I Amplitude of the maximums in single slit diffraction

AI Thread Summary
The discussion focuses on calculating the amplitude of secondary maxima in single slit diffraction, highlighting discrepancies between two methods. One method uses a formula leading to maxima at x=±4.493409 and x=±7.72525, while another method based on phasors suggests slightly different values. Participants note that both approaches yield results that are approximations, with the exact intensity distribution involving the sinc function. The inaccuracies arise from the assumption that phasors align perfectly along a circular arc, which is not entirely accurate. The conversation emphasizes the need for careful interpretation of the equations and methods used in diffraction calculations.
jaumzaum
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Hello!

I was trying to calculate the amplitude of the secondary maximums in the single slit diffraction.
When I use the formula:
$$ I = I_0 (\frac {sin(\Delta \phi /2)} {\Delta \phi /2})^2 $$
If I take ## x = \Delta \phi /2 ## and derivate I get that the maximum occurs when:
$$ x = tan(x) $$
The first 2 solutions are x=± 4.493409 and x=± 7.72525
which gives ##I_0/I##= 21.19 and 60.68 respectively.

However, this site gives a more direct way of calculating the maximums.
They say the first secondary maximum occurs when the phasors make 1 and a half loop, and the second secondary maximum when the phasors make 2,5 loop. However, that gives a slight different answer, 22.21 and 61.68 respectively
View attachment 323616
sinint8.png

Why are the values different? Which one is right?

Thanks!
 

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jaumzaum said:
Why are the values different? Which one is right?
Not my area but I believe neither answer is correct! (Though they’re accurate enough for most purposes.)

I think the equations you are using are (pretty good) approximations. But there will be some small inaccuracies.

The ‘exact’ intensity distribution (if Wikipedia is to be trusted) is:$$I(\theta) = I_0 \left[ sinc \left( \frac {\pi a}{\lambda} \sin \theta \right) \right]^2$$(The ‘##sinc##’ function is defined as ##sinc(x) = \frac {\sin(x)}x##.)

You’d have to differentiate that to find the angles of the maxima. (Note the plural of ‘maximum’ is ‘maxima’!)

https://en.wikipedia.org/wiki/Diffraction_from_slits#Single_slit

Maybe someone with a more in-depth knowledge will be able to provide more detail.
 
Steve4Physics said:
Not my area but I believe neither answer is correct! (Though they’re accurate enough for most purposes.)

I think the equations you are using are (pretty good) approximations. But there will be some small inaccuracies.

The ‘exact’ intensity distribution (if Wikipedia is to be trusted) is:$$I(\theta) = I_0 \left[ sinc \left( \frac {\pi a}{\lambda} \sin \theta \right) \right]^2$$(The ‘##sinc##’ function is defined as ##sinc(x) = \frac {\sin(x)}x##.)

You’d have to differentiate that to find the angles of the maxima. (Note the plural of ‘maximum’ is ‘maxima’!)

https://en.wikipedia.org/wiki/Diffraction_from_slits#Single_slit

Maybe someone with a more in-depth knowledge will be able to provide more detail.

Thanks @Steve4Physics

That is the exact same equation I am using.

Where ##\Delta \phi = \frac {2\pi a sin\theta}{\lambda} ##
 
jaumzaum said:
That is the exact same equation I am using.

Where ##\Delta \phi = \frac {2\pi a sin\theta}{\lambda} ##
Aha! I misinterpreted the meaning of ##\Delta \phi##. (It wasn't defined in Post #1.)

So intensity as a function of ##\theta##, expressed without the '##sinc##' and ##\Delta \phi## is:$$I(\theta) = I_0 \left[
\frac
{\sin \left( \frac {\pi a}{\lambda} \sin \theta \right)}
{\left( \frac {\pi a}{\lambda} \sin \theta \right)}
\right]^2$$To find a maximum intensity we need the value of ##\theta## which makes ##\frac {dI}{d\theta} = 0## and ##\frac {d^2I}{d\theta^2} < 0##.

Assuming your solution for this is correct, we only need to explain why the 'phasor' method is slightly inaccurate.

It appears that when doing the phasor-addition, it is only an approximation (but a good one) to assume the phasors line-up along a perfectly cicular arc. But it turns out the arc is not perectly circular.

A quick search produced this (concerning the 1st two maxima). See link below for context:

“These two maxima actually correspond to values of ϕ slightly less than 3π rad and 5π rad. Since the total length of the arc of the phasor diagram is always NΔE0, the radius of the arc decreases as ϕ increases. As a result, E1 and E2 turn out to be slightly larger for arcs that have not quite curled through 3π rad and 5π rad, respectively. “​
About halfway down https://phys.libretexts.org/Bookshe...on/4.03:_Intensity_in_Single-Slit_Diffraction
 
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