caused by imperfections in the mirror is a convolution of the true image with the "point spread function" created by the imperfections.

Convolution of the image is mathematically equivalent to multiplying the Fourier transform of the image with the Fourier transform of the point spread function. Wherever the transform of that point-spread function is zero or close to it, the image information at the corresponding frequencies is destroyed.

The point-spread function created by random imperfections looks like a Gaussian distribution. The Gaussian distribution has the interesting property that its Fourier transform is also a Gaussian, centered on 0-frequency with bandwidth inversely proportional to the width of the point-spread function.

In other words, the point-spread function is a low-pass filter. It destroys the high-frequency information in the image, where all the fine detail lives. The wider the point spread, the worse it is.


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### Answer by username_13 on 2021-05-16. Score: 6

They do corrections in computers. But it's not a perfect thing. Just how you can't tell the difference between 1 + 2 + 1 and 1.3 + 1.4 + 1.3, the equations we solve don't have just one solution. They have many. So we have to do guesses in the algorithm. Given that science is exploring that which is just outside of our knowledge, its useful to get actual raw data rather than having to manipulate it first.

I think my favorite example of this is the [famous black hole picture](https://www.jpl.nasa.gov/edu/news/2019/4/19/how-scientists-captured-the-first-image-of-a-black-hole/) from 2019. It was, of course, a computer reconstruction of the data, using software from a team headed by [Katie Bouman](https://www.bbc.com/news/science-environment-47891902). She has a wonderful [TED talk](https://www.youtube.com/watch?v=BIvezCVcsYs) on the technology where she points out the challenges with biasing algorithms to detect what we expect to see. One of the tools they relied on to gain confidence that the algorithm wasn't just showing them what they expected to see was to train the algorithm on pictures of her friends and faculty. If the resulting image processed after training on these pictures was basically the same as the one from generated from processing after training on pictures of known stellar objects, they were reasonably confident it wasn't bias.

But its always easier to just have better raw data.

> **Comments:**
> **username_2**: Yes, that's done with radio astronomy; just as I described in [my answer](https://astronomy.stackexchange.com/a/43817/7982). But the question was about correction optical mirrors and at 500,000 GHz, so the analogy to radio astronomy doesn't work.  
> **username_13**: @uhoh The equations may be different, but if computerized sharpening could prevent the need for quality measurements, then we would have certainly replaced our telescopes with computers.  It's still a game of imperfect information.  
> **username_2**: We don't have analog to digital converters that work at 500,000 GHz nor the rest of the electronics, so even you wanted to try to run the world's largest supercomputer to deal with the rate, light frequencies are just too high to treat like radio waves given the current state of the art in electronics. It may be possible some day in the not-too-distant future, but not yet.  
> **username_14**: 1+3+1 = 5. 1.3+1.4+1.3 = 4. I think I can tell the difference.  
> **username_13**: @WaterMolecule ... I need to stop posting math when I'm sleepy!  Fixed!  
> **username_15**: The maths example makes no sense to me. Yes, they both add up to 4, but so does -2, 8, -2. How does this look illustrate the point you’re trying to make? Or are you trying to demonstrate the effect of rounding? If so, 1.4 rounds to 1, not 2.  
> **username_13**: @ChappoHasn'tForgottenMonica If all you have is 4, you cannot determine which of those two cases was the true un-blurred data.  You can get somewhere with a convolution to invert a point spread function, but eventually it gets to a point where there are simply more than one possibility for what the image could be.  This is especially true in the presence of noise.  
> **username_11**: Minor nitpick: there were actually four different teams (one with Katie Bouman) working independently on the reconstruction of the M87 black hole image, testing different methods. The final result was the combination of all four teams' work. https://physicstoday.scitation.org/do/10.1063/PT.6.1.20190411a/full/  

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### Answer by username_16 on 2021-05-16. Score: 0

In situations where one is using a camera to look at a brightly glowing or brightly illuminated object at relatively short range, and one can observe the object for a long time without it changing, the total amount of information present in the incoming photons will be much greater than the amount of useful information one would want in a final image. Using more photons will make it possible to reduce the level of noise in a picture, but halving the amount of noise will require roughly quadrupling the number of photons to be collected. When photons are plentiful, it's possible to produce images with far less noise than when photons are scarce, but reducing noise levels requires having a lot of photons.

It's possible to construct a digital filter to undo the effects of precisely modeled optical defects or limitations that would cause images to be blurry. Unfortunately, no matter how perfectly such a filter is constructed, using one to sharpen blurry images tends to amplify noise as much as it would amplify useful signal. The more photons one can exploit to reduce noise, the less of a problem this will be. Unfortunately, in astronomical photography, it's necessary to make due with a relatively limited number of photons. While it might theoretically be possible to use a suitably constructed digital filter to sharpen images when performing astronomical photography, the purpose of using large mirrors is to collect as many photons as possible, and much of the reason for doing that is to minimize noise. If use of a filter would double the amount of noise, and the increased noise wasn't acceptable, one would need to increase the mirror area by a factor of four to collect enough photons to compensate. Conversely, if one would be willing to tolerate a factor-of-two increase in the amount of noise present in an image, that would imply the mirror was four times as big as needed to meet the noise targets.

In short, when photons are scarce, trying to use digital sharpening filters to correct blurriness caused by imperfect optics may be possible, but achieving a certain image quality with imperfect optics would require increasing the size of the optics so much that it's less practical than simply using higher quality optics to start with.

> **Comments:**
> **username_17**: I believe this answer is beside the point. The optics determine the image. But sharpening a blurry image will never and under no circumstances reveal more details which are already lost in the optics (it may make better visible small differences otherwise overlooked - but not bring back those which are not there). One may try to rescue some part with applying the inverse PSF, but that only goes so far  
> **username_16**: @planetmaker: If one had absolutely precise (infinite resolution) measurements of the optical defect, and absolutely precise measurements of the image produced with the defective optics, and absolutely precise knowledge of the distances to the object being observed, then one could reconstruct the original image perfectly.  The difficulty is that small uncertainties in one's measurements of the defect, the image, or--for nearby objects--the distances, often translate into larger uncertainties in the reconstructed image.  Astronomical distances are large enough that uncertainties there...  
> **username_17**: you describe the knowledge of the PSF in your comment which then can be used to deconvolve the obtained image with. Your answer only talks about sharpening the obtained image and dealing with noise - both of which will do no good to improve the data quality.  
> **username_16**: ...don't really matter, but uncertainties in the observed images are significant even before one tries to correct for optical defects.  Sorry my answer failed to properly describe the type of filter I had in mind.  I'll edit it and let you know when I'm done.  
> **username_16**: @planetmaker: Okay, tell me if that's better.  My intended point was that there are times when such filtering may allow one to extract more visual information from a scene than could be resolved directly with one's optics, but the number of photons one would need to grab to achieve a given level of detail would vastly exceed the number of photons one would have to grab if using better optics.  Consider, for example, using a camera whose sensor resolution is four times the highest special frequency of interest in an object which is "at infinity", but the focus...  
> **username_16**: ...is a bit short of infinity, thus effectively applying a Gaussian filter.  If one took enough pictures while the object was stationary, but with variations in lighting of +/- a few pixel values, gamma correcting and averaging them would yield an image which contained information at the frequency of interest.  If the area of interest was a checkerboard whose spacial frequency is 1/4 the captured resolution, the averaged captured image would have a checkerboard of gray on very very slightly lighter gray.  The amount of information one would need to capture would be much larger than the...  
> **username_16**: ...useful amount of information that could be extracted, possibly by multiple orders of magnitude.  Perhaps not a problem if one can capture an essentially unlimited amount of optical information, but not so good if every photon is precious.  

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