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Calculating visibility of interference fringe

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A photomultiplier (PMT) is arranged to detect the photons from a double-slit experiment. It is placed at a point P in the detection plane and makes an angle [tex]\theta[/tex] with the horizontal of one of the slits. Assume that the two slits have different widths and that the widths are much less than the wavelength of light, [tex]\lambda[/tex].

    The probability amplitude for a single photon of wavelength [tex]\lambda[/tex] to strike the PMT from one of the slits is [tex]\sqrt{2}[/tex] more than for the other slit. Calculate the visibility of the interference fringes:

    [tex]V=\frac{P_{max} - P_{min}}{P_{max}+P_{min}}[/tex]

    Where [tex] P_{max}[/tex] is the maximum probability and [tex]P_{min}[/tex] is the minimum probability that a photon is detected.

    2. Relevant equations

    [tex]A = A_1 + A_2[/tex] (amplitudes are summed)

    [tex]P = (Abs(A_1 + A_2))^2[/tex] (probability is sum of amplitudes squared)


    3. The attempt at a solution

    I let [tex]A_1[/tex] be the greater amplitude, so it is equal to [tex]A_2 + \sqrt{2}[/tex]. Then I use the equation for probability above to solve for [tex]A_2[/tex].

    First problem:

    I get different results if I solve it by hand or if I use Mathematica. Solving by hand I find that [tex]A_2 = A_1 = \frac{1-\sqrt{2}}{2}[/tex], but Mathematica tells me that [tex]A_2 = -1.20711 < 0[/tex] or [tex]A_2 = -0.207107 < 0[/tex]

    The answer Mathematica gives is in decimal format...but the fact that it comes up negative is really confusing me...I feel like the amplitude must be positive...?

    Second problem:

    If I take the answer I calculate by hand and add the amplitudes and square to find the probability I get:

    [tex]P = A_1 + A_2 = (Abs(\frac{1-\sqrt{2}}{2} + \frac{1-\sqrt{2}}{2}))^2 = 3 - 2\sqrt{2} = 0.171573 < 1[/tex]

    The probability should sum to one, but it is much less than one.

    What am I doing wrong?
     
  2. jcsd
  3. Feb 17, 2009 #2
    Should I post this somewhere else?
     
    Last edited: Feb 18, 2009
  4. Aug 13, 2010 #3
    This work is intended as a help to show, using very simple geometrical models, how http://www.visibilityacceleration.com [Broken] is built into the Young's interference patterns produced by incoherent sources.
     
    Last edited by a moderator: May 4, 2017
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