What do these submaxima represent - Interference Pattern

[APUGYael]

Hey all,

Here's a neat picture:

The green arrow points towards one of the submaxima that I am curious about. It's the submaximum of the curve labeled 1.

1. represents the interference pattern
3. represents the diffraction pattern
2. represents the irradiance pattern

I guess that those maxima are created by waves not being 100% constructive or destructive (but rather somewhere in between, e.g. slightly destructive). Is that the right explanation?

While I am on this subject, what is the difference between all the different curves. Is curve 2 the result of combining curve 1 and 3?

EDIT: Picture not working?
Here's a direct link. Remove the space next to the . of .com
https://imgur. com/a/qMfzW8E#btBKyBH

EDIT: Solved.
https://phys.libretexts.org/Bookshe..._Interference/3.3:_Multiple-Slit_Interference

 

[BvU]

Not what I call a neat picture:

 

[APUGYael]

Not what I call a neat picture:

View attachment 243508

Weird.
Here's a direct link. Remove the space next to the . of .com
https://imgur. com/a/qMfzW8E#btBKyBH

 

[BvU]

Same result. And I refuse to accept.

 

[APUGYael]

Same result. And I refuse to accept.

Oh. How can I bypass imgur? Like you did with your picture.

EDIT: Nevermind. I think I did just that now.

 

[BvU]

A simple copy/paste perhaps ? If necessary via a snipping tool.

Anyway, hyperphysics has (most of) your answers ...

 

[APUGYael]

A simple copy/paste perhaps ? If necessary via a snipping tool.

Anyway, hyperphysics has (most of) your answers ...

Should be fixed now. Thanks for the link. It does indeed answer some questions. The primary reason for making this post is still unanswered though. :-) How come secondary maxima exist in interference patterns with multiple slits?

 

[BvU]

Unanswered in what sense ?

Under the Fraunhofer conditions, the light curve (intensity vs position) is obtained by multiplying the multiple slit interference expression times the single slit diffraction expression. The multiple slit arrangement is presumed to be constructed from a number of identical slits, each of which provides light distributed according to the single slit diffraction expression. The multiple slit interference typically involves smaller spatial dimensions, and therefore produces light and dark bands superimposed upon the single slit diffraction pattern.

 

[APUGYael]

Unanswered in what sense ?

My question is:
For the interference pattern there is clear primary maxima (100% constructive interference) and minima (100% destructive interference). There is also secondary maxima. How come these exist? Is it because of partly (so not 100%) constructive/destructive interference?

I don't seem to find an explanation anywhere (other than: look at this formula). Did I read over the answer on hyperphysics?

 

[Ibix]

The paragraph BvU quoted answers you. They are related to diffraction from the individual slits. If you can vary the slit width while keeping the centre-to-centre distance of the slits constant then you will see them change.

 

[APUGYael]

The paragraph BvU quoted answers you. They are related to diffraction from the individual slits. If you can vary the slit width while keeping the centre-to-centre distance of the slits constant then you will see them change.

I don't think the paragraph does answer my question to be honest. Although I could be wrong because the last sentence I can't comprehend and that might just be the sentence you're referring to.

"The multiple slit interference typically involves smaller spatial dimensions, and therefore produces light and dark bands superimposed upon the single slit diffraction pattern."

 

[BvU]

I think I see your question. The five slit diffraction pattern in hyperphysics also shows three minimaxima between two maximaxima. It's a bit corny to say: well, that's the formula (for the Fourier transform) doing it for you. Intuitively I'd say that's subsets of the five slits showing constructive interference, but my intuition could easily be proven wrong .

 

[APUGYael]

I think I see your question. The five slit diffraction pattern in hyperphysics also shows three minimaxima between two maximaxima. It's a bit corny to say: well, that's the formula (for the Fourier transform) doing it for you. Intuitively I'd say that's subsets of the five slits showing constructive interference, but my intuition could easily be proven wrong .

Yep, you're on the right track with what I am looking for :-).

I know those minimaxima, as you call them, exist as predicted by a bunch of formulas.
The amount of minimaxima is directly related to the amount of slits, yada yada. I am wondering, intuitively, what they represent though.

 

[Ibix]

The exact shape of the diffraction pattern depends on the width of the slits, the slit spacing, and the illumination profile.

Modelling the diffraction grating as an infinite array of infinitely narrow slits gives you an infinite array of delta functions as the illumination pattern. Replacing the infinitely narrow slits with identical slits of finite width gives you the convolution of the single-slit pattern and the comb of delta functions, which is just a sum of $(\sin x)/x$ functions centred at each delta function.

Depending on the ratio between the width of the slits and the centre-to-centre distance, the scale of the $(\sin x)/x$ oscillations may be larger or smaller than the scale of the comb. The pattern you've shown would be consistent with the grating having fairly wide slits with narrower dark bands, which hyperphysics is calling "not typical".

That's off the top of my head. You can easily generate a sum of $(\sin x)/x$ functions (note that $(\sin 0)/0)=1$) and vary the scale compared to the spacing and see if you can reproduce the minor maxima.

 

[APUGYael]

The exact shape of the diffraction pattern depends on the width of the slits, the slit spacing, and the illumination profile.

Modelling the diffraction grating as an infinite array of infinitely narrow slits gives you an infinite array of delta functions as the illumination pattern. Replacing the infinitely narrow slits with identical slits of finite width gives you the convolution of the single-slit pattern and the comb of delta functions, which is just a sum of $(\sin x)/x$ functions centred at each delta function.

Depending on the ratio between the width of the slits and the centre-to-centre distance, the scale of the $(\sin x)/x$ oscillations may be larger or smaller than the scale of the comb. The pattern you've shown would be consistent with the grating having fairly wide slits with narrower dark bands, which hyperphysics is calling "not typical".

That's off the top of my head. You can easily generate a sum of $(\sin x)/x$ functions (note that $(\sin 0)/0)=1$) and vary the scale compared to the spacing and see if you can reproduce the minor maxima.

Thanks for your time but I think you don't quite understand what I am asking.
My question has nothing to do with the diffraction pattern. It's completely about the interference pattern labeled as 1. Again: appreciate the time you took for writing this though. But it's not very helpful in the sense that it doesn't answer my question.

 

[APUGYael]

I think I see your question. The five slit diffraction pattern in hyperphysics also shows three minimaxima between two maximaxima. It's a bit corny to say: well, that's the formula (for the Fourier transform) doing it for you. Intuitively I'd say that's subsets of the five slits showing constructive interference, but my intuition could easily be proven wrong .

I think I found the answer as to why those secondary maxima exist.
"When this condition is met, 2d sin θ is automatically a multiple of λ, so all three rays combine constructively, and the bright fringes that occur here are called principal maxima. But what happens when the path length difference between adjacent slits is only λ/2? We can think of the first and second rays as interfering destructively, but the third ray remains unaltered. Instead of obtaining a dark fringe, or a minimum, as we did for the double slit, we see a secondary maximum with intensity lower than the principal maxima. "

 

[BvU]

Which is what I mean with subsets showing constructive interference -- but I grant you the wording in #16 is better.

 

[BvU]

yada yada one can even see that the middle minimaximum in the five slit pattern is weaker than its neighbours: It's from a single subset (1,3,5) of three slits with double the spacing. The two neighbours are from four subsets (1,4 ; 2,5 of a maxmax and idem of its neigbouring maxmax) of two slits with three times spacing .

[edit] i don't even believe myself here - would have to play around with Fourier transforms to say something sensible

If you think you get it, try to answer this: what about the (1,4 ; 2,5) with four times the spacing ?

 

[vanhees71]

I'm no sure whether I understand what the problem is. In the picture in #1 everything is nicely drawn: Curve 3 is the interference pattern of a single slit. It's amplitude is a sinc function. The central peak of the single-slit curve gets wider and wider the smaller the single slit becomes. In the limit of 0 single-slit width it converges to a constant.

Then there's the interference pattern indicated by curve 1, which is the interference pattern of some number of zero-slingle-slit-width openings. The true pattern with finite single-slit width is given by the product.

Mathematically in Fraunhofer observation the amplitude of the interference pattern is given by the Fourier transform of the geometry of the openings, and the interference pattern is the modulus squared.

 

[BvU]

Hi @vanhees71 ,
we try to find a pedestrian explanation of the smaller peaks between the spikes. positioning and relative height. Just curiosity.

 

[jtbell]

There is an elegant graphical construction for multiple slit interference, that uses addition of phasors. Try a Google search for something like “multiple slit phasor diagram”.

I’ll look for something when I get home, if nobody has found anything first.

[later...]

It turns out that the search phrase above gives some interesting hits among the Google image-search results. Going to the documents from which they come, here are a few links:

https://www.kshitij-iitjee.com/phasor-diagrams-of-waves/
http://www.jick.net/skept/Phasors/(note in particular the diagram at the very bottom of this page!)

http://www.unistudyguides.com/wiki/Interference_and_Diffraction#Phasors
A couple of slide-presentations look like they might be pretty good but you have to register or subscribe in order to see them.

 

[APUGYael]

Hi @vanhees71 ,
we try to find a pedestrian explanation of the smaller peaks between the spikes. positioning and relative height. Just curiosity.

Hardly pedestrian.
Feynmann once said that if you can't explain something simply then you don't understand it.

 

[BvU]

Thereby logic dictates I haven't ... see the edit in #18

I decided to postpone playing around with the FT until after retirement when I find a nice tool in a fancy programming language and need a decent topic ​

 

[ZapperZ]

Hardly pedestrian.
Feynmann once said that if you can't explain something simply then you don't understand it.

https://www.physicsforums.com/threads/if-you-cant-explain-it-to-your-grandmother.765734/
Zz.

 

[Charles Link]

Are you familiar with the formula $I(\theta)=I_o \frac{\sin^2(\frac{N \phi}{2})}{\sin^2(\frac{\phi}{2}) }$, where $\phi=\frac{2 \pi d \sin{\theta}}{\lambda}$? $\\$ This formula gives the entire graph of the function that you have in your first graph with $N=5$. The primary maxima are found in the limit when the denominator goes to zero, when the numerator is also zero. (You will find $m \lambda=d \sin{\theta}$ for these primary maxima, where $m=$ integer. In this limit, $I(\theta_{max})=N^2 I_o$). Meanwhile the numerator goes to zero $N-1$ places between primary maxima where the denominator is not zero. The secondary maxima necessarily occur between these zeros of the intensity function, because $I(\theta)$ is always greater than or equal to zero. Otherwise, from a mathematical standpoint, I don't think there is anything special about these secondary maxima.

 

[Charles Link]

And the above formula for $I(\theta)$ is very useful in diffraction grating spectroscopy, where the diffraction grating (typically 2" wide) may have an $N=30,000$ and more "lines" or grooves. The primary maxima are where light of a given wavelength winds up from the grating,(different $\lambda$ has a different $\theta_{max}$), and yes, you do get several (rainbow) spectra produced with different $m's$ from a diffraction grating spectrometer. The formula for $I(\theta)$ even allows one to calculate the theoretical line width ( $\Delta \theta_{max} \rightarrow \Delta \lambda$= the width of the primary maximum=a single wavelength produces a spectrum that is a "line") from a diffraction grating spectrometer, which basically tells the theoretical resolution limit $\Delta \lambda$ of a spectrometer with a grating with $N$ grooves or "lines". Two separate sources of wavelengths $\lambda_1$ and $\lambda_2$ that are separated by less than $\Delta \lambda$ will be unresolvable from that of a single wavelength. $\\$ Note: We used the word "line" here for two different meanings: The (diffraction grating) "line"=($N$ "lines" or grooves) here has a completely different meaning than the "linewidth" ($\Delta \lambda$) of the spectral "line" of the observed spectrum. $\\$ And for $N=30,000$, the secondary maxima have very little intensity. The primary maxima get almost all of the energy. I think the first secondary maxima on either side of each of the primary maxima might get about 5% (or less) of the energy that the primary maximum has, but the other ones are basically a wash.

 

[APUGYael]

https://www.physicsforums.com/threads/if-you-cant-explain-it-to-your-grandmother.765734/
Zz.

I enjoyed reading that. Thanks.
I do however feel like most things should be explainable in simple words.
Simply saying: "x happens because that's what the formula predicts" doesn't suffice most of the time.

Again though: thanks for the link. Interesting point of view.

 

[ZapperZ]

I do however feel like most things should be explainable in simple words.
Simply saying: "x happens because that's what the formula predicts" doesn't suffice most of the time.

That has nothing to do with making an explanation understandable. I can explain many things without invoking formulas, and yet, they can still be understandable for many people.

You need to stop perpetuating the fallacy that just because someone is unable to explain something in understandable form to another person, it automatically means that he/she doesn't understand the subject. It is why I wrote that article.

Zz.

 

[vanhees71]

Hi @vanhees71 ,
we try to find a pedestrian explanation of the smaller peaks between the spikes. positioning and relative height. Just curiosity.

The most simple way is just to calculate the Fourier transform of the slits. This you can even do in high-school (at least we did at the time).

The most simple example is the single slit. The opening is described as $f(x)=Theta(-b/2<x<b/2)$. The Fourier transform is
$$A(x) =C \int_{-b/2}^{b/2} \mathrm{d} x' \exp(\mathrm{i} k x x'/L)= C' \text{sinc}(b k x/2L),$$
where $b$ is the width of the slit, $k=2 \pi/\lambda$ the wave number of the light, $L$ the distance between the slit and the screen, and $x$ the position on the screen. The Intensity is gus given by
$$I(x)=|A(x)|^2=I_0 \text{sinc}^2(b k x/2L).$$
When you now have $N$ such slits with the centers in equal distances of $a$ from each other, you have to sum a (finite) geometric series, leading to
$$I(x)=\frac{I_0}{N^2} \frac{\sin^2(N a k x/2L)}{\sin^2(ak x /2L)} \text{sinc}^2(b k x/2L).$$
The phasor method is just a geometrical method for summing up the exponential functions. I never found this geometrical method more simple than the algebraic method, but that's because I'm more algebraically inclined the geometrically.

If you just plot the function, you find the pattern with the minima and maxima as in #1, as shown for various $N$ in my (German) E&M manuscript,

https://th.physik.uni-frankfurt.de/~hees/publ/theo2-l3.pdf
on page 161.

A good description of Fraunhofer diffraction is also given in the Wikipedia,

https://en.wikipedia.org/wiki/Fraunhofer_diffraction_equation
To quote another genius of physics (Einstein): One should explain things as simple as possible, but not simpler ;-)).

 

[Charles Link]

@vanhees71 has very much what I have in my formula in post 25. His also includes the term for the finite slit width, while mine assumes narrow slits. One correction I have for his post is that $I(x)=I_o ...$ without the $N^2$ in the denominator. ($I_o$ is the intensity that would be caused by a single slit).

 

[vanhees71]

In my notation $I_0$ is the intensity at $x=0$. Just expand around $x=0$, which gives
$$I(x) = I_0 + \mathcal{O}(x^2).$$
That's why I've put the $1/N^2$ factor in my formula.

See also the figures on p. 161 of my manuscript, cited already above.