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 at 5:59

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.