How to Calculate Intensity and Probability for Double Slit and Diffraction?

grkm
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Homework Statement


upload_2015-1-5_22-16-56.png


Homework Equations


upload_2015-1-5_22-38-51.png

slit width = a , slit separation = 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//

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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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.
 
Can I say |I(θ)|^2~P ? and thanks.
 
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?
 
so interval can be taken from h to h+d ? I am not sure but at Imax 's probability is 1/2 I am trying to find on detector.
 
grkm said:
so interval can be taken from h to h+d ?
Well, you'll need to find the angles θ1 and θ2 that correspond to the distances h and h+d.
I am not sure but at Imax 's probability is 1/2 I am trying to find on detector.
Sorry, I don't understand this comment.
 

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