In the Bragg's diffraction the diffraction angle theta is equal to the incident angle because for simplicity it is convenient to derive the Bragg's law considering a straight line (representing parallel diffracted rays) which passes from a point P at infinity and which forms a theta angle (=incident angle) with the surface of the crystal?

EDIT: considering the following figure:

enter image description here

Bragg's condition (constructive interference) says that the difference of path must be a multiple of the wavelength in order to get constructive diffraction. In this case:

$$d \space \left(\sin(theta)+sin(alpha)\right)=m \lambda$$

with m integer. Is it possible to show that the constructive interference occurs when alpha=theta?

  • $\begingroup$ Constructive interference will occur when alpha = theta ONLY if $2d(sin(alpha)) = m\lambda$. $\endgroup$ – S. McGrew Feb 8 '19 at 5:59

Only when scattering angle = incident angle will the scattered waves from all the atoms in the scattering plane be in phase.

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