Intensity of light in Fraunhofer diffraction pattern

[ProPatto16]

Homework Statement

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

I=I₀(sin$\gamma$/$\gamma$)² where

$\gamma$=$\pi$asin$\theta$/$\lambda$


Show that the equation for the vaules of $\gamma$ at which I is maximum is tan$\gamma$=$\gamma$


well, intensity is approxiamately max as follows

I_m=I₀/[(m+1/2)²$\pi$²]

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

but i have no clue how to "show" that tan$\gamma$=$\gamma$ at the max intensity??

 

[ehild]

How do you find the maximum points of a function?

ehild

 

[ProPatto16]

differentiate!

 

[ProPatto16]

but what?

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

differential of $\gamma$ is $\gamma$'=$\pi$acos$\theta$/$\lambda$

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 $\gamma$=0?

but that comes up with zero?

 

[ehild]

but what?

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

differential of $\gamma$ is $\gamma$'=$\pi$acos$\theta$/$\lambda$

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

 

[ProPatto16]

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
...

 

[ehild]

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

ehild

 

[ProPatto16]

eh of course,
sin(Y)/Y derivative, let g=sinY then g'=cosY
h=Y and h'=1
y'=(YcosY-Y)/Y²

 

[ProPatto16]

then = (Y(cosY-1))/Y²

so = (cosY-1)/Y

 

[ProPatto16]

then subbing back into I'

I'=2I₀*(sin(Y)/Y)*((cosY-1)/Y)

but tan = sin/cos.

i can see its getting closer...

 

[ProPatto16]

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?

 

[ehild]

eh of course,
sin(Y)/Y derivative, let g=sinY then g'=cosY
h=Y and h'=1
y'=(YcosY-Y)/Y²

y'=(γcosγ -sinγ )/γ².

ehild

 

[ProPatto16]

Oh god. Way to feel dumb -.- thanks mate

 

[ProPatto16]

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...

 

[ehild]

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

 

[ProPatto16]

oh. of course. had absolutely no vision for this question. thanks for your help. very much.