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Photon information after passing the pupil

  1. Jan 20, 2016 #1
    Hello,

    I have a question: lets assume that some photons are passing ~10m pupil, like in largest telescopes. Are they EXACTLY identical AFTER passing, than before? Or is there some lost of infromation becouse most part of their wavefront is cutted off and we only register ~10m part of it (from what I know in fact it is infinite)?

    Best regards,
    Alex
     
  2. jcsd
  3. Jan 20, 2016 #2

    A. Neumaier

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    No. Any interaction (except for special non-demolition arrangements) changes the state. For this reason, nothing in physics is EXACT, except if defined to be EXACT (like the speed of light).

    But in many cases, one can ignore the change because it is inessential for what one is trying to model.
     
  4. Jan 21, 2016 #3
    Just to be clear, the photons entering your eye are not from the stars or even the ones that go through the telescope!
    The path from the star (or any other thing) comprises a series of segments of travel by different photons being emitted and absorbed. The average distance between emission and absorption is called the mean free path. The mean free path for photons in air at ground level is very short; the ones entering your eye are being emitted right in front of the eye... and because of the aqueous humour in the front cavity before the eye's lens and the vitreous humour between the lens and retina, the photons that ultimately initiate the sequence resulting in a depolarization down the optical nerve are emitted immediately in front of the retina; so all the light you have ever seen was emitted inside the eye itself... ! :)
     
  5. Jan 21, 2016 #4
    This is not exactly the point. Im wondering, does the photon aquire momentum uncertainity due to reflection from the mirror ( should we apply Heisenberg principle to this case)? And if yes --- if we would amplify this photon, i.e. by laser medium crystal, the resulting photons would also be affected by additional delta p?
     
  6. Jan 22, 2016 #5
    That sounds wrong to me. If the mean free path of photons were short, then air would be opaque, not transparent. Now, there is some scattering going on (responsible for the blue sky), but it isn't strong enough to wipe out our visibility of things like the moon. The mean free path must be on the same order of magnitude as the height of the atmospheric layer. If we look at images of Earth taken from space, we can see the ground (in contrast to images of Saturn).

    The problem with this question is that it depends on what "EXACTLY identical" means. Physics can answer questions about what result a particular hypothetical measurement might make, but it can't answer a question simply about some semantic labeling.
     
  7. Jan 27, 2016 #6

    jfizzix

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    By passing through an aperture of finite size, we are imposing an upper cutoff to the spatial frequencies (i.e., transverse momentum) that can comprise an image of the thing we're looking at. The image as a result would necessarily have a little bit of blurring (the amount of which would be inversely related to the size of the aperture). This is part of what is known as the diffraction limit for imaging. There are ways to mitigate this blurring while still getting a good image, though:
    See:
    http://www.pas.rochester.edu/~jschneel/Howland_DirectXPmeasCS_PRL2014.pdf


    Tangentially, the magic number for the human eye's diffraction limit is approximately 10,000.
    That is, the Rayleigh diffraction criterion (with a minimum wavelength of 400nm and a maximum pupil width of 5 mm) gives a smallest resolvable angular width of about (1/10,000) radians.

    At a given distance away from you, the smallest thing your eyes are capable of resolving would be a width of 1/10,000 of that distance.
    For example, with modern smartphones having a resolution of as much as if not better than 500 pixels per inch, holding such a phone more than 20 inches away would make it literally impossible (without additional magnification) to discern individual pixels with the naked eye.
     
    Last edited: Jan 27, 2016
  8. Jan 27, 2016 #7
    I find one estimate (absorption), restricted to the mean free path of photons being absorbed by carbon dioxide, of 33 meters.
    "This means that a quantum/wave shifts about 33 meters to hit onto a molecule of carbon dioxide in the atmosphere."

    Another estimate (scattering), including water vapor in clouds, of 10 meters.
    "At visible wavelengths, the scattering mean free path in clouds is of order 10 m."
     
  9. Jan 29, 2016 #8
    10 meters sounds right for clouds. That's why we can't see through clouds. The 33m number for carbon dioxide must be for infrared or something. Carbon dioxide is invisible.
     
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