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Why integer in Bragg's law

  1. Aug 11, 2012 #1
    Hi there,

    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
     
  2. jcsd
  3. Aug 11, 2012 #2

    Redbelly98

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    Welcome to Physics Forums.

    If the extra path 2dsinθ is not an integer multiple of the wavelength, then you don't get constructive interference, hence no wave for that angle θ.

    img_full_47305.gif
     
  4. Aug 11, 2012 #3
    Ah, yes! Sorry just also came up with that haha. Thanks!
     
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