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Physics - Single Slit Diffraction

  1. Nov 28, 2011 #1
    Hi Guys, my textbook mentioned that how the equation for the minima for single slit diffraction was derived:

    Consider a slit of width W with 2 light rays, one emitting from the edge, one emitting from the center. Their path difference is W/2 sin T . If the path difference is 1/2 lambda, then they will experience destructive interference. Same can be said for light rays spaced apart by W/3, W/4, so on. Hence the general equation for the minima, W sin T = m (lambda).

    This is perfectly reasonable. However, I say:

    Consider 2 light rays emitting from the single slit at each edge. Path difference will be W sin T. If W sin T were an integral multiple of wavelength, there should be constructive interference. So W sin T = m lambda can be the equation for constructive interference too.

    Why is it that light rays spaced apart by the entire width of the slit aren't counted?
     
  2. jcsd
  3. Nov 28, 2011 #2

    atyy

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    A less cheating derivation would take the interference between all pairs of rays into account. Usually this is done using Huygen's construction, eg. eq 7.2 in http://phyweb.phys.soton.ac.uk/quantum/lectures/waves7.pdf [Broken] has an integral over the entire slit width.
     
    Last edited by a moderator: May 5, 2017
  4. Nov 28, 2011 #3

    jtbell

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    Staff: Mentor

    In other words, if the positions across the slit are x=0 through x=W, then the rays coming from x=0 and x=w/2 interfere destructively. So do the rays coming from x=d and x=w/2+d where d is a small increment. So do the rays coming form x=2d and x=w/2+2d. Etcetera. The rays coming from all positions across the slit can be put into pairs like this, each with the path difference (w/2)sin(theta). Each pair cancels destructively, so the net result is complete destructive interference.

    Because there is only one such pair of light rays. None of the other light rays emerging from the slit can be put into such a pair, that cancels destructively.
     
  5. Nov 28, 2011 #4
    Thanks for the advice.
     
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