Picture of Bragg's diffraction sounds confusing

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SUMMARY

The discussion centers on the confusion surrounding Bragg's Law of Diffraction, specifically regarding the interference of wave beams C and C' on a detector screen. It is clarified that while these beams hit different points on the screen, they can still undergo constructive interference due to their phase alignment, provided the path difference equals an integer multiple of the wavelength. The explanation emphasizes that the incoming and outgoing rays are nearly parallel, and although they may intersect at the detector, calculations can assume parallelism within the crystal structure for simplicity.

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  • Understanding of Bragg's Law of Diffraction
  • Familiarity with wave interference principles
  • Basic knowledge of x-ray diffraction techniques
  • Concept of path difference in wave mechanics
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  • Study the derivation of Bragg's Law in detail
  • Explore the concept of constructive and destructive interference in wave physics
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  • Investigate the geometry of wave propagation in crystalline structures
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Students and researchers in physics, particularly those focused on crystallography, material science, and wave mechanics, will benefit from this discussion.

kasha
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This may be a little of a stupid question.

But I was looking at a diagram describing Bragg's Law of Diffraction.

2Ltz4.jpg


and I was like...how can an interference happen if wave beam C and wave beam C' are hitting different points on the detector screen?! they are not hitting the same point on the detector.

Per Wikipedia: The two separate waves will arrive at a point with the same phase, and hence undergo constructive interference, if and only if this path difference is equal to any integer value of the wavelength...but how would they arrive at the same point? I think the graph is very confusing.

Any help to clear things up would be very appreciated...
 
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The beams are much wider than the plane spacing. You actually have many many parallel incoming rays and many many parallel outgoing rays. Try shifting one of the rays in the diagram horizontally so that the outgoing rays coincide. With this new diagram, deriving the path difference to be 2d sin θ is more complicated, but it can be done.
 
I actually found the answer for that...
First: the lines are not perfectly parallel, they are almost parallel. over the very short distance (d in Angstrom) the paths of the reflected x-rays C and C' are pretty much parallel, but over the distance of the screen of the detector let's say it is L...the lines are not parallel. it is safe to assume the lines are parallel in our calculations inside the Crystal Latex...but they do intersect at some point on the screen.
 

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