# Bragg condition for transmission: Why is the full diffracted angle Two times Theta? Or isn't it?

On a Bragg reflection with incomming angle Theta the total diffraction angle of the incomming wave is 2*Theta, of course.

But I have Bragg transmission with electrons on a graphite crystal (experiment about de-Broglie-wavelength and wave-particle-duality). The Bragg interference condition is still the same. But do the diffracted patterns behind the crystal appear under Theta or under 2*Theta? And why?

All I found was the pure statement "They appear under 2*Theta", but completly without explanation why this should be the case for transmission. What I think here: I can't apply Snell's law (incoming = outgoing angle) here, because I have transmission and not reflection. So I'd assume that the diffracted electrons will appear also under Theta (and not 2*Theta). Because they enter the crystal under Theta, do their interfering and exit in the same direction as they came in (no reflection, but transmission).

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Sorry, I would comment if I could. Wikipedia has a rather nice image which explicitly mentions $2 \theta$ - as compared to the reflection, i.e. $\theta$ as compared to the ‘lower’ surface of the crystal. This might be the source of your confusion?

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Thanks, but I already solved my problem. I have reflection in my case, too, contrary to what I assumed from the explanation I got with the experimental setup. And this explanation about the setup is rather incomplete. What confused me is this: The setup consists of a graphite crystall that again consists of crytsallites on which the Bragg reflection happens. And I wrongly assumed that it has one big homogenous graphite crystal that transmits the electrons. That's the case here, too, but the actual physics happens on the smaller crystallites which actually do reflection. –  Foo Bar Oct 10 '12 at 9:49