Have a look at this example:

In addition, the Keck I and Keck II telescopes can work together as the Keck Interferometer. The 85-metre (279 ft) separation between the two telescopes gives them the effective angular resolution in one direction of an 85-metre (279 ft) mirror

So, If you focus in on one image, is combining two telescopes (through optical interferometry) really equivalent to having a larger mirror?

It's a follow-up question to an answer I got at here...


It rather depends on points of definition, but it's not generally equivalent - light losses in interferometry are significant, for example, while processing and data reduction is something of a mystery to most astronomers (i've been tangentially involved with a VLTI project and it certainly is to me). On the VLT the interferometers also operate in the infra-red (1-2.4μm for AMBER, and 8 to 13μm for MIDI), I don't know about Keck interferometry as i've not used it.

To give an example of the limitations, the documentation for MIDI says "On the UTs [i.e. the 8-meter telescopes] it is possible to reach an H-band magnitude of 7", which is pretty bright - a 'real' large telescope would go far deeper. But, on the other hand, the documentation says "AMBER is able to resolve features between 2mas (milli-arcsecond) and 50mas with the UTs" whereas NACO, the adaptive-optics assisted near-IR imager on UT4 (i.e. a single telescope), can reach a best resolution around 50mas.

So best to think of optical/IR interferometry as allowing you to achieve the resolving power of a (much) larger telescope under certain, possibly quite severe, restrictions, rather than as a general equivalent

  • $\begingroup$ The magnitude limit isn't just form the loss of collecting area compared to a single dish - it's because there is the question of how much power in each resolution element. If you image an object into a single point you can obviuosly go much deeper than if you are looking at the surface brightness in mas regions. Even with an interferometer you can observe point-like objects much fainter than ones with structure - but there wouldn't be much point! $\endgroup$ – Martin Beckett Apr 11 '12 at 16:14

It's also not quite the same as an image taken with a single dish. The interferometer doesn't produce an image, it produces the fringe pattern you would get if you projected the image through a double slit.
Essentially what you have is a constraint on how much structure there is on various scales for a slice through the object at a certain set of angles.

You can't unfortunately convert this directly into an image. What you do is try and generate an the image of the object, using your best guess about the shape, calculate the fringe pattern you would get, compare that to the measured pattern, adjust the guess ....and repeat...

Depending on how good your software (and your guesses) you get certain artifacts in the resulting image. Of course you also get artifacts in a 'normal' image.

  • $\begingroup$ +1, good answer. The details of the iterative process of matching the fringe pattern to the model is the bit that I find very hard to get my head around. $\endgroup$ – strmqm Aug 10 '11 at 9:18
  • $\begingroup$ @strmqm - the interferometer gives you a fourier transform of the image. But not the entire UV plane, just a few lines, points and curves in it (depending on how many baselines you have). You try to recreate an image that matches this transform. The main algorithms are called "memsys" and "clean" $\endgroup$ – Martin Beckett Aug 10 '11 at 15:22
  • $\begingroup$ Is the double slit fringe pattern the result for any interferometer, or only a two element one? If the latter, how are the patterns for more complex designs described? $\endgroup$ – Dan Neely Apr 11 '12 at 15:56
  • $\begingroup$ @DanNeely - you get a double slit pattern for each pair of dishes - all superimposed (in time or space - depending on your detector system). You normally arrange the dish pattern so that each pairwise combination is at a different spacing and/or angle so you can capture more of the fourier plane in each 'shot' and con unravel the individual pairs $\endgroup$ – Martin Beckett Apr 11 '12 at 16:11

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