4 slit interference, unequal separation

[Monarch]

Hi all

I can't find any guidance on this on the internet, so hopefully someone here can help!

Basically, I want to work out the intensity profile produced by a multiple slit experiment, with 4 slits. No problem there, but I need the separation between slits to be unequal! I've found mention of it, but no actual help on how to construct it, shots anyone have any idea on how to go about working it out?

Cheers!

 

[Simon Bridge]

Work it out the same as for even separation... Sum the phases.

 

[jtbell]

Yep, go back to the beginning of the derivation for two slits (or N equally spaced slits) and generalize it so you're adding four sinusoidal terms with the required phase differences. You'll probably have to calculate the intensity for each point on the screen by brute force (computer program or spreadsheet), instead of trying to come up with a nicer formula, unless the slit spacings have some simple relationship.

 

[Monarch]

Thanks for the help!

I've tried following some derivations through, and I've struggled to make the adaptation to uneven slit separations. I've come up with something though, but through a slightly different method. It seems to give roughly the right pattern for equal spacings, and an unequal one that was easier to model, but I'm not convinced it's technically right.

The problem I seem to be having is that there are a lot of very wishy washy derivations on the internet that don't fully explain what's going on, if either of you (or anyone else) has a link to a good one then that might help me.

Cheers!

 

[Simon Bridge]

The correct derivation is just to make a vector sum of the phases: represent the light wave using phasors.
Treat the light as being in-phase at each of the slits - then the light from each slit will have different phases at the screen depending on the path length to get there. The intensity is proportional to the square of the magnitude of the final phase vector.

Consider 4 slits at x=0 positioned at $y_1$, $y_2$, and $y_3$: $y_3 > y_2 > y_1$ and $|y2-y1|\neq |y_3-y_2|$.
Light arrives at these slits with the same amplitude and frequency and phase.

We want to know the intensity of the light arriving at a screen a distance $x=D >> |y_3-y_1|$ away.
The distance from the nth slit to point $(D,y)$ on the screen is given by $r_n^2=D^2+(y-y_n)^2$ ... get the idea?

Usually $|y2-y1| = |y_3-y_2| = d$ so you just substitute this in the derivation.
You have to decide which approximations to get rid of. Probably best not to make many to start with.
You'll end up having to do a vector sum of the four phase vectors.

http://www.jick.net/~jess/hr/skept/Phasors/
http://web.mit.edu/8.02t/www/materials/StudyGuide/guide14.pdf

 

[sophiecentaur]

The correct derivation is just to make a vector sum of the phases: represent the light wave using phasors.
Treat the light as being in-phase at each of the slits - then the light from each slit will have different phases at the screen depending on the path length to get there. The intensity is proportional to the square of the magnitude of the final phase vector.

Consider 4 slits at x=0 positioned at $y_1$, $y_2$, and $y_3$: $y_3 > y_2 > y_1$ and $|y2-y1|\neq |y_3-y_2|$.
Light arrives at these slits with the same amplitude and frequency and phase.

We want to know the intensity of the light arriving at a screen a distance $x=D >> |y_3-y_1|$ away.
The distance from the nth slit to point $(D,y)$ on the screen is given by $r_n^2=D^2+(y-y_n)^2$ ... get the idea?

Usually $|y2-y1| = |y_3-y_2| = d$ so you just substitute this in the derivation.
You have to decide which approximations to get rid of. Probably best not to make many to start with.
You'll end up having to do a vector sum of the four phase vectors.

http://www.jick.net/~jess/hr/skept/Phasors/
http://web.mit.edu/8.02t/www/materials/StudyGuide/guide14.pdf

Why do you make that stipulation?
Edit - oh yes, sorry, he wants unequal spacing - but the special case of equal spacing is no different really and you can extend this to any number of slits, spaced any way you like.

 

[Simon Bridge]

;) Yeah - I was just translating the asked-for situation into maths.
Technically I don't need to number the slits in any particular order either, or require the slit-screen separation to be large compared to the extent of the slits. But if I do the diagram gets tidier. It is also implicit that the slit width is comparable to the wavelength and small compared to the slit spacing.

I think the key message to get across here is the bit about the phasors.

 

[sophiecentaur]

As long as the 'throw' is reasonably long, you can always modify the initial pattern to take into account the diffraction pattern of the individual (finite width) slits, assuming the widths are all the same. You only need to multiply the array factor by the element factor in any particular direction. (Variable separable situation)

 

[Monarch]

The correct derivation is just to make a vector sum of the phases: represent the light wave using phasors.
Treat the light as being in-phase at each of the slits - then the light from each slit will have different phases at the screen depending on the path length to get there. The intensity is proportional to the square of the magnitude of the final phase vector.

Consider 4 slits at x=0 positioned at $y_1$, $y_2$, and $y_3$: $y_3 > y_2 > y_1$ and $|y2-y1|\neq |y_3-y_2|$.
Light arrives at these slits with the same amplitude and frequency and phase.

We want to know the intensity of the light arriving at a screen a distance $x=D >> |y_3-y_1|$ away.
The distance from the nth slit to point $(D,y)$ on the screen is given by $r_n^2=D^2+(y-y_n)^2$ ... get the idea?

Usually $|y2-y1| = |y_3-y_2| = d$ so you just substitute this in the derivation.
You have to decide which approximations to get rid of. Probably best not to make many to start with.
You'll end up having to do a vector sum of the four phase vectors.

http://www.jick.net/~jess/hr/skept/Phasors/
http://web.mit.edu/8.02t/www/materials/StudyGuide/guide14.pdf

Cheers for this; I found another page just after posting that mentioned a similar method for deriving 2 slits and I think I've managed to extend it to 4. I'll run through what you've said and make sure it all gives the same results.

As you said, the phasors part of it was most important, which I hadn't seen anywhere else. It's much more intuitive than the other methods I'd seen around though, and made it a much easier process.

 

[Simon Bridge]

For some reason everyone thinks that phasors are hard ... there's a nice description of using phasors by Feynman:
http://vega.org.uk/video/subseries/8

 

[jtbell]

I agree that phasors are a good way to approach this. Back when I taught an optics course, both of the textbooks that I used, used the phasor method for multiple-slit interference, so I assumed it would be all over the Web.

Another approach would use the equations for adding two oscillations of the same frequency to produce a resultant with the same frequency but different amplitude and phase: $$A_1 \cos(\omega t + \phi_1) + A_2 \cos (\omega t + \phi_2) = A \cos (\omega t + \phi)$$ where $$A^2 = A_1^2 + A_2^2 + 2 A_1 A_2 \cos (\phi_2 - \phi_1) \\ \tan \phi = \frac {A_1 \sin \phi_1 + A_2 \sin \phi_2} {A_1 \cos \phi_1 + A_2 \cos \phi_2}$$ First add waves 1 and 2, then add wave 3 to the resultant, then add wave 4 to the new resultant.

 

[Monarch]

Hi again

So I worked on this for a bit and have put it to the side for a while because I was happy with it; now I've come back and I'm short of confidence!

Essentially what I've ended up with is the magnitude of the resultant pattern (phi) to be equal to...

(some constant) * sqrt( (sin(d1) + sin(d2) + sin(d3))^2 + (1 + cos(d1) + cos(d2) + cos(d3))^2 ) (sorry don't know how to use the equation thing)

where d1 = 2*pi*a1*sin(theta) / wavelength where a1 is the separation between the first slit and second

d2 = 2*pi*a2*sin(theta) / wavelength where a2 is the separation between the first slit and third, and the same for d3.

The constant doesn't really matter to me as it just pushes the scale up and down, but does that look right?! It just looks to have simplified itself down a little too far considering how much hassle it was giving me to start with.

Sorry if this is a bit on the simple side of things, it's been a long time since I've done anything like thisEDIT: The above gives the right patterns for some known scenarios (e.g. equally spaced slits)...hence why I'm tempted to say it works.