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(D) When $X$-rays are incident on parallel atomic planes with spacing $d$ at a glancing angle $	heta$, the path difference between the rays reflected

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$2 d \sin 	heta = n \lambda$

Vedclass · Answer

(D) When $X$-rays are incident on parallel atomic planes with spacing $d$ at a glancing angle $	heta$, the path difference between the rays reflected from successive planes is given by $\Delta x = 2 d \sin 	heta$.For constructive interference (maxima), the path difference must be an integer multiple of the wavelength $\lambda$.Therefore, the condition for maxima is $2 d \sin 	heta = n \lambda$, where $n = 1, 2, 3, \dots$ is the order of the interference fringe. This is known as Bragg's Law.

$X$-rays of wavelength $\lambda$ are reflected from parallel planes of atoms in a crystal with spacing $d$ between two planes, as shown in the figure. If the two reflected beams interfere constructively, then the condition for maxima will be (where $n$ is the order of the interference fringe):

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