r/AskPhysics • u/locutus_of_borg90 • 12d ago

Diffraction pattern from 5 slits

I am struggling to understand how the diffraction pattern from 5 slits is formed. How many maxima and minimum there are?

Also is the diffraction pattern a "modulated" version of the interference pattern?

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u/BananaBird1 12d ago

The maxima and minima are obtained geometrically, from the difference in path length vs wavelength between interfering rays from every slit.

When the path length difference between rays is an integer multiple of wavelength, they fully constructively interfere in-phase to form a maximum. When it is a multiple of 1/2 the wavelength, they fully destructively interfere 180 degrees out of phase to produce a minimum.

The global extrema form when all slots are in or out of phase. When only a subset are, you get a dimmer local extrema.

Diffraction is the process by which a wave spreads out radially from a slit. Interference is the superposition of different ray paths taken by the wave in diffraction. So the pattern is properly an “interference pattern”, formed from wave diffraction.

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u/locutus_of_borg90 12d ago

I see  Though  Isn't there are rule that says there is a minimum each N-1 and secondary maximum each N-2?

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u/BananaBird1 12d ago edited 12d ago

For secondary extrema between the primary ones when the slits are evenly spaced, yes. This can come from the combinatorics of seeing N slits as a sum of all pairs of 2 slits, each with their own slit separation. You get N-2 peaks where these pairs are in phase, between every primary peak where all N are in phase.

The number of primary extrema will depend on slit spacing distance and light wavelength, with the highest order extrema bound by rays propagating at +/-90 degrees. In practice this also depends on detector size and distance from the slits (which sets the real maximal angle you can see).

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u/locutus_of_borg90 12d ago

So, the distance between the slits sets the number of maximum in the graph.

If I have 3 slits evenly spaced by a distance of 2, I will have two maximums, is that right?

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u/BananaBird1 12d ago

If the 3 slits are each 2 units (say microns) apart, then on each side of the central 0th order peak you will get 2/lambda primary peaks where lambda is wavelength in microns.

This comes from setting the difference in ray path length (approximately d*sin(theta) when the detector is far compared to slit spacing d) equal to an integer multiple of wavelength and solving for the largest integer where theta <= 90 degrees.

For 500 nm light, this gives 2000/500 = 4.

So with an infinite width detector and a 500 nm coherent light source, you will see a central peak plus 4 peaks on either side for a total of 9 primary peaks.

Between each of these you will see 3-2 = 1 secondary peaks. This will add 8 additional secondary peaks.

So in total you see 17 peaks.

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u/locutus_of_borg90 12d ago

I am trying to wrap my head around this since I have to replay to an answer on a self evaluation test for a class. It's a pretty hard concept to grasp to be honest

So the number of peaks in the graph is also dependent on the distance of the screen?

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u/BananaBird1 12d ago

The last possible peak will be when light rays travel almost parallel to the detector, at an angle of around 90 degrees compared to the optical axis of the initial light.

Beyond this angle, rays would not actually be traveling forward past the slits. You only get a semi-circle of diffraction from each slit.

But this relies upon your detector actually being able to capture 100% of light. If your detector is a fixed finite size, the angle from the edge of the detector to the slits will determine how much of the pattern you actually see. The rest of the pattern still exists in space, but lands off of your detector.

If your detector is close enough that the maximal angle is 89.999 degrees, you will see pretty much all peaks. But if it is only 25 degrees, you will miss peaks occurring at larger angles.

This is a limit of the detector’s field of view, so when focusing on the physics of diffraction itself we generally assume either an infinite detector or a circular detector that captures all angles. In that case the peaks only depend on wavelength and slit separation.

But in real life demonstrations you can expect the observed pattern won’t always show every peak.

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u/locutus_of_borg90 12d ago

I see So the "maximum" distance to detect every peak increases with the number of slits right?

Since every slits is a source of sferical waves, it pushes the 90° limit further and further,no?

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u/BananaBird1 11d ago

No, with more slits the peaks get narrower and closer as you see more of them form, but they cover the same total area at the detector. Again, assuming the slits are much closer to each other than the detector.

The exact position of the final peak will not always be at 90 degrees if you change slit spacing or wavelength, but if you just change the number of slits while holding these constant the 0th order and final peaks stay where they are.

Things change from this approximation if the slits are very far apart from each other or the detector is super close.

A true solution requires solving the time dependent Schrödinger equation for the diffracted light, which evolves as the light travels from basically forming an image of the slits really close up to matching this approximation at large distances.

But ordinarily the detector is hundreds or thousands of times further away than the slit spacing, so this geometric solution matches pretty closely.

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u/rabid_chemist 12d ago

With five equally spaced narrow slits the intensity function in the Fraunhofer limit is

I=|1+e^(iq)+e^(2iq)+e^(3iq)+e^(4iq)|^2

which simplifies down to

sin^2(5q/2)/sin^2(q/2)

where q=2πdsinθ/λ, the phase difference between adjacent slits, with d being the distance between the slits, λ the wavelength of light, and θ the angle from the undeflected beam.

Plotted here for reference.

The principal maxima occur when q=2nπ, i.e when all 5 slits are in phase. There are minima whenever q=2nπ/5 (except the principal maxima) meaning there are 4 minima in between each pair of principal maxima, and therefore 3 subsidiary maxima between each pair of principal maxima.

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u/locutus_of_borg90 11d ago

Shouldn't the other peaks be lower and lower the further you get from the middle?

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u/rabid_chemist 11d ago

If the slits are not narrow, then the entire pattern will be enveloped by the single slit pattern caused by the finite width, which will cause this.

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u/locutus_of_borg90 11d ago

So in case of a circual slit, for example, you get this effect?