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Help with single slit diffraction (fraunhoffer)

  1. Dec 2, 2007 #1
    urgent Help with single slit diffraction (fraunhoffer)

    why is it a minimum at a*sin ([tex]\theta[/tex]) = m*[tex]\lambda[/tex],

    a = slit separation

    I would have thought if the path difference is equal to a whole wavelength then there would be constructive interference??

    See http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2

    Doc Al

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    Staff: Mentor

    Don't confuse the double slit interference condition (http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html#c1").

    In the single slit condition, when the light from opposite ends of the slit have a full wavelength phase difference, that means that half the light from the slit is exactly one half wavelength out of phase with the other half. Which mean destructive interference.
    Last edited by a moderator: Apr 23, 2017
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