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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.

enter image description here

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.

EDIT: a reply to dmckee comment: If an integral number of wavelenghts $n\lambda$ can fit into the distance $2l$ then the two rays contribution to the scattered wave front would be in phase and we would obtain maximum diffraction at angle $\phi$

enter image description here

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We usually assume that the screen, $L$, is much further away than the distance $d$ between the lattice planes. Then the lines which converge on a point are NEARLY parallel, although they are not quite parallel. Then we make the approximation that they ARE parallel. If you work through the math carefully, you'll see this approximation only changes the length of the lines at second order in $\frac{d}{L}$. Thus, since $d<<L$, this approximation is valid.

In simpler words: The paths aren't parallel, they converge to a point. However, in the limit that the screen is much further away than the distance between lattice planes, we can approximate the paths as parallel without changing their lengths much. The amount we're wrong is proportional to $(\frac{d}{L})^2$, which goes to zero quickly when $d$ is much less than $L$.

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    $\begingroup$ This really isn't complete without mentioning that thought we draw rays as lines, Huygens' principle is in place and each of the ray should be envisioned at outgoing wave-fronts with small but non-trivial angular breadth. It would be even better if you mentioned the size of the detector elements. $\endgroup$ – dmckee Jul 29 '15 at 3:27
  • $\begingroup$ That's not really relevant when computing the difference in path length. $\endgroup$ – Jahan Claes Jul 29 '15 at 20:03
  • $\begingroup$ No, but it is important is making it clear why "parallel" doesn't mean non-intersecting. There is nothing wrong with your answer, of course, but the poster was looking at the diagram we draw and thinking about Euclid and non quite seeing how they come together. Huygens' principle is the answer. $\endgroup$ – dmckee Jul 29 '15 at 20:48
  • $\begingroup$ But we DO want the lines to intersect. They're only APPROXIMATELY parallel. $\endgroup$ – Jahan Claes Jul 29 '15 at 23:26

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