Timeline for How does the diffraction pattern of a 2-dimensional grating/lattice relate to the reciprocal lattice?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 9 at 11:02 | history | bounty ended | CommunityBot | ||
| Jun 8 at 10:07 | vote | accept | SalahTheGoat | ||
| Jun 3 at 6:48 | comment | added | user34722 | That's correct! | |
| Jun 3 at 6:42 | comment | added | SalahTheGoat | Okay I think I understand most of it now. To summarize: the diffraction pattern produced on a screen by shining monochromatic light onto a 2D lattice is in fact the Fourier transform of the projection of the 2d lattice onto the screen and not the transform of the actual lattice. When the incident light is perpendicular to the lattice, then the projection is simply equal to the lattice. But when the lattice is rotated, the projection compresses and hence by the Fourier Scale Theorem, the diffraction pattern expands? | |
| Jun 3 at 1:22 | comment | added | user34722 | Yes the generalization to 2D is as you say. My scaling transformation was wrong at first. $f(x) \rightarrow (f(x/\cos \theta)$ compresses the pattern. Then do a few change-of-variables until you get the $\cos \theta$ on the right-hand side with $y$. Or better yet, look up the relevant Fourier transform identities to be sure. Contraction in real-space expands in Fourier space, because contractions increases spatial frequencies. | |
| Jun 3 at 1:16 | history | edited | user34722 | CC BY-SA 4.0 |
scaling fix
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| Jun 2 at 10:26 | comment | added | SalahTheGoat | ... incident ray to the object. I can also see how the size of the object is effectively contracted by $x\cos(\theta)$. What do we do with this $x\cos(\theta)$ though? Do we use $f(x\cos(\theta))$ in the integral as well as substitute $d(x\cos(\theta))$ for $dx$ and sub $x\cos(\theta)$ for $x$ in the complex exponential as well? I can't see how this would result in an expansion of the pattern by $1/\cos(\theta)$ . | |
| Jun 2 at 10:24 | comment | added | SalahTheGoat | Thanks for the great response. Just two issues I need to clear up. The first is simple: For the case of a 2d square lattice, the f(x) in your integral would become $f(x,y)$ where $f(x,y)$ is the convolution of a 2d square dirac comb and a electron density function $\rho(x,y)$ for the unit cell right? So that $ \tilde{f}(y / \lambda d,x/\lambda d)$ would be the product of the reciprocal lattice and $\tilde{\rho}(x,y)$ ? Secondly, I can see how the change in distance from object to the screen, $x\sin(\theta)$, is cancelled out by the equal and opposite change in the distance from the ... | |
| Jun 2 at 6:52 | history | answered | user34722 | CC BY-SA 4.0 |