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Optics - Double Slit Irradiance of Fringes

  1. Apr 8, 2009 #1
    [Solved] Optics - Double Slit Irradiance of Fringes

    1. The problem statement, all variables and given/known data
    A double slit diffraction pattern is formed using mercury green light at 546.1nm. Each slit has a width of 0.100mm (= b), slit separation is 0.400mm (=a). The pattern reveals that the fourth-order interference maxima are missing from pattern.

    What is the irradiance of the first three orders of interference fringes, relative to the zeroth-order maximum?


    2. Relevant equations
    I = 4Io (sin([tex]\beta[/tex])/[tex]\beta[/tex]))^2 * (cos([tex]\alpha[/tex]))^2
    [tex]\beta[/tex] = 0.5kbsin([tex]\theta[/tex])
    [tex]\alpha[/tex] = 0.5kasin([tex]\theta[/tex])
    k = (2(pi)/[tex]\lambda[/tex])


    3. The attempt at a solution
    I've been trying to figure this one out for days and the book never gave any practice problems or simple explanation on what to do here.

    First I tried to solve for the first order interference fringe (m=1):

    sin([tex]\theta[/tex]) = (m[tex]\lambda[/tex]/a), then use this result to solve for alpha and beta in the two equations above and then finally solve for the sinc and cos^2 functions in the irradiance to find the ratio to the zeroth order fringe.

    The answers in the back of the book for the first three orders are (0.8106, 0.4053, 0.09006) but I am not getting these.

    The book I'm using is "Introduction to Opitics 3rd edition by Pedrotti"

    Thanks
    Best
     
    Last edited: Apr 8, 2009
  2. jcsd
  3. Apr 8, 2009 #2
    Nevermind, I solved it by using the RAD mode on my calculator rather than DEG.
     
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