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Intensity of light in Fraunhofer diffraction pattern

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    The intensity of light I in the fraunhofer diffraction patter of a single slit is

    I=I0(sin[itex]\gamma[/itex]/[itex]\gamma[/itex])2 where

    [itex]\gamma[/itex]=[itex]\pi[/itex]asin[itex]\theta[/itex]/[itex]\lambda[/itex]


    Show that the equation for the vaules of [itex]\gamma[/itex] at which I is maximum is tan[itex]\gamma[/itex]=[itex]\gamma[/itex]


    well, intensity is approxiamately max as follows

    Im=I0/[(m+1/2)2[itex]\pi[/itex]2]

    and for m=1... Im=0.0472I0
    m=2....I0=0.0165I0

    but i have no clue how to "show" that tan[itex]\gamma[/itex]=[itex]\gamma[/itex] at the max intensity??
     
  2. jcsd
  3. Sep 25, 2011 #2

    ehild

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    How do you find the maximum points of a function?

    ehild
     
  4. Sep 25, 2011 #3
    differentiate!
     
  5. Sep 25, 2011 #4
    but what?

    differential of I is I'=I0(-2sinx2/x3)

    differential of [itex]\gamma[/itex] is [itex]\gamma[/itex]'=[itex]\pi[/itex]acos[itex]\theta[/itex]/[itex]\lambda[/itex]

    for general functions the process is differentiate function and sub in zero's for critical points.

    so in this case would i use I' and then sub in [itex]\gamma[/itex]=0?

    but that comes up with zero?
     
  6. Sep 25, 2011 #5

    ehild

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    No. The problem asks the places of maximum in terms of gamma. Differentiate the intensity with respect to γ. Apply chain rule for the square of sin(γ)/γ, then you have a fraction, and you have to differentiate sin(γ), too. No need to differentiate gamma with respect to theta.

    ehild
     
  7. Sep 25, 2011 #6
    With u=sin(Y)/Y
    I=Io*u^2
    u'=-sin(Y)/Y^2
    I'= Io*2u
    So
    I' with respect to gamme is =Io*2sin(Y)/Y*-sin(Y)/Y^2

    =Io*-2[(sinY)^2]/Y^3
    ...
     
  8. Sep 25, 2011 #7

    ehild

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    sin(γ)/γ is a fraction. What is the derivative of f/g?

    ehild
     
  9. Sep 25, 2011 #8
    eh of course,
    sin(Y)/Y derivitive, let g=sinY then g'=cosY
    h=Y and h'=1
    y'=(YcosY-Y)/Y2
     
  10. Sep 25, 2011 #9
    then = (Y(cosY-1))/Y2

    so = (cosY-1)/Y
     
  11. Sep 25, 2011 #10
    then subbing back into I'

    I'=2I0*(sin(Y)/Y)*((cosY-1)/Y)

    but tan = sin/cos.

    i can see its getting closer...
     
  12. Sep 26, 2011 #11
    not sure what to do next, and the only way i can see to get cos on the bottom of the fraction is to change the two functions for quotient rule but then it wouldnt match the rule.. help?
     
  13. Sep 26, 2011 #12

    ehild

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    y'=(γcosγ -sinγ )/γ2.

    ehild
     
  14. Sep 26, 2011 #13
    Oh god. Way to feel dumb -.- thanks mate
     
  15. Sep 28, 2011 #14
    So I'= 2I0*(sinY/Y)*((YcosY-sinY)/Y2)

    thats with u=sinY/Y so then I'=2I0*u*du/dY which is from above.

    i tried subbing in tanY for Y but it doesnt work out to zero?

    to "show" that tanY=Y is a max, then I'=0...
     
  16. Sep 28, 2011 #15

    ehild

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    You do not need that "u". I'=0 either when sinγ/γ=0 or γcosγ-sinγ=0, that is, γcosγ=sinγ, divide both side with cosγ, ... Do you know how tanγ is related to sinγ and cosγ?


    ehild
     
  17. Sep 28, 2011 #16
    oh. of course. had absolutely no vision for this question. thanks for your help. very much.
     
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