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A Modified Bragg's Law

  1. Oct 2, 2016 #1
    Hi guys, the x-ray data booklet gives a modified Bragg's Law that seems to be a combination of Snell's and Braggs. I'll post a picture of what this looks like. I've tried combining the two equations and coming up with their answer but can't get a solid derivation. Any help or a point toward a derivation would be awesome. Thanks!

    http://imgur.com/a/CvUGz

    http://imgur.com/a/CvUGz
     
  2. jcsd
  3. Oct 2, 2016 #2
     
    Last edited by a moderator: May 8, 2017
  4. Oct 2, 2016 #3

    Charles Link

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    I get something similar, but a couple of corrections. I don't know if my calculations are correct, but I can show you what I got.
    One problem with the Bragg equation is the ## \theta ## is not measured from the normal to the surface. In the following derivation, I will use ## \theta ## as from the normal, and ## \theta ' ## as measured from the surface. The index "n" is assumed to be approximately 1 but is assumed to be ## n=1+\delta ##. (This is one of two places where I don't agree completely with what they stated.) ## \\ ## Beginning with ## 2nd cos(\theta_r)=m \lambda ## for constructive interference, and using Snell's law ## n sin(\theta_r)=sin(\theta_i) ##, then ## sin(\theta_r)=sin(\theta_i)/n ##. Also ## sin(\theta_i ')=cos(\theta_i) ## which will be used momentarily. We have ## cos(\theta_r)=(1-(sin(\theta_i)/n)^2)^{1/2} ##so that ## n cos(\theta_r)=(n^2-sin^2(\theta_i))^{1/2}=(n^2-1+1-sin^2(\theta_i))^{1/2}=(n^2-1+cos^2(\theta_i))^{1/2}=(n^2-1+sin^2(\theta_i '))^{1/2} ##Now expand with ## n^2-1=2 \delta ## (approximately)and ## sin(\theta_i ') ## being the larger term. This gives ## n cos(\theta_r)=sin(\theta_i ')(1+2 \delta/sin^2(\theta_i))^{1/2}=sin(\theta_i ')(1+\delta/sin^2(\theta_i ') ) ##. Now we have that ## 2d sin(\theta_i ')=m \lambda ## (Bragg's law without correction).So that ## sin^2(\theta_i ')=(m \lambda)^2/(4 d^2) ## . Putting it all together: ## \\ ## $$ 2d sin(\theta_i ')(1+4 d^2 \delta/(m \lambda)^2)=m \lambda $$. I will try to proofread my response carefully, but I think I have done it correctly. Note that I get a "+" sign for the correction part, not in concurrence with the attachment in the OP.
     
    Last edited: Oct 2, 2016
  5. Oct 2, 2016 #4
    Oh wow, thanks a ton! This looks great.

    So far I had this:

    http://imgur.com/a/xJsHR

    Lol not even close. This was my latest attempt at least. Tried a ton of different ways to make sense of this. I didn't consider the geometry enough.

    Also, from your first form of the Bragg eqn, what happened to the 'n' term? Clearly, it's not part of the answer although I don't see where it disappears.
     
    Last edited: Oct 2, 2016
  6. Oct 2, 2016 #5

    Charles Link

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    I edited it just a moment ago, (a minor change), but I might continue to update it if I see any additional typos, etc., so please look at my original post once more, etc.
     
  7. Oct 2, 2016 #6
    Cool, will do; Thanks again man!
     
  8. Oct 2, 2016 #7

    Charles Link

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    The "n " term is multiplying ## cos(\theta_r) ##. It multiplied the parenthesis of ## (1-sin^2(\theta_i)/n^2)^{1/2} ## to give ## (n^2-sin^2(\theta_i))^{1/2} ##.
     
  9. Oct 2, 2016 #8
    I see. Great! I'll try to talk to someone about the sign discrepancy... it's listed other places with the negative sign as well.
     
  10. Oct 2, 2016 #9
    I think the sign error comes from the original form of Braggs law used. I think if we start with $$m\lambda=2nd\sin(\theta_r)$$ then we can remedy the sign error.
     
  11. Oct 2, 2016 #10

    Charles Link

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    My equations assume a constructive interference between each of the atomic layers throughout the material. I think I did it correctly. If I got a wrong sign for some reason, it wouldn't be the first time. I'm assuming a positive correction ## \delta ## for the refractive index...
     
  12. Oct 2, 2016 #11
    Ahhh okay, fair enough.
     
  13. Oct 2, 2016 #12

    Charles Link

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    A google just now, I think, supplies the answer. The article stated, in talking about x-rays, that the index of refraction is just slightly less than 1. Thereby they are using a positive ## \delta ## in your textbook, but use the definition ## n=1-\delta ##. Looks like we are now in concurrence with the textbook result. :-) :-)
     
  14. Oct 2, 2016 #13
    Beautiful, so then we can say $$n^2-1\approx-2\delta$$ Do I have that right? Then we'll get the negative sign in the expansion. Thank you.
     
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