# Fraunhofer's multiple slits versus atomic scatteres (diffraction theory)

• A
Hey, I am currently busy with studying solid state physics and looking at diffraction theory. Following link explains Frauenhofer diffraction pretty good: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/mulslid.html#c3

Let's assume a N=6 multiple slits. Its diffraction pattern depends on slit width w and slit distance d.

But when you have N=6 atomic scatteres (scattering centers) you only have the distance d between them. There is no slit width. But is there a atomic width or something similar?

So how differs the diffraction pattern of 6 slits and 6 atomic scatteres? (1D observation only for now)

EDIT: I meant Fraunhofer, sorry. Maybe a mod could edit the title please.

<mentor edited title>

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Homework Helper
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The Fraunhofer single slit diffraction is much simpler than the crystal scattering that you previously did. (Recall https://www.physicsforums.com/threads/diffraction-on-periodic-structures.952210/#post-6033221 ) . The crystal scattering involves reflections off of crystal planes with angle of incidence=angle of reflection, along with the atoms in adjacent planes needing to have the optical path difference be an integer number of wavelengths. ## \\ ## The interference principles are the same in multi-slit diffraction as for scattering by the atoms of a crystal, but the Fraunhofer case for multiple slits is much simpler than scattering by a crystal. Here is a recent homework that came up on this topic that you might find of interest: https://www.physicsforums.com/threa...tion-pattern-ratio-of-power-densities.956805/ This is quite a detailed examination of the intensity of the peaks as a function of angle, but if you can follow this, you understand the multi-slit case quite well........... In particular, the multi-slit case uses a very specialized formula (see post 2 of the "link" ) that is quite exact. It is a very useful formula, and is used very often in the multi-slit case. It is really important to be able to work this formula if you really want to have a good handle on the multi-slit case. This homework problem is a good exercise in working with that formula, and further on in the thread, I describe in detail how the formula is used in the calculation when the denominator goes to zero, which happens at the primary maxima of the diffraction pattern. Other than this slightly complex formula, the multi-slit case is quite straightforward. ## \\ ## And see also the "link" found in post 7 of this very detailed thread. In that "link" they give the correction factor that occurs for a slit of finite width. That is just the single slit diffraction factor. The formula discussed above is the interference factor resulting from multiple slits of very narrow width. If the slits are finite in width, the result is simply the inclusion of a single slit diffraction factor to get the complete result.

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• Gamdschiee
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One additional comment: In quite a number of the write-ups on multi-slit interference, they assume normal incidence and have ## \Phi=\frac{2 \pi \, d \sin(\theta)}{\lambda} ##. In the more general case where the incident angle differs from zero (normal incidence), the equation reads ## \Phi=\frac{2 \pi \, d (\sin(\theta_i)+\sin(\theta_r))}{\lambda} ##. ## \\## And an additional item: I believe this may have been discussed in the previous thread a couple months ago about crystal scattering: (see e.g. post 20 of that discussion) For the derivations with the diffraction grating, the phase factor ## \phi=(\vec{k}-\vec{k}')\cdot \vec{r} ## is replaced by ## \phi=\frac{2 \pi \, x (\sin(\theta_i)+\sin(\theta_r))}{\lambda} ## where ## x ## is the position on the plane of the slit, and the integral over ## \vec{r} ## is replaced by an integral over ## x ##. ## \\ ## The derivation with the multi-slit case is quite similar to the scattering by a crystal, but actually a little simpler, and with equally spaced slits, the result can be readily computed by summing the geometric series. The intensity ## I ## is then found by taking the square of the amplitude of the resultant ## E ##, and gives the formula as provided in post 2 of the second "link" above. With the integration over ## x ## , the ## \phi ## of significance is ## \Phi=\frac{2 \pi \, d (\sin(\theta_i)+\sin(\theta_r))}{\lambda} ##, and it is this last ## \Phi ## that appears in the intensity formula of post 2 of the second "link" above. ## \\ ## We will repeat that formula here: ## I=I_o \frac{\sin^2(\frac{N \Phi}{2})}{\sin^2(\frac{\Phi}{2})} ##.

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• Gamdschiee
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Thanks you!

But you can say in general that the diffraction pattern from N=6 scatteres looks like the one from N=6 slits?

Like this (example with 5 slits): http://www.physics.louisville.edu/cldavis/phys299/notes/lo_msdiffraction_fig2.jpg

So you can say N slits = N scatteres.
I don't think that works. If you have 6 point sources in a line , the pattern will in some ways be similar to that of 6 slits, but not identical. ## \\ ## Most often, whether it is a bunch of slits, or a diffraction grating, there is some illumination that extends in the ## y ## direction. If you are referring simply to atoms, I think you would get a completely different result. 6 scatterers might produce a similar pattern, but the signal would be lost in the noise. The closest analogy might be if 6 crystal planes were involved in the scattering. The Bragg condition still includes the condition that angle of incidence equals angle of reflection, so that scatters from the same plane constructively interfere with a ## m=0 ## type maximum. In general, the Bragg peaks involve constructive interference from perhaps 1,000, 000 or more scatterers, and perhaps 1,000,000,000 or more. ## \\ ## (Edit: And on further thought, these numbers are even very conservative). And in crystal scattering, the wavelength needs to be much shorter, because atomic distances are on the order of 2 Angstroms, thereby, the scattering is done using x-rays.

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• Gamdschiee

What about 2D then? E.g. we take this six atom scatteres and form it into a hexagon. For startes I read this here: https://www.doitpoms.ac.uk/tlplib/diffraction/diffraction3.php

So could it be that the diffraction pattern will just look like a hexagon without the lines between the "intensity points" ?