Q.State and obtain Bragg's Law.

Bragg's law gives the condition for constructive interference of X-rays reflected from successive parallel planes of atoms in a crystal, $2d\sin\theta = n\lambda$, derived from the geometric path difference between rays reflected from adjacent planes.

A crystal can be regarded as made up of a set of parallel, equally spaced atomic planes separated by an interplanar spacing $d$. When a monochromatic beam of X-rays of wavelength $\lambda$ falls on these planes at a glancing angle $\theta$ (measured from the plane, not the normal), each plane reflects a small fraction of the incident beam, obeying the ordinary law of reflection (angle of incidence = angle of reflection, both $=\theta$).

Derivation: Consider two parallel rays, one reflected from the upper plane at point $A$ and the other from the next plane (a distance $d$ below) at point $B$, directly below $A$. Draw perpendiculars from $A$ to the incident and reflected rays of the lower ray, meeting them at points $C$ and $D$ respectively. In the right triangle $ACB$, $AB = d$ and $\angle ABC = \theta$, so $AC = d\sin\theta$; similarly $AD = d\sin\theta$. The extra distance travelled by the ray reflected from the lower plane, compared with the ray reflected from the upper plane, is

$\text{Path difference} = CB + BD = AC + AD = 2d\sin\theta$

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Using Bragg's law with the maximum possible glancing angle ($\sin\theta = 1$), the longest analysable wavelength in the first order equals $2d$.

1. Bragg's law of X-ray diffraction from a crystal: $n\lambda = 2d\sin\theta$, where $d$ is the interplanar spacing, $\theta$ the glancing angle, and $n$ the order of diffraction.
2. For a crystal to diffract a wavelength $\lambda$ in order $n$: $\sin\theta = \dfrac{n\lambda}{2d}$, and since $\sin\theta \le 1$ always, this places an upper limit on $\lambda$ for a given order. …