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Confused about Imaging in Astronomy

  1. Nov 28, 2008 #1

    cepheid

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    I know that the bigger the telescope aperture, the more photons you can collect, and the fainter the objects you can see. I also know that the bigger the telescope, the smaller the diffraction disc of a point source imaged with it, and therefore the greater the angular resolution (i.e. the sharper the images and the better your ability to see detail in them). When we give advice to prospective telescope buyers in this subforum we emphasize that magnification is nothing, and aperture is everything (along with a sturdy mount :wink:). I would certainly agree that the ability to zoom in on an object by some ridiculous amount is useless if there are no features to see due to inadequate resolution, the image is too dim, and the field of view is too narrow to provide comfortable (and sensible) viewing of the object.

    However, I am interested in the role magnification plays in imaging with professional research-grade telescopes. You don't hear that word magnification used much. All that seems to matter is that you have a large enough primary mirror. We see Hubble images of galaxies in which individual stars are resolved. Surely this depends upon more than just the diffraction limit. For if two point sources are nicely resolved by the Rayleigh criterion, but their physical separation on the image plane (i.e. the detector) is so small that our eyes cannot determine that they are distinct, then what use is that? Sure, one could always "zoom in" on the image itself, but only to a limit. If two point sources have a subpixel physical separation on the detector, then that suggests you've made a poor choice of plate scale. The plate scale is determined by the effective focal length of the telescope optics. You want to choose it such that your object takes up a decent amount of the detector area. Isn't that what angular magnification does? You are choosing your effective focal length such that you're "zoomed in" on this object by a sensible amount (i.e. the angular size of the image is suffcient). So these telescopes do have to provide a certain amount of "magnification*." Am I right?

    *The only danger is in talking about that to the layperson and implying (incorrectly) that that magnification is the primary purpose of a telescope and is what allows us to see these amazing Hubble images (e.g. of a galaxy with individually-resolved stars), without pointing out that we could just as easily have a crappier telescope that shows an equally magnified image of said galaxy that looks like nothing more than a blurry mess due to insufficient angular resolution. Again, am I right? Comments and corrections welcome.
     
    Last edited: Nov 28, 2008
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  3. Nov 28, 2008 #2

    cepheid

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    I just did some more thinking about it, and I realized that I completely forgot to consider that it's not just the focal length of the telescope itself that is of concern. There's more optics in the mix. It totally depends upon what instrument you decide to throw in the way of the telescope beam. My failure to mention that probably made me seem like a bit of a newb. Since I've been using Hubble as an example, I'll stick with that. I imagine that the field of view provided by the ACS is different from that provided by the wide-field planetary camera (or whatever it's called). I imagine that "wide-field" here must have some specific meaning. The image on the detector must correspond to a portion of the sky that is x number of square arcseconds. So I guess it's a matter of choosing the right instrument for the right job. Need the image of a distant galaxy to take up the entire area of your CCD? I guess you'll choose a different instrument than the one you'd use if said object was, say, Saturn. So I guess it's matter of taking into consideration the angular size of the object in question, the telescope focal length, and the optics of your instrument. Still, I would wonder whether creating a telescope + suite of instruments that is as versatile as Hubble is a difficult optical design problem, or whether people have basically got it nailed?
     
  4. Nov 28, 2008 #3
    Hubble and other means of relatively detailed imaging don't always necessarily rely on a snap-and-go image but rather use long exposure times or composites of images taken in the same reference frame at different times.
    This would allow more photons to be collected from the target leading to better detail, as you know. (Think Hubble Ultra Deep Field)
    This doesn't answer your question but touches on it since having the ability to see very distant and faint cosmological bodies is going to be futile with just a quick peek in any hardware setup.
     
  5. Nov 28, 2008 #4

    mgb_phys

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    Magnification is meaningless for an imaging system such as a camera.
    The focal ratio of a large telescope primary is generaly as fast as possible in order to keep the tube length short and so the costof the dome down.
    This is a problemwith 8m scopes because the plate scale becomes very large, so a galaxy (or whatever) covers an inconveniantly large number of mm at the focal plane of the camera.
    It also becomes more difficult to acheive an optically corrected focal plane with large telescopes and so their field of view is reduced.
     
  6. Nov 29, 2008 #5

    cepheid

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    Can you elaborate upon this? Let's forget about the telescope for a second, and just consider me taking a picture with my digital camera. My understanding is that if I zoom in or out, then I change the effective focal length of my optics, thereby changing the field of view. How is this not some form of "magnification?"
     
  7. Nov 29, 2008 #6

    cepheid

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    If what you mean is that you want the focal length to be as short as possible relative to the aperture size, which means that you want a low f number, then this makes perfect sense to me...

    Side note: the use of the word "fast" in this context, however, doesn't make much sense to me. In the context of a camera, where it is the aperture that is changing, and the focal length that is held fixed (for a given shot anyway), the word fast makes sense, because the lower the f number, the bigger the aperture, and the less exposure time you need to correctly expose your image. For the case of the telescope, where it is the opposite i.e. aperture is held fixed, and focal length is varied (by the designer anyway, until he decides upon it), what does the word "fast" mean? It seems like an inappropriate insertion of optics jargon. Why don't we just say short focal length or small focal ratio? Isn't that much clearer?
     
  8. Nov 29, 2008 #7

    cepheid

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    See now here you are referring to plate scale. Isn't that just a manifestation of the magnification of your optical system?
     
  9. Nov 29, 2008 #8

    cepheid

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    What do you mean by optically corrected? Do you mean that the shorter the focal length, the more severe the Seidel aberrations? (That seems to ring a bell to me). In which case...you would be unable to use off-axis portions of your image. Is THAT why the field of view would be reduced?
     
  10. Nov 29, 2008 #9

    mgb_phys

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    An afocal instrument (one which doesn't make an image ) like a telescoep with an eyepiece or a pair of binoculars has a magnification based on the relatie sizes of the input and output beams. Or on the angluare extent of the object and image (same thing).
    An instrument which makes an image doesn't have a magnification in the same way, instead the angular size of the object makes a certain linear size on the film.
    You can't say a camera makes an image 20x the original all you can say is that Xdeg in the image is Ymm on the film - this angular magnification is called the plate scale. In telescopes it's generally quoted the other way up - so plate scale is the number of arc-sec/mm, higher 'magnification', ie smaller field of view is smaller plate scale.

    What limits a camera telescope field of view is the plate scale, the size of the imager and what angle field can be corrected.
    As the telescope gets faster it's more difficult to correct the field of view off axis. To get a large flat field of view you need a lot more optical elements in the corrector, which since the focal plane is large - gets big and expensive. Or you can have a curved focal plane like in a Schmidt.
     
    Last edited: Nov 29, 2008
  11. Nov 29, 2008 #10

    turbo

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    Here is an object lesson. Greg Parker has a hyperstar now that reduces the focal ratio of his SCT, but most of these images were made with a 90mm (aperture) Takahashi Sky 90 and LOTS of exposure time. He emails the images to Noel Carboni, who processes them, and they have a KILLER book in the works. Having been a long-time member of the Our Dark Skies forum, I have gotten front-row seats to these images as they are captured, built up over time, added to, etc. Noel and Greg are amazing!

    http://www.newforestobservatory.com/

    ODS link:
    http://forum.ourdarkskies.com/

    You can lurk if you want on ODS, but if you join, and ask Noel, he will send you a free set of Adobe Photoshop/CS actions with which to process astrophotos - a very generous gift, indeed, available only to members of ODS, I believe. BTW, you can join ODS for free.
     
    Last edited: Nov 29, 2008
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