Recent content by Gamdschiee

Ok, I see your point. 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...

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.

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...

@SpinFlop I see, thank you for the accurate answer. That made me to understand it way more than before. But I still have a few questions below. 1. I know that the energy gap is just the energy difference from the energies of both standing waves ##\psi(+)## and ##\psi(-)## at that exact zone...

@Pablo L Martin Thanks for your answer! But I do not get the origin from such standing wave. Does it come from the waves with wave vector ##\vec k## and ##\vec k - \vec G## ? Or can you only say in a 1D crystal that a wave traveling to the right with wavevector k = \pi/a (zone edge in...

I know that Bragg reflection in solid states at the edge of e.g. the first Brillouin Zone causes standing waves at these edges, which creates a gap between the energy bands. In this picture below you can see the probability density of a symmetric (+) and anti-symmetric (-) standing wave. The...

Thank you, I think I get the principal, but what do you mean with the angle ##\phi##? And could you explain again please, when to use a + sign before the cos() and a - sign before it? I can't imagine that part how I should work out this.

So you basically can say that the cases (1) ##\theta_i \neq \theta_r## and (2) ##\theta_i = \theta_r## both appear upon diffraction, but the first one can be ignored due to low intensity in comparison with the second case? Why is there a single Bragg Peak for m=0, when you only consider a...

Thank you. What do you mean by that? That ONLY in a single plain with ##\theta_i \neq \theta_r## you can't have constructive interference at all? What about parallel planes then? What do you mean with ##\Delta## here btw? And also what do you mean that there is no sufficient scattering in an...

But what is when the incident angle is not equal the reflection angle, when we see the crystal as parallel planes? How does that change any diffraction pattern? Because the light is scatterecd in all directions ##\vec k'## we said, which means the angle of the reflected light isn't always the same.

I see so the x-ray diffraction pattern must be the same for 3D crystals: Let's consider the peak (110) at the bcc crystal above. (110) also called miller indices and describes a simple plane in the crystal. (110) has many planes parallel to each other. So that parallel planes are like a...

Ahh, thank you I understand the Fraunhofer example now! 1. So the so-called Bragg Equation comes from ##\theta_i = \theta_r = \theta##, which results in ##m\lambda = d(sin \theta + sin \theta)=2dsin\theta##. But I also want to understand the picture I posted. So ##cos \varphi## is negative...

Thank you for your help, I thought about it and I understand it a bit better now! I also found another German book "Festkörperphysik" by Gross and Marx. There I found a picture for which explains the so-called Laue Equations: You described this earlier as ##m\lambda = d(sin(\theta_i)\pm...

Thanks for the accurate answer. So when I have an observation point ##\vec x## in the far-field, I always write it like that, when I want to observe in the far-field and I got a scatterer(or origin that "produces" a wave) at x=b? Which we can write in 3D as you showed: ##D=Ae^{i\vec k (\vec x...

Thank you. I guess ##a## is a typo, right? I understand that when a wave is coming form our Origin it will have some certain amplitude at your scatterer ##\vec x '##. I know acoustic waves radially decrease their amplitude from their origin, but why is this in our case only at the scatterer and...