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The trouble with Hubble

  1. Mar 2, 2009 #1
    I'm trying to calculate the diffraction limit/angular resolution for the Hubble Space Telescope. I know this can be found using the formula:

    [tex]\theta = 1.22 \frac{\lambda}{D}[/tex]

    Where [tex]\lambda[/tex] is the wavelength of the light being observed and [tex]D[/tex] is the diameter of the objective lens (2.5 m on Hubble).

    Now since Hubble is able to observe a wade range of wavelenghts from the ultraviolet to the visible to the infrared spectrum I would describe the diffraction limit as an interval depeding on this range of wavelenghts.

    However, in all examples I've been able to find, even http://www.nasa.gov/missions/highlights/webcasts/shuttle/sts109/hubble-qa.html" [Broken], the diffraction limit of optical telescopes is calculated using a single wavelength of 500 nm (~cyan). What's the justification for using this particular wavelength for Hubble (and other optical telescopes).
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 2, 2009 #2
    Some later thought...

    500 nm coincides with:

    1. The irradience top of sunlight (and thus any G-type star). See: http://en.wikipedia.org/wiki/File:EffectiveTemperature_300dpi_e.png
    2. The sensitivity top of the human eye.
    3. The approximate middle wavelength of the visible spectrum.

    However (respectively):

    1. G-type stars aren't all that common, comprising only 7.6% of all stars, thus the usage is limited and this point cannot be applied to all visible starlight.
    2. The HST employs CCD sensors and not direct human observation, thus point 2 is irrelevant.
    3. This is the only point that seems relevant. However the HST would theoretically achieve greater angular resolution by using say the 380-450 nm range.

    So what's the point of using 500 nm when calculating the diffraction limit?
     
  4. Mar 3, 2009 #3

    Chronos

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    I see no trouble with hubble. TheMan112 partiially covered your objection. If you have a point, I will address it.
     
  5. Mar 3, 2009 #4

    Borek

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    TheMan answered his own post, so he probably doesn't feel like he knows WHY 500 nm was selected... :wink:

    Honestly, I have no idea as well. Is there any reasoning behind selecting 500 nm instead of some other wavelength? TheMan has listed some possible arguments, but they don't seem decisive.
     
  6. Mar 3, 2009 #5

    jtbell

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    This equation is derived by considering two fuzzy diffraction disks from point sources getting closer and closer to each other, and asking, "at what separation do they merge into a single indistinguishable blob?"

    The diffraction patterns of the two sources merge in a continuous fashion as their separation decreases. By convention, we use the separation at which the first minimum of one source's diffraction pattern coincides with the maximum of the other source's pattern. However, this is by no means a sharp cutoff. It's a "fuzzy" criterion, enough so that if we're dealing with visible light (400-700 nm), it suffices to pick a wavelength in the middle of that range. That would give 550 nm, but 500 nm is a bit easier to calculate with, and the edges of the visible range are "fuzzy" to begin with.

    As you can see, there's a lot of fuzziness here. :smile:
     
  7. Mar 4, 2009 #6

    Chronos

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    I can't disagree with the 'fuzzy' objection. I am tempted to push the envelope, but, will refrain.
     
  8. Mar 4, 2009 #7

    mgb_phys

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  9. Mar 5, 2009 #8
    Thanks for your thoughts guys, I talked to my professor who agreed on the whole fuzzy criterion thing.
     
  10. Mar 5, 2009 #9

    mgb_phys

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    I wish more people knew that - I had a proposal to do speckle imaging rejected once by a technical referee who told me that the image contains no information below the diffraction limit!
    In fact if you know what you are looking for you can model the scene, calculate back what the diffraction pattern should look like, compare and adjust the model. You can do a lot better than the diffraction limit.
     
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