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Seeing around corners

  1. Sep 11, 2006 #1
    Take an object with well defined features, illuminated by a full spectrum source and located around a sharp, 90 degree vertical corner. How well can an observer hidden behind the corner from the object and at the same height reconstruct its image from the diffracted light?
     
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  3. Sep 12, 2006 #2

    Meir Achuz

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    As well as I can see you now.
     
  4. Sep 12, 2006 #3

    Danger

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    :rofl:
    To some extent, it might depend upon ambient conditions. Loren seems to be alluding to the mirage effect, which is very real. I've never heard of it working around a 90 degree bend, though.
     
  5. Sep 12, 2006 #4

    LURCH

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    I would say it's entirely dependant on the mechanism used to collect the data (the diffused light). For example, if the observer used a mirror, he could tell a lot about the object. If he used no instrument, he could tell nothing at all. If the experiment took place in a vacuum, he could tell nothing at all, because light doesn't diffuse in a vacuum.

    In a perfectly dust-free "clean room", almost no data. In a smoke-filled bar, color and size perhaps (if the source of illumination were bright enough).
     
  6. Sep 12, 2006 #5
    Folks,

    Pardon, I was referring to diffraction than diffusion. Would that change your answers? Consider the phenomenon in vacuo.
     
  7. Sep 12, 2006 #6

    Claude Bile

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    If you are observing right on the wall, you would get no information whatsoever. This is because light takes a finite amount of time (and distance) for the diffraction angle to evolve from 0 to whatever the far-field diffraction angle may be.

    Claude.
     
  8. Sep 12, 2006 #7
    In one-slit diffraction, all interference fringes (with the information they represent) but one are not along a straight (classical) path from the light source, through the slit and to the screen. The same argument holds for a object, corner and observer.

    All that remains is for the observer to reconstruct the information about the object from the fringes, but can it be done in practice? (Using some kind of Fourier transform?)
     
  9. Sep 13, 2006 #8
    Remember that this one-slit example is illuminated by a full-spectrum source. The interference displayed on the screen arises from those wavelengths interacting with the geometry of the slit itself. This experiment, as explained above, is analogous to the "corner" situation.
     
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