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Bragg's Law in Crystallography

  1. Jan 10, 2013 #1
    I took a course on protein crystallography last year and there's one thing I couldn't figure out then, and still can't figure out now. My understanding of Bragg's law hinges on the fact that in-phase scattered waves constructively interfere, and the requirement to be in-phase is met only when the extra distance traveled is an integral multiple of the photon's wavelength. With this premise and a diagram of two atoms scattering a photon, it becomes apparent that nλ = 2dsin(θ).

    What I don't understand is why only maximal-intensity waves are observed. Unless the two waves are π radians out of phase, there should still be some constructive interference that gives rise to amplitude at the detector plate. Therefore, there should be some intensity at almost any value of θ. The wikipedia page on Bragg's law acknowledges this fact:

    Unfortunately, I don't see a clear and unbroken line of reasoning in the explanation. So many atomic planes means more complex interference... why is that guaranteed to produce mostly destructive interference? For help, I turned to Gale Rhodes' Crystallography made Crystal Clear. He explains it as follows:

    This is starting to approach an answer but I don't think it's complete. Why must there exist some parallel plane that produces a 180 degree phase shift? Why do such cancelling planes not exist for the set of planes that comply with Bragg's equation?

    Thanks for any help!
     
  2. jcsd
  3. Jan 11, 2013 #2

    DrDu

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    If there is a phase shift ##\phi## between rays reflected on adjacent planes (distance d) then there will be a phase difference ##n\phi## between planes with distance nd.
    Hence if there are many planes you will observe constructive interference between all the reflected waves only if ##\phi< n/N## where N is the total number of planes in the crystal, i.e. D/d, where D is the diameter of the crystal (or better that of domains in the crystal). As D is many orders of magnitude larger than d, you get very sharp reflections.
     
  4. Jan 14, 2013 #3
    I might be extremely obtuse but I still don't get it. I drew a picture to visualize what you're saying:

    diffraction_zps24bdbbf9.png

    Unfortunately, I don't see the relationship between

    and

    From my picture, it becomes clear that if d##\phi## = λ/2, then every neighboring plane should destructively interfere. Likewise, if d##\phi## = λ/4, then every second plane should destructively interfere. What you seem to be saying is that if there are very many planes in the crystal, it is statistically likely that the phase shift from one of them would be ##\phi##, and from another would be ##\phi## + λ/2, or very close to it, resulting in destructive interference and consequent loss of amplitude. However, wouldn't this apply to the planes spaced to give the Bragg angle as well? Isn't it just as likely that there is some plane between them such that its diffracted photons are of phase ##\phi## + λ/2? Why is the diffraction from the Bragg planes "special", and not wiped out from the diffraction pattern?

    Sorry again if I'm being stupid; I just don't see it clearly yet.
     
  5. Jan 15, 2013 #4

    DrDu

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    First I wonder why you are asking this question in the biology forum and not in the general physics forum where you could calculate with much more answers.
    Now to your question.
    The fields of the reflected waves sum up to the total field whose square is proportional to the intensity.
    Now this sum is something like ##E=E_0\sum_{j=-N/2}^{N/2} \sin(2jd\sin(\theta)/\lambda)##. The latter sum can be calculated exactly. The point is that it will be nearly zero as long as ##2d\sin(\theta)/\lambda## is not a multiple of 2π because only for this value all ##\sin(2jd\sin(\theta)/\lambda)=1## and no two planes will interfere destructively.
     
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