# 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?