# What would be the maximum theoretically possible angular resolution?

I have studied some basic astronomy, but I have difficulty conceptualizing the physics of luminosity and optics. We use electromagnetic spectra to detect existence and properties of distant planets, but I am curious about the extent of what could theoretically be optically captured.

(I am unsure if optical is the correct term, but I'm referring to what could be directly captured vs. computed.)

Assuming the use of technology far beyond what we currently are capable of, what is the maximum possible optical angular resolution we could theoretically obtain of a distant sky object? Or, for example, of the surface topography of an outer planet/moon/dwarf planet?

What would be the limiting factor at the extreme end of possible angular resolution?

• I don't really understand what's going on here, but the space-Earth radio interferometer RadiAstron/VLBI maybe has hit such an astrophysical limit for resolution in the form of interstellar media scattering the photons. Commented Nov 10, 2015 at 8:57
• duplicate of the unanswered astronomy.stackexchange.com/questions/8675/… Commented Nov 10, 2015 at 10:15
• There are no limits imposed by science. Build an arbitrarily large system, you get arbitrarily sharp resolution. Commented Nov 10, 2015 at 19:03
• There are limits imposed by science, but they are quantum mechanical so for astronomic purposes as good as no limits! Commented Nov 14, 2015 at 12:53
• Along Florin's lines, can't you basically just create telescope "binoculars"? If we're getting really far-fetched-but-kinda-plausible-over-really-long-timescales, install a series of super-Hubbles or whatever throughout the galaxy. Basically a galactic-scale VLA/LBT. Alternatively, you could just launch any ol' telescope (your own eyes and body in cryogenic hibernation or something, say) out into space and wait for it to get close enough to achieve the desired resolution. Commented Jan 2, 2018 at 7:05

The answer that you are looking for rests on two main pillars. Are you limited in your ability to resolve objects because of the atmosphere like most ground-based telescopes, or because of the fundamentals of the wave nature of light (as per space telescopes)? I am assuming the latter in your question.

One of the biggest rule-of-thumb equations when it comes to astronomy is the equation that you are looking for in your answer. Astronomers define angular resolution with regard to the Rayleigh criterion: two point sources are regarded as just resolved when the principal diffraction maximum of one image coincides with the first minimum of the other (Wikipedia). See Airy Disk functions and the Wikipedia article on Angular Resolution for a better idea as to what this means.

The accurate, but still approximate, equation for the resolution of an object $\theta$ (radians), for a given wavelength of light $\lambda$, over some telescope's diameter $D$, is given below so long as the wavelength of light and the diameter is measured as the same length unit (i.e meters, feet, angstroms).

$$\theta = \frac{1.22 \lambda}{D}$$

Obviously, the atmosphere is a pain, so, adaptive optics in most famous telescopes try and correct to this limit. To exceed this limit, major breakthroughs would need to happen in the wave nature of photons themselves. If your "technology far beyond" includes this, then, I don't think your question can be answered as you would hope aside from the standard "who knows". Currently, only space-based telescopes operate at this limit.

For the most part, this is really all you need. It does not matter what object you are observing, this still always applies.

Edited: Included unit specifications.

• what are the units for each of the terms in your equation? Perhaps you can edit the answer to add them. Commented Jan 2, 2018 at 17:54
• This is the limit for an unobstructed aperture, but more complex aperture functions can have somewhat higher or lower resolution. For example astronomy.stackexchange.com/a/24182/7982 where a carefully constructed aperture can achieve the smaller Michelson criterion, rather than the Rayleigh criterion described here.
– uhoh
Commented Jan 3, 2018 at 3:27