# Why array of telescope is used?

To increase the resolution of an instrument, smaller wavelength and larger aperture is desirable. It is mentioned in some textbooks that the "effective" diameter of a telescope can be increased by using arrays of smaller telescopes. I just wonder why it is possible because every telescope is separated.

Picture yourself looking into a large mirror on the wall. Now picture the mirror is made up of smaller, tiled mirrors. You will still see your reflection. If you begin to remove the tiles, so that there are only a few left, you can still use them to reconstruct the image of your face that was given by the original mirror. This is what is happening with an interferometer. Astronomers are constructing an image measured by the "full mirror" (the longest baseline) based on the information they get from a few tiles (individual antennas).

• I think as they are using arrays of telescopes they certainly get better resolutions, I am just not quite sure about the effect of diffraction on smaller apertures. If I look through a circular tube with diameter D which consists of many smaller tubes inside, like a bundle of drinking straws, and if the diameter of these drinking straws are small compared to the wavelength of visible light, can we still get the same resolution as looking through the big tube with diameter D but no drinking straws inside? – Kelvin S Aug 27 '15 at 4:34

Greatly rewritten based on feedback in comments
In order to understand this issue, it is worth considering what a telescope (or any optical / radio imaging system) really does.

Taking a simple parabolic mirror, the shape is chosen such that the total path length for all rays "from infinity" to the focal point is the same. By making the path lengths the same, the signals will al ibe in phase when they arrive at the focal point, which results in constructive interference. We see this as a "bright spot" at the focal point. I explained this in detail in this earlier answer.

Now if you consider a second set of rays that is at a slight angle to the first set of rays, the path lengths will no longer all be the same. There will come a point where the rays from one side are so far out of phase with the rays from the other side, that when you add them all together they cancel out exactly. For a circular aperture, this happens at an angle $$\alpha = \frac{1.22 \lambda}{d}$$ which is the usual expression for the angular resolution of a circular aperture (while the expression tells you where the zero is on one side, this is a pretty good approximation of the width of the central peak as shown in this diagram from source - where of course 0.61/R = 1.22/d :

"Resolution", then, is interpreted as "how far off axis does the signal from a point source decay to zero" which is another way of saying "how close together can two points be and be seen as distinct" (it's not exactly the same thing, but it's usually "good enough" to equate these concepts). The resolution of the individual component doesn't matter too much - while its signal will be "virtually unchanged" as the source moves off axis, the phase relationship with the signal from the next (small) element will change much more rapidly. So while the components by themselves don't get better resolution, the combined signal does.

This has a few implications. First - any imaging system needs to maintain the phase relationship between the incident beams to much better than a wavelength - this is why an optical telescope has really smooth surfaces, but a radio antenna can be made out of a "roughly shaped" surface. Second - it means that we don't need to have a continuous circular lens/mirror in order to do imaging. Any set of reflectors that result in the detected signals remaining in-phase will behave in the same way. Of course the larger the total area, the more signal is detected - but if you are interested in angular resolution and you have enough signal, you need to increase the width of your optics; you don't necessarily need to increase the area.

And so it becomes possible to dream up telescopes that have unusual shapes, but that are particularly good at resolving along a particular axis. This is the principle behind very long baseline interferometry but it works on any scale. The key is that you have to maintain the phase relationship between the signals from different parts of your "mirror" to within a fraction of a cycle - the better you do this, the greater the resolution of the system will be. At optical wavelengths this quickly becomes really, really hard - at the wavelengths used in radio astronomy (meters down to millimeters) it is quite achievable.

A helpful description of the process (and the difficulty of doing this optically) is found in the Wikipedia article on aperture synthesis which states:

Aperture synthesis is possible only if both the amplitude and the phase of the incoming signal are measured by each telescope. For radio frequencies, this is possible by electronics, while for optical lights, the electromagnetic field cannot be measured directly and correlated in software, but must be propagated by sensitive optics and interfered optically. Accurate optical delay and atmospheric wavefront aberration correction is required, a very demanding technology which became possible only in the 1990s. This is why imaging with aperture synthesis has been used successfully in radio astronomy since the 1950s and in optical/infrared astronomy only since the 2000 decade. See astronomical interferometer for more information.

• It seems like you're trying to describe very long baseline interferometry, but without using any of the language... – Kyle Kanos Aug 26 '15 at 20:56
• @Floris, Thanks all for answering. I can understand that every small telescope can be seen as a small tile of a big mirror/telescope. But I still wonder why this kind of arrangement can increase resolution. For a single aperture you need to have big diameter so that the diffraction effect is small (smaller angle for the first dark ring). For an array of telescope each telescope still has relatively small aperture and diffraction would still affect the image. – Kelvin S Aug 27 '15 at 3:34