X-Ray diffraction - The Von Laue's equation

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shayaan_musta
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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.

Textbook figure of X-ray scattering geometry: incident beam n0 hits point O, scattered beam n1 observed at a distant point, with vector r from O to P and perpendicular drops from P onto the incident and scattered wavefronts.


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?

Textbook figure 2.5 showing a diamond-shaped diagram of incident and diffracted beams, the scattering normal N, and the scattering plane, with labelled angles theta and n_0, n_1.


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.
 
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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
 
shayaan_musta said:
Textbook page with a geometric diagram of incident and diffracted beams, the scattering normal N and the scattering plane, captioned Figure 2.5, above equations (2.2-6) and (2.2-7) for the Laue equations.

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

Thank you all.

Thank you for answer. I am sorry, I lost my connection therefore I have late in reply. So can you answer this one too?