Intensity of light in Fraunhofer diffraction pattern

  • #1
ProPatto16
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0

Homework Statement



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??
 
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  • #2
How do you find the maximum points of a function?

ehild
 
  • #3
differentiate!
 
  • #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?
 
  • #5
ProPatto16 said:
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]

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
 
  • #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
...
 
  • #7
sin(γ)/γ is a fraction. What is the derivative of f/g?

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

so = (cosY-1)/Y
 
  • #10
then subbing back into I'

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

but tan = sin/cos.

i can see its getting closer...
 
  • #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 wouldn't match the rule.. help?
 
  • #12
ProPatto16 said:
eh of course,
sin(Y)/Y derivitive, let g=sinY then g'=cosY
h=Y and h'=1
y'=(YcosY-Y)/Y2


y'=(γcosγ -sinγ )/γ2.

ehild
 
  • #13
Oh god. Way to feel dumb -.- thanks mate
 
  • #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 doesn't work out to zero?

to "show" that tanY=Y is a max, then I'=0...
 
  • #15
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
 
  • #16
oh. of course. had absolutely no vision for this question. thanks for your help. very much.
 
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