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Double slit and Diffraction

  1. Jan 5, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-1-5_22-16-56.png

    2. Relevant equations
    upload_2015-1-5_22-38-51.png
    slit width = a , slit seperation = b = d (at photo),
    tanQ=h/L m.λ=b.sinQ
    λ.b=sinQ
    B=π.d.sinQ/λ
    α=π.a.sinQ/λ
    Iq=Imx(cosB)^2 x (sinα/α)^2//
    3. The attempt at a solution
    My first move was to find intensity but I have no idea about the probability of finding one photon on the detector since It is not a point and It has own length.Should I think the detector as a point ? Location can be relevant with fringes but I'm not sure.
     

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  3. Jan 5, 2015 #2

    TSny

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    Finding the intensity is a good first move. Next, think about how the intensity function I(θ) is related to the probability of a photon ending up in some angular interval between θ1 and θ2 after passing through the slits.
     
  4. Jan 5, 2015 #3
    Can I say |I(θ)|^2~P ? and thanks.
     
  5. Jan 5, 2015 #4

    TSny

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    You wouldn't square the intensity. (The intensity, I(θ), already contains the square of the electric field.) When thinking of the probability of where a photon will go, you need to include some interval of angles. For example, if you consider an infinitesimal interval from θ to θ + dθ, then a photon will end up in that interval with a probability that's proportional to I(θ)dθ. For a finite interval from θ1 to θ2, how do you think you would calculate the probability that a photon ends up between θ1 to θ2?
     
  6. Jan 6, 2015 #5
    so interval can be taken from h to h+d ? im not sure but at Imax 's probability is 1/2 im trying to find on detector.
     
  7. Jan 6, 2015 #6

    TSny

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    Well, you'll need to find the angles θ1 and θ2 that correspond to the distances h and h+d.
    Sorry, I don't understand this comment.
     
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