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Bragg condition reflection vs. diffraction

  1. Dec 23, 2011 #1
    I've been asking myself this question that is not entirely clear to me: the Bragg condition used to describe the constructive interference of waves on a crystal is based on the assumption of reflection. However, x-rays diffract on a crystal rather than reflect, so theoretically there is always some ray which refracts at an angle that would yield a path difference just right to produce constructive interference. Why then does the Bragg condition based on a reflection model work?

    Thanks for any answers!
     
  2. jcsd
  3. Dec 23, 2011 #2

    nez

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    You are right...but that takes you back to Laue refraction equation, a few months before Bragg's equation and its more complicated.
    It was Bragg's assumption to treat diffraction as reflection from lattice planes which led him to the simpler well known condition.
     
  4. Jan 3, 2012 #3
    Well ok, I understand that the full description the underlying physics is given by the Laue equations. But what I don't understand is why does the Bragg condition work at all since it is based on an "unphysical" assumption of reflection. Is there some simple reasoning I am missing why the reflection description, even if not correct, nevertheless accurately produces the condition for constructive diffraction?
     
  5. Jan 5, 2012 #4

    nez

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    "it is based on an "unphysical" assumption of reflection"

    Not like that...There is real reflection, in the sense that the combined scattering of x-rays from crystal planes can be looked upon as reflections from these planes.

    At certain glancing angles, these reflections are in phase and produce maximum intensity. By measuring theses angles for different set of planes, knowing the wavelength, we can calculate interplanar spacing from Bragg's eq.
     
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