# Diffraction angle - Bragg's Law

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:

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?

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