Is there a limit to the resolving power of a mirror telescope?

Like, if you hammered out the asteroid 16 Psyche into a 1 mm thick iron foil disc telescope mirror with 2.4x the radius of the Sun, could you resolve details on the surface of an exoplanet? At what resolution? Could you make the mirror arbitrarily bigger and continue to get better resolution? What are the formulas for calculating this and what are the limits, if any?

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BTW--This would have been a fine candidate for Astronomy.SE which is still in beta, though it is certainly on-topic here as well. –  dmckee Dec 6 '11 at 0:56
Wow. Another really good optics question. Somebody do my day job so I can answer all of them? –  Colin K Dec 6 '11 at 2:31
The answer is yes. –  Ron Maimon Dec 7 '11 at 7:05
@endolith: I meant, no limit. I couldn't see one. You could go on forever. Did you calculate the size you need to see details on an exoplanet? I assume this is photon number limited. –  Ron Maimon Dec 7 '11 at 19:06
It would be unlimited in theory, but depending on what degree of practicality you hold yourself to, there are some interesting limits. –  Colin K Dec 14 '11 at 17:34

Some theoretical answers were provided, but here's a practical answer from an astronomer's point of view.

(First off, the resolving power is given by diameter, not surface area. So we will talk about diameter here.)

For visible light, the practical resolving power of a 100mm diameter mirror is 1 arcsec. A 200mm mirror: 0.5 arcsec. And so on. This is the rule used by astronomers.

A mirror "with 20x the surface area of the Sun" (apparent area, I assume) would have approx 4.5x the diameter of the Sun, which is 6.3 x 10^6 km. Such a mirror would have a resolving power of 1.6 x 10^-11 arcsec.

At 10 light years distance, such a mirror would resolve details as small as 7 meters.

Please note this discussion is entirely theoretical, as there are no known technologies to manufacture such a big mirror with the required precision for visible light astronomy optics - which means the surface error cannot be bigger than 100 nanometers - in fact, for a good mirror, the acceptable error is 4x ... 5x smaller. There's no way to maintain such precision across millions of kilometers of reflective surface.

Currently, the biggest monolithic mirror is 6m in diameter and it never performed very well. The biggest well-performing monolith mirror is 5m. The biggest segmented mirrors are 10m in diameter, with a 40m project having had its initial funding approved very recently.

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"There's no way to maintain such precision across millions of kilometers of reflective surface." Are you sure? We'd have the technology to flatten asteroids. en.wikipedia.org/wiki/… I'm imagining the foil would be stabilized by a scaffolding and electric or magnetic (since it's iron) fields, but that's probably not enough precision. Or the image could be gathered computationally despite the aberrations in the mirror? –  endolith Dec 16 '11 at 19:45
There's no way - currently. In the future, all bets are off, so we may well be able to do it. If we manage to do it, it will probably be some kind of active system, indeed not unlike present-day composite telescopes, which monitor the position of each mirror segment in the honeycomb and actively correct differences. But we haven't built any kind of structure of stellar size, let alone one that could maintain a 100 nm precision across its entire surface. So we're talking far future technology. –  Florin Andrei Dec 16 '11 at 20:36
Computer-based corrections are doable, provided you at least know the exact shape of the mirror with great precision, and then you could apply algorithms that fix the image. To some degree, anyway. If it's a blurry piece of junk then it's less likely you could fix it. –  Florin Andrei Dec 16 '11 at 20:37
Computer based corrections can do a lot better than you suggest. In fact, there are lensless imaging techniques that reconstruct diffraction limited images from "a blurry piece of junk." –  Colin K Apr 9 '12 at 19:44

The resolving power of a telescope with a circular apertre is given by $$\theta = 1.22\frac{\lambda}{D}$$ Where $\lambda$ is the wavelength, $\theta$ is the smallest resolvable angular size and $D$ is the diameter of the aperture (lens/mirror) of the instrument.

The larger $D$ gets the larger the fourier component you get to investigate, this represents the finer angular detail resolvable. I guess at some point you might run into some qm type issues but I'm not qualified to talk about that!

To resolve details on exoplanets would actually require a modest few 100's of km's of diameter.

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Can you show your calculations for hundreds of km? –  endolith Dec 14 '11 at 17:45
Using $\theta\approx\tan(\theta)\approx R/d$ where R is the Earth radius and d is the distance to the exoplanet's star (from en.wikipedia.org/wiki/List_of_extrasolar_planets), I get D=7km for Lalande-21185 (d=8ly) and D=1800km for Kepler-11 (d=2000ly). So you need anything from about 10 km to 2000 km and more, depending on how far your exoplanet is. –  Arnoques Dec 16 '11 at 5:46

The resolving power of a device is limited by the so called Rayleigh criterion (commonly known as the diffraction limit). This is applicable for "classical" states of light (i.e light that has a coherent state representation). For non-classical light (squeezed light or entangled light), you can beat the diffraction limit.

A few interesting articles What Diffraction Limit?, Resonant Lithography,Quantum Limits on Optical Resolution,Quantum Imaging--pdf file. An interesting experimental overview provided by Boyd's group at Rochester.

The point is, for non-classical light the diffraction limit is $\lambda(N)=\frac{\lambda(1)}{N}$ where N is the number of photons in an entangled state (say a two mode state such as: $|N,0>-|0,N>)$ and $\lambda(1)$ is the wavelength dependence you would expect for classical light.

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It will continue to work as long as the light remains coherent. It's perhaps easier to consider two small telescopes a distance D apart, the light from both telscopes are combined to obtain a resolution of order of $\frac{\lambda}{D}$. The observed photons coming from a single point will not take a single path through either one of the telescopes, instead they are in a superposition of these two paths. This gives rise to an interference pattern (which depends on the wavelength). If photons from some point end up on some pixel of the sensor, then due to the interference, it will be mostly photons coming from within an angle of order $\frac{\lambda}{D}$ around that point that will hit that pixel.

If we take D of the order of millions of kilometers, then one would need to correct for the bending of light by gravity that leads to the path length difference changing on short enough time scales to cause a shifting interference pattern during the exposure. If the phase difference between the paths shifts then the combined telescope is effectively looking in a different direction. So, you'll get the analogue of motional blurring if this isn't corrected for.

The larger you make D the more such effects you'll need to correct for but in principle you could keep up playing this game. However, you'll eventually get into trouble when the two paths start to make an appreciable angle from the original source. As long as D is not too large, the angle between the two paths is small and the atoms that orginally emitted the photons will be with very large probability in the same quantum state. When this is no longer the case, the interference pattern will start to get washed out gain but this time you can't correct for that.

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