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Fringes of a perfect beam?

  1. Sep 3, 2010 #1
    Hi All,

    Just got to thinking about interference, if we have two sources (say a laser beam split in two), and introduce a phase shift between the two by making them travel different path lengths, then recombine them, we see fringes in the wavefront right. (interferometry)

    My problem is this, if we have two perfectly monochromatic and perfectly collimated (i.e. no divergence as they propagate out to infinity) beams, the fringes would not be visible right? as the transverse pattern occurs due to the fact that the beams are indeed slightly divergent. You would however see a spot of light whose brightness in proportional to the phase difference between the two beams.

    so the only reason we have interferometry is because we cant make perfect beams.
    Does this make sense?

    Thoughts please.
     
  2. jcsd
  3. Sep 3, 2010 #2

    Andy Resnick

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    If I understand you correctly, it is true that in order to use interfereometry as a tool, there needs to be some bias present (some fringes) to increase the accuracy.

    But I'm not sure I understood your question.
     
  4. Sep 3, 2010 #3
    I am not saying I agree, and I am not saying I disagree. What I am sure of is that Heisenberg's principle prevents both a perfect monochromatic or a perfectly collimated beam. Both concepts may be useful for our understanding, but neither is physically possible.

    In this sense, it's not really a matter of "our ability" to make them, it's just that close analysis reveals that the idea of making such a perfect beam is as absurd as building an infinitely long rod or something.
     
  5. Sep 3, 2010 #4
    yes, this is what I was getting at.

    The fact that there are transverse modes in any beam means that interfering two identical beams will cause an interference pattern.

    If there are no transverse modes, then there will be no spacial variance of interference in the plane of propogation
     
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