# When one can make telescope that can distuingish 1 mas?

When one can make telescope that can distuingish(angular resolution) 1 mas object by angular diameter? If already possible, tell the name of telescope/observatory. How to calculate angular resolution $\theta$ of Telescope from the diameter $D$ of its mirror? Just in case I'm understood wrong; is it through Rayleigh equation; $\theta \approx 1,22 \lambda/D$?

Indeed the angular resolution (smallest resolvable angle) is typically given as $$\theta \approx \sin\theta \approx 1.22 \frac{\lambda}{D},$$ where $\lambda$ is the wavelength of interest and $D$ is the diameter of the aperture.

This applies in the ideal case (the Fraunhofer limit), in which the image formed is the Fourier transform of the wavefront at the aperture. The Fourier transform of a point source "at infinity" (i.e. a source whose wavefront is essentially a plane wave with no curvature over the size of the aperture) is an Airy pattern.

That is, think of the aperture as being a function on the plane $\mathbb{R}^2$ that is $1$ on the disk of radius $D$ and $0$ everywhere else. Take the 2D Fourier transform of this, inserting $\lambda$ in the right places to make the units work. The result is an Airy pattern in which the distance from the center to the first ring of minimum intensity is approximately $1.22\lambda/D$.

If you have two point sources, you can't reasonably expect to tell them apart if they are closer together than this (the central peaks of their respective Airy patterns will be mostly overlapping).

Now $1\ \mathrm{mas} \approx 4.85\times10^{-9}\ \mathrm{rad}$, so at a visible wavelength of $550\ \mathrm{nm} = 5.5\times10^{-7}\ \mathrm{m}$, the aperture needs to have a diameter $D \gtrsim 140\ \mathrm{m}$. This is quite a bit larger than the two individual Keck telescopes, which are essentially the largest optical telescopes in the world at a $10\ \mathrm{m}$ diameter.

However, one can overcome this size limitation with interferometry - take two or more telescopes and combine the light they receive before taking an image. Angular resolution in a given direction basically depends only on the furthest separation along that axis. The Keck telescopes, separated by $85\ \mathrm{m}$, can operate together, but only in infrared wavelengths for an angular resolution of about $5\ \mathrm{mas}$.

Wikipedia has a list of interferometers you can browse. The longest baseline for an up-and-running array is with CHARA, whose baseline can go up to $330\ \mathrm{m}$ and which can apparently work in the visible. (SUSI is not up to its listed specification yet.) Note for that list that visible bands are B (blue), V (visible/green), and R (red), while most of the other letters are for wavebands in the infrared and beyond.

• hmmm...I think there a more factors here. If this worked as you describe it, why did they put the Keck twins "only" 85 m apart? Why not 1 on Mauna Kea and 1 on La Silla, for example? And why are concepts like these then needed, if you could just build multiple, simpler telescopes? – Rody Oldenhuis Sep 2 '13 at 11:52
• Interferometry brings it's own engineering limitations, such as every optical surface must be accurate to $\lambda$/2 , ie a few hundred nm. That's not easy over any distance and pushes engineering to the limits at CHARA. – Nic Sep 2 '13 at 12:11
• And re 'concepts like this': Single dish telescopes will have a massive advantage in light collection and therefore sensitivity if lacking in resolution to interferometers – Nic Sep 2 '13 at 12:12
• @Nic: Agreed. But do you agree that it is not true that an interferometer is equal to a single mirror/dish the size of the baseline? I guess my problem here is that the answer makes it sound like that... – Rody Oldenhuis Sep 2 '13 at 12:27
• If you want to put twin telescopes on Mauna Kea and on La Silla, you have to transmit the information received from one telescope to the other while keeping quantum coherence. There is a proposal to do this using quantum repeaters, which can be implemented using ideas from quantum information, but this is really hard. It's much easier for radio telescopes, because of the longer wavelength. – Peter Shor Sep 2 '13 at 13:01