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I X-Ray scattering in crystal

  1. Mar 23, 2017 #1
    This question is as dumb and simple as it can be: Why is the path difference 2dsinθ? Or better said: Why x-ray scatters like in the image? (Couldn't there be another scattering angle for both atoms other than θ?)
    640px-Bragg_diffraction_2.svg.png
     
  2. jcsd
  3. Mar 23, 2017 #2

    sophiecentaur

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    Not dumb, if you are relatively new to interference theory. Start with a large number of sources in a regular row (2 - Young's Slits - would be a way into the problem) and feed them in phase,. They will create an interference pattern with the angles of the maxima set by the wavelength and the spacing (the 2dsinθ formula). If you use an incident beam to illuminate a similarly spaced row of scattering points to try to get the same result, the only incident direction that will cause the scatterers to have a phase that's a whole number of wavelengths (equivalent to being in-phase), will be when the angle of arrival is a mirror image of the diffraction maximum. No other condition will provide the appropriate phasing of the scatterers. This is just like a mirror reflection but only for the correct angle I and R.. The deeper and wider the crystal, the sharper and better defined are the maxima. If the angle of incidence is wrong, then the scatterers do not add up coherently.
    They glibly refer to the Bragg Reflection Formula but they never seemed to put it in the above way, when I learned about it. (Perhaps I just wasn't listening!!)
     
  4. Mar 23, 2017 #3
    The atoms scatter at all angles, sure.
    But the scattered intensity has maxima (peaks) just for some specific angles. This diagram is drawn to calculate the angles for these peaks.
    It is more like an aid to finding Bragg's formula than a depiction of what really happens. But it gives the same answer as the more complete theory so it is still used.
    You can look up "Laue's theory of x-ray diffraction" for a model which does not assume any specific scattering direction in the beginning. The result applies not only to x-ray diffraction but to other scattering processes as well.
     
  5. Mar 23, 2017 #4

    sophiecentaur

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    Lucky that Waves of all wavelengths work much the same. Radio engineers use similar calculations.
     
  6. Mar 23, 2017 #5
    And people into ultrasound array transducers too.
     
  7. Mar 23, 2017 #6

    sophiecentaur

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    One stupid omission was to point out that, to get the beam pointing in a particular direction, there has to be a steady phase 'tilt' across the row and not have them cophased. Everything else follows.
     
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