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So in bragg's law [itex]2d\sin \theta =n\lambda[/itex], n needs to be an integer. Can anyone explain why? I mean, what if the extra path 2d that the 'second beam' has is not dividable by a wavelength?

Not sure if this is asked before but could not find it!

Cheers,

Adnan

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# Why integer in Bragg's law

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