I know that radio telescopes can be used in "interferometric fashion" to "emulate" telescopes sevearl kms large. Is something similar possibile for "visual" telescopes? Which are the physycal limits to interferometry? I mean: can two 100mm telescopes 10 meters apart emulate a 10000 mm. telescope?
Yes, I think, but not sure. In fact even the two halfs of one telescope can be (and are) used to get measure of stellar diameters via interence effects. This is hard to understand classically as each photon interfers with its self - sort of like each "goes thru" both slits of the two slit interferometer. The reason I qualified my "yes" is that photons do have a "coherence length." The interference fades out if one of the two interfering paths is longer than the other by the "coherence length" or more. Thus with telescopes 10,000 meters apart most stars would not have the same path length and it is not easy to redirect their outputs to one common screen or detector. I have measured the "coherence length" of some photons with a two path Mach/Zelnor interferometer (the most common one with two half silvered mirrors and two full silvered ones.) My photons were about 30 cm long in that when paths differed in length by more than this there was zero interference posssible. later by edit: Hell yes, in your case. I misread that as 10m telescopes 10,000 m apart. I am almost sure that some of the largest telescopes have many separate smaller mirrors now days and a few are bigger than 10 meters or at least under construction. - I am a retired physiscist, never much of an expert on astronomy and telescopes so may be wrong on parts of this.
I don't think a 10,000 meters distance would mean such a path-length difference, being any observed object more than 1,000,000,000 meters away! Please Register or Log in to view the hidden image! But I may be wrong, don't know almost anythong about optical physics...:shrug: Any web page to suggest? Please Register or Log in to view the hidden image!
Here is a site that I have had bookmarked for awhile. It lists many of the latest papers published, and links to specific experiments. Most of the emphasis is in the infrared wavelengths. http://olbin.jpl.nasa.gov/papers/2006.html
You do not understand how inteference works. EACH PHOTON INTERFERES WITH ITS SELF. In ways impossible for humans to really understand, when a photon has two different paths it can travel for source to the detector (human eye, cardboard screen, photo grapic film, even a rock - anything where it ceases to be a photon and transforms the energy it had into some other form.) IT GOES BY ALL POSSIBLE PATHS. (EACH AND EVERY ONE SEPARATELY!) If there are only two paths, A & B, For inteference to occur, it makes no difference if the photon has been traveling for million years to reach the detector. What is important is the DIFFERENCE in the path lengths (La -Lb) which for convenience I will call "D." Now each photon effectively has a length which I will call "L" Interference can occur IFF* L>D. (note how far the photon travels total does not enter into this requirement.) Think of the photon as a string of electric (and magnetic, but I will drop this aspect in this explanation) oscillations for example 50cm long (Billions or trillions of cycles of the E field oscillations within the length L and none before or after.) Now consider an instant in time (like a snap shot) when the positve E field peak, of cycle 2372 from start of L, going via path "A," is just falling on the detector. If the peak of the E field oscillation, of cycle 5233 from start of L, for going via path "B" is also having the peak of a positive E field, then "constructive interference" is occuring at that point on the detector. (If path B is with an negative E field peak at this time, then "destructive interference" is occuring.) (In this numeric example case, D = 5233-2372 wavelengths long.) Lets assume (only consider) point "X" on the detector where there is this constructive interference but now consider a slightly later time (by half the oscillation period) Then the path A is producing max negative E field at that point on the detector, but so is the path B. Thus the two path's E fields go up and down together ("in phase" is how this is described). I.e. at point X, constructive interference is CONTINUOUSLY occuring, because if the La differs at all from Lb, it does so by precisely integer number of "wavelengths." Now let us consider a near-by point "Y" on the detector where (because the paths A' & B' are slight different than paths A & B such that now the max negative E field of path A' occurs simulatneously with the max positive E field of path B'. Here at Y on the detector there is destructive interference. If the detector is a simple cardboard screen, there is bright light light at X and darkness at Y on the screen PROVIDED THAT D< L If D> L then the screen is uniformly bright (no interference pattern). This is because the photon via one path has not yet arrived at the screen until THE SAME PHOTON, (its "tail") via the other path has already been extinguished by hitting the screen. I know this makes no sense to a human with his every day experience as for humans it it not possible that THE SAME PHOTON travels by two entirely different paths form source to the detector, but no one has told the photon that this is "impossible" and in fact many "impossible" things are the way physics actually is when dealing with quantized small energy packets etc. I do not expect you to get a "warm /fuzzy feeling" of comprehension. If you have >50 years of living with this knowledge, as I have, at least it ceases to be "strange." I like to teach. - Think I do a reasonable good job of it, but if you want more discussion of this, just ask specific questions. Almost certain any you have will be based on the fact that this all SEEMS to be "impossible." - That is because it is not within the realm where humans have any direct experience. (Nature /Physics covers a greater range of things than humans directly experience.) ---------------------- *IFF is a reasonably standard way to quickly say: "If and only if" When D=L/2 the strength of the interference pattern will be half of what it is when D = 0. I.e. an equal mix of uniform illumination on the screen and a D = 0 max interference pattern, which may not have any "no light null" locations if there is more absorption in one path than the other.
thanks for the long explanation.... but unfortunately it does not answer at all to my question! I know about constructive/destrucive interference, but I think this is a different issue. What I am trying to understand are the quantities really involved in calculations about visual intererometry. As I said: if two radiotelescopes 1000 km away correspond to a single radiotelescope 1000 km wide, is it the same for visible telescopes? If no, why? Which are the limits? What is the correlation among mirrors widths and mirrors distance?
Only two things matter: (1) How close can you keep D = 0 (or any interger number of wavelengths if less than complete nulls are acceptable in your interference pattern) over the parts of the detector (for example a screen) where you desire inteference "nulls" to occure. (2) The total energy collected and delivered (via the two different telescopes) to the common (single) detector should be as equal as possible, for maximiun null darkness. That is it! (very simple in theory, not easy in practice.) Thus, the telescopes should be the same diameter (in most cases). How far apart they are will determine how big a region of the sky they can use to make its stars produce an interference pattern. (I.e. must keep D <<<< L if you want good deep nulls in your interference pattern.) The farther apart the telescopes are , the harder this will be to achieve and the smaller will be the region of sky* in which stars can be used to produce interference, even when you acieve it. When you really understand how optical interference is produced, (1) & (2) above will be obvious to you and you will not need to ask anything more. --------------------- *If D>L, for all of the sky, you could at least in principle, jack up the more distant from star telescopes so that the path lengths to the star via each telescope are more equal. If you bother to think about this, you will understand why only a small part of the sky will satisfy the D is approximatel zero condition for good nulls, understand the geometry of the problem etc. BTW, post 5 does answer all your questions - you just do not understand the nature of the requirements yet.
I hope so; the Very Large Telescope Array is doing this right now. http://en.wikipedia.org/wiki/Very_Large_Telescope#Interferometry_and_the_VLTI but only to see bright sources in detail, as it loses a lot of light in the process.
You should also check out the twin Keck telescopes here on the Big Island. http://www.keckobservatory.org/about.php Regards, Walter L. Wagner
"The next phase of the W. M. Keck Observatory is underway as teams of scientists and engineers continue work on improving the Keck Interferometer. The Keck-Keck Interferometer combines the light of both Keck telescopes to obtain a tenfold increase in resolution. It is a significant cornerstone of NASA?s "Origins" program, which ultimately seeks to identify and characterize planets around Sun-like stars. The interferometer will also help astronomers detect giant gas planets, measure and characterize planet-forming dust around stars, and obtain extremely high-resolution images of protoplanetary disks. It has already produced significant results, including observation of a supermassive black hole in the center of a galaxy (NGC 4151) more than 40 million light years away."
