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X-Ray diffraction - The Von Laue's equation

  1. Aug 15, 2012 #1
    Hello experts!

    I have some questions related to the X-Ray Diffraction using Laue's treatment.

    I have attached some images. I have marked in red what I want to ask.

    Kindly tell me where did these equations come? Is there any low level physics included too, if so then kindly guide me through.

    1.jpg

    How do you write PA as r.n[itex]_{o}[/itex]?
    Where does it come, ∅[itex]_{r}[/itex]=[itex]\frac{2π}{λ}[/itex](r.N). I know this is path difference. But where does the (r.N) come?

    2.jpg

    How path difference become equal to 2πh[itex]^{'}[/itex]=2πnh
    and,
    How aNcosα=2asinθcosα ?
    and similarly,
    How 2asinθcosα=h[itex]^{'}[/itex]λ=nhλ

    Thank you all.
     
  2. jcsd
  3. Aug 17, 2012 #2
    Hi, with only high-school knowledge you can answer your questions. PA=OP.cos(PA,OP)=OP.PA/PA.cos(PA,OP)=OP.n0
    [itex]\phi[/itex]r is phase difference, not path difference
    [itex]\phi[/itex]r=wavevector×pathdifference
    wavevector=2π/λ
     
  4. Aug 22, 2012 #3
    Thank you for answer. I am sorry, I lost my connection therefore I have late in reply. So can you answer this one too?
     
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