<p>Good optics work in the diffraction limit. The diffraction limit is a hard physical limit and relates the <em>angular resolution</em> <span class="math-container">$\theta$</span> to the diameter of the optics <span class="math-container">$D$</span> and the wavelength you are observing in <span class="math-container">$\lambda$</span>:
<span class="math-container">$\theta = 1.22\frac{D}{\lambda}$</span> for a small angle approximation with <span class="math-container">$\sin\theta \approx \theta$</span>.</p>
<p>The factor 1.22 is somewhat empirical and comes from actually looking at the image of a point source: it creates a circular diffraction pattern and 1.22 describes the position of the first dark ring around your point source. If you are satisfied with different sensitivity and confidence of separation one can choose somewhat different values there; one can distinguish two different sources, if you can make them out as separate sources. From the <a href="https://en.wikipedia.org/wiki/Angular_resolution" rel="nofollow noreferrer">wiki page on the airy disk</a>, the comparison shows the transition from separated to not distinguishable anymore:
<a href="https://i.sstatic.net/xZxx1xiI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xZxx1xiI.png" alt="Diffraction limit" /></a></p>
<p>Given the optics you usually want to choose your imaging sensor such that you don't needlessly cover the central area of a point source with several pixels, but at the same time enough so that you can still distinguish close sources.</p>


CC BY-SA 4.0
2024-12-22T19:55:02.300Z
2024-12-23T18:00:10.090Z
2024-12-23T18:00:10.090Z
planetmaker
<blockquote>
<p>It is implied Dawe's limit may be the smallest a telescope could 'see'. Is that right?</p>
</blockquote>
<p>Generally, yes. As <a href="https://astronomy.stackexchange.com/a/59137/7982">planetmaker explains</a> we have to be careful how we define what 'see' means, but yes, larger apertures mean better resolution, and Dawe's limit is a helpful rule of thumb. You have to do a full physical optics simulation (electromagnetic waves, diffraction, etc.) to get a more accurate, detailed answer.</p>
<blockquote>
<p>Does it become a single pixel on a detector?</p>
</blockquote>
<p>You can put any detector you want at your focal plane. It is common to have pixel sizes around this limit or sometimes a bit smaller (oversampling) to avoid pixelation making the resolution slightly worse.</p>
<blockquote>
<p>Is there some other limit to determine the smallest?</p>
</blockquote>
<p>Although you specify &quot;...for visual wavelengths (562nm)...&quot;, for other readers I'll point out that of course, choosing shorter wavelengths helps resolution! <a href="https://en.wikipedia.org/wiki/Dawes%27_limit" rel="nofollow noreferrer">Dawe's limit</a> requires a wavelength <span class="math-container">$\lambda$</span> so if you use blue or UV light and ignore red and infrared, you'll get better resolution. In space you can use <a href="https://en.wikipedia.org/wiki/Ultraviolet#Subtypes" rel="nofollow noreferrer">vacuum ultraviolet</a> (VUV) or <a href="https://en.wikipedia.org/wiki/Extreme_ultraviolet" rel="nofollow noreferrer">extreme ultraviolet</a> (EUV) but that cuts down on the visible objects; many stars will be quite dim out there.</p>
<p>You could go shorter to X-rays but the imaging technology is more complicated since normal telescope configurations won't work; there are no convenient lenses or broadband mirror surfaces<sup>1</sup>.</p>
<blockquote>
<p>Does that change by other factors such as exposure time?</p>
</blockquote>
<p>Theoretically, yes, a bit. If you have really high quality, low (relative) noise images, <a href="https://en.wikipedia.org/wiki/Deconvolution" rel="nofollow noreferrer">deconvolution</a> of the telescope's <a href="https://en.wikipedia.org/wiki/Point_spread_function" rel="nofollow noreferrer">point-spread function</a> might help. But you have to do that carefully.</p>
<blockquote>
<p>Would an interferometer the size of the spacecraft change that figure?</p>
</blockquote>
<p>There is no fundamental difference between interferometry and a single aperture telescope of the same size as the baseline of the interferometer, except that because the interferometer <em>undersamples</em> the wavefront, you have to work harder to recover the image and there are a lot of artifacts that you have to learn how to ignore.  But it works! See for example the <a href="https://en.wikipedia.org/wiki/Very_Large_Telescope" rel="nofollow noreferrer">Very Large Telescope</a> and other ground-based interferometers.</p>
<blockquote>
<p>Would the answer reasonably get better in the future?</p>
</blockquote>
<p>Yes, maybe.</p>
<p>Also, your space ship could simply have a few drone ships that artificially extend its aperture!</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/57727/7982">Do we have the technology to create a satellite telescope constellation that uses interferometry?</a></li>
</ul>
<p>See also:</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/36633/7982">Has optical interferometry been done at radio frequency using heterodyning with a laser in a nonlinear material?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/35837/7982">Can the interferometer called &quot;Gravity&quot; measure &quot;a few centimeters on the Moon&quot;?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/44591/7982">Reference request (explaining) how optical correlators combine light from multiple telescopes to produce ultra-high resolution interferometric images?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/33914/7982">Will the Magdalena Ridge Optical Interferometer be able to image extended objects like the surface of the Moon?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/12327/7982">What makes small interferometers useful? Like NIRISS on JWST</a></li>
<li><a href="https://astronomy.stackexchange.com/q/29335/7982">Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically?</a></li>
</ul>
<hr />
<p><sup>1</sup>while multilayer mirror coatings tend to have reflectivities for only a narrow energy range, with great and <em>inconvenient</em> difficulty you can make &quot;depth-graded multi-layer&quot; coatings that sacrifice peak reflectivity for a more broadband response, cf. Koglin et al. (2009) <a href="https://www.researchgate.net/publication/228674474_NuSTAR_hard_x-ray_optics_design_and_performance" rel="nofollow noreferrer">NuSTAR hard x-ray optics design and performance</a></p>

CC BY-SA 4.0
2024-12-23T20:57:35.737Z
" ... there are no ... broadband mirror surfaces". Not true. The NuSTAR x-ray mirrors cover 3-79 keV, almost five octaves! But they are not diffraction limited. The problem in the x-ray is that it's not practical to make a large curved mirror with an atomically precise figure.
John Doty
CC BY-SA 4.0
2024-12-23T23:04:28.433Z
@JohnDoty you omitted the word "convenient" when you quoted my sentence to make it appear not true. :-( I've added a footnote for you.
uhoh
CC BY-SA 4.0
2024-12-23T23:43:11.360Z
If you don't want to go above 10 keV, you don't need multilayers, just grazing incidence. But you still can't make an atomically precise figure: Chandra's optics (the best ever made) are about three orders of magnitude short of diffraction limited.
John Doty
CC BY-SA 4.0
2024-12-24T06:17:15.920Z
@JohnDoty instead of leaving comments to me, why not write a short answer addressing the likelyhood of getting better resolution at wavelengths where good-old aluminum-coated mirrors no longer work? I probably won't dig deeper, but it seems there's more to be said.
uhoh
CC BY-SA 4.0
2024-12-24T18:54:39.727Z
Such an answer would not address the question, which is specifically about optical wavelengths.
John Doty
CC BY-SA 4.0
2024-12-24T21:45:16.107Z
@JohnDoty Indeed! Wow I read right past "...for visual wavelengths (562nm)" *Thanks!*
uhoh
CC BY-SA 4.0
2024-12-25T01:08:05.333Z
@JohnDoty as an aside, the answer to [Plot of best available resolution vs wavelength - radio through gamma rays?](https://astronomy.stackexchange.com/q/18905/7982) is pretty cool
uhoh
CC BY-SA 4.0
2024-12-25T20:12:48.083Z
Thanks uhoh! Great answer that hit all my questions with great detail and even more. I liked the drone idea and all the further reading. In the sim, full throttle means 100m/s^2, so not exactly drone speed but getting extra extra arcseconds while sitting idle (not accelerating) and deploying drones makes for a great trade-off and mechanic.
Crispin Bivans
CC BY-SA 4.0
2024-12-22T22:55:31.660Z
2024-12-24T21:47:46.330Z
2024-12-24T21:47:46.330Z
uhoh
<p>For a spacesim, I'm looking to understand what is the smallest arcseconds a spacecraft could see in space based on its telescope size, for visual wavelengths (562nm). So far if I go by Dawe's limit, that would result in a resolution of 115.8 / Telescope size (mm) for arcseconds to resolve if something seen was two objects or just one.</p>
<p>It is implied Dawe's limit may be the smallest a telescope could 'see'. Is that right? Does it become a single pixel on a detector? Is there some other limit to determine the smallest? Does that change by other factors such as exposure time? Would an interferometer the size of the spacecraft change that figure? Would the answer reasonably get better in the future?</p>
<p>Thank you.</p>

CC BY-SA 4.0
2024-12-22T18:33:37.997Z
Dawe's limit is an absolute physical limit determined by the wave nature of light; it's not a rule of thumb or anything based on the telescope construction or detector.  You can think about it in terms of the Heisenberg uncertainty principal: the more you know about a photon's position (so the smaller the aperture) the less you can learn about it's momentum (precise direction angle).  An interferometer can get you a much larger effective aperture for determining astrometric position.
antlersoft
CC BY-SA 4.0
2024-12-22T22:37:24.753Z
@antlersoft note also [this answer to *Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically?*](https://astronomy.stackexchange.com/a/29337/7982)  but also see [this answer to *What makes small interferometers useful? Like NIRISS on JWST*](https://astronomy.stackexchange.com/a/24182/7982) If you know ahead fo time something about the pattern that needs to be resolved, aperture masking does have a niche application where it gives "better" results than a clear aperture.
uhoh
CC BY-SA 4.0
2024-12-25T17:19:24.927Z
To me, for one, it's obvious that the 'smallest' anything any eye or mechanism could see must be a single pixel. Can you say why that doesn't work for you?
Robbie Goodwin
<space-telescope><photometry>
Smallest arcseconds viewable by perfect conditions (i.e. space-based telescope)
Crispin Bivans
# Smallest arcseconds viewable by perfect conditions (i.e. space-based telescope)

(Asked by: username_0 on 2024-12-22)

For a spacesim, I'm looking to understand what is the smallest arcseconds a spacecraft could see in space based on its telescope size, for visual wavelengths (562nm). So far if I go by Dawe's limit, that would result in a resolution of 115.8 / Telescope size (mm) for arcseconds to resolve if something seen was two objects or just one.

It is implied Dawe's limit may be the smallest a telescope could 'see'. Is that right? Does it become a single pixel on a detector? Is there some other limit to determine the smallest? Does that change by other factors such as exposure time? Would an interferometer the size of the spacecraft change that figure? Would the answer reasonably get better in the future?

Thank you.

**Tags:** space-telescope, photometry
> **Comments:**
> **username_1**: Dawe's limit is an absolute physical limit determined by the wave nature of light; it's not a rule of thumb or anything based on the telescope construction or detector.  You can think about it in terms of the Heisenberg uncertainty principal: the more you know about a photon's position (so the smaller the aperture) the less you can learn about it's momentum (precise direction angle).  An interferometer can get you a much larger effective aperture for determining astrometric position.  
> **username_2**: @antlersoft note also [this answer to *Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically?*](https://astronomy.stackexchange.com/a/29337/7982)  but also see [this answer to *What makes small interferometers useful? Like NIRISS on JWST*](https://astronomy.stackexchange.com/a/24182/7982) If you know ahead fo time something about the pattern that needs to be resolved, aperture masking does have a niche application where it gives "better" results than a clear aperture.  
> **username_3**: To me, for one, it's obvious that the 'smallest' anything any eye or mechanism could see must be a single pixel. Can you say why that doesn't work for you?  

---

## 2 answers

### Answer by username_2 on 2024-12-22. Score: 7 | Accepted answer

> It is implied Dawe's limit may be the smallest a telescope could 'see'. Is that right?

Generally, yes. As [planetmaker explains](https://astronomy.stackexchange.com/a/59137/7982) we have to be careful how we define what 'see' means, but yes, larger apertures mean better resolution, and Dawe's limit is a helpful rule of thumb. You have to do a full physical optics simulation (electromagnetic waves, diffraction, etc.) to get a more accurate, detailed answer.

> Does it become a single pixel on a detector?

You can put any detector you want at your focal plane. It is common to have pixel sizes around this limit or sometimes a bit smaller (oversampling) to avoid pixelation making the resolution slightly worse.

> Is there some other limit to determine the smallest?

Although you specify "...for visual wavelengths (562nm)...", for other readers I'll point out that of course, choosing shorter wavelengths helps resolution! [Dawe's limit](https://en.wikipedia.org/wiki/Dawes%27_limit) requires a wavelength $\lambda$ so if you use blue or UV light and ignore red and infrared, you'll get better resolution. In space you can use [vacuum ultraviolet](https://en.wikipedia.org/wiki/Ultraviolet#Subtypes) (VUV) or [extreme ultraviolet](https://en.wikipedia.org/wiki/Extreme_ultraviolet) (EUV) but that cuts down on the visible objects; many stars will be quite dim out there.

You could go shorter to X-rays but the imaging technology is more complicated since normal telescope configurations won't work; there are no convenient lenses or broadband mirror surfaces1.

> Does that change by other factors such as exposure time?

Theoretically, yes, a bit. If you have really high quality, low (relative) noise images, [deconvolution](https://en.wikipedia.org/wiki/Deconvolution) of the telescope's [point-spread function](https://en.wikipedia.org/wiki/Point_spread_function) might help. But you have to do that carefully.

> Would an interferometer the size of the spacecraft change that figure?

There is no fundamental difference between interferometry and a single aperture telescope of the same size as the baseline of the interferometer, except that because the interferometer *undersamples* the wavefront, you have to work harder to recover the image and there are a lot of artifacts that you have to learn how to ignore. But it works! See for example the [Very Large Telescope](https://en.wikipedia.org/wiki/Very_Large_Telescope) and other ground-based interferometers.

> Would the answer reasonably get better in the future?

Yes, maybe.

Also, your space ship could simply have a few drone ships that artificially extend its aperture!

* [Do we have the technology to create a satellite telescope constellation that uses interferometry?](https://astronomy.stackexchange.com/q/57727/7982)

See also:

* [Has optical interferometry been done at radio frequency using heterodyning with a laser in a nonlinear material?](https://astronomy.stackexchange.com/q/36633/7982)
* [Can the interferometer called "Gravity" measure "a few centimeters on the Moon"?](https://astronomy.stackexchange.com/q/35837/7982)
* [Reference request (explaining) how optical correlators combine light from multiple telescopes to produce ultra-high resolution interferometric images?](https://astronomy.stackexchange.com/q/44591/7982)
* [Will the Magdalena Ridge Optical Interferometer be able to image extended objects like the surface of the Moon?](https://astronomy.stackexchange.com/q/33914/7982)
* [What makes small interferometers useful? Like NIRISS on JWST](https://astronomy.stackexchange.com/q/12327/7982)
* [Under what situations can an aperture mask improve the resolution of a small/medium amateur telescope? Is this demonstrable mathematically?](https://astronomy.stackexchange.com/q/29335/7982)

---

1while multilayer mirror coatings tend to have reflectivities for only a narrow energy range, with great and *inconvenient* difficulty you can make "depth-graded multi-layer" coatings that sacrifice peak reflectivity for a more broadband response, cf. Koglin et al. (2009) [NuSTAR hard x-ray optics design and performance](https://www.researchgate.net/publication/228674474_NuSTAR_hard_x-ray_optics_design_and_performance)

> **Comments:**
> **username_5**: " ... there are no ... broadband mirror surfaces". Not true. The NuSTAR x-ray mirrors cover 3-79 keV, almost five octaves! But they are not diffraction limited. The problem in the x-ray is that it's not practical to make a large curved mirror with an atomically precise figure.  
> **username_2**: @JohnDoty you omitted the word "convenient" when you quoted my sentence to make it appear not true. :-( I've added a footnote for you.  
> **username_5**: If you don't want to go above 10 keV, you don't need multilayers, just grazing incidence. But you still can't make an atomically precise figure: Chandra's optics (the best ever made) are about three orders of magnitude short of diffraction limited.  
> **username_2**: @JohnDoty instead of leaving comments to me, why not write a short answer addressing the likelyhood of getting better resolution at wavelengths where good-old aluminum-coated mirrors no longer work? I probably won't dig deeper, but it seems there's more to be said.  
> **username_5**: Such an answer would not address the question, which is specifically about optical wavelengths.  
> **username_2**: @JohnDoty Indeed! Wow I read right past "...for visual wavelengths (562nm)" *Thanks!*  
> **username_2**: @JohnDoty as an aside, the answer to [Plot of best available resolution vs wavelength - radio through gamma rays?](https://astronomy.stackexchange.com/q/18905/7982) is pretty cool  
> **username_0**: Thanks uhoh! Great answer that hit all my questions with great detail and even more. I liked the drone idea and all the further reading. In the sim, full throttle means 100m/s^2, so not exactly drone speed but getting extra extra arcseconds while sitting idle (not accelerating) and deploying drones makes for a great trade-off and mechanic.  

---

### Answer by username_4 on 2024-12-22. Score: 6

Good optics work in the diffraction limit. The diffraction limit is a hard physical limit and relates the *angular resolution* $\theta$ to the diameter of the optics $D$ and the wavelength you are observing in $\lambda$:
$\theta = 1.22\frac{D}{\lambda}$ for a small angle approximation with $\sin\theta \approx \theta$.

The factor 1.22 is somewhat empirical and comes from actually looking at the image of a point source: it creates a circular diffraction pattern and 1.22 describes the position of the first dark ring around your point source. If you are satisfied with different sensitivity and confidence of separation one can choose somewhat different values there; one can distinguish two different sources, if you can make them out as separate sources. From the [wiki page on the airy disk](https://en.wikipedia.org/wiki/Angular_resolution), the comparison shows the transition from separated to not distinguishable anymore:
[![Diffraction limit](https://i.sstatic.net/xZxx1xiI.png)](https://i.sstatic.net/xZxx1xiI.png)

Given the optics you usually want to choose your imaging sensor such that you don't needlessly cover the central area of a point source with several pixels, but at the same time enough so that you can still distinguish close sources.


---

