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Intensity in a Double-Slit problem?

  1. Dec 10, 2009 #1
    I've been looking over an old exam for one of my classes, and cannot decipher the train of thought used in solving (or method used) the problem on the last page:


    Specifically, for the first part, what is going on...? Firstly, I have the equation that I = I0cos^2(phi). So if I want the intensity doubled then I end up with cos^2(phi) = 2, which isn't possible? Clearly I am thinking about this in the entirely wrong way and could use some help, thanks in advance.
  2. jcsd
  3. Dec 10, 2009 #2
    I think it means just where there is positive interference. Which would be approximately y = m*lamba*L/d. m=1,2,3...
  4. Dec 10, 2009 #3


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    The intensities add up, so if you get intensity I_0 from one beam and the same from another beam you get 2*I_0 at the point where there is maximum constructive interference.
  5. Dec 10, 2009 #4
    Nevermind, I figured it out.
  6. Dec 10, 2009 #5
    That was my first thought, but turns out it was in correct. The answer happens to be what you posted above for m=1 but all divided by 4 since the intensity equals the sum of the individual intensities and the point one fourth of the way from the center point of the interference pattern to the center point of the first bright spot - the formula you gave. if you plot the intensity function, this becomes quite clear. but thank you for your reply.
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