Derive Bragg's equation .

Explanation

When a beam of X - rays strikes a crystal, the constituent particles of the crystal scatter or deflect some of the X - rays from their original path. The scattered X - rays can be detected on a photographic plate. From the diffraction pattern, the distance between constituent particles in a crystal can be studied.
In the above figure X - rays with wave-length `lambda` strike a crystal face at an angle- `theta` and, then reflect at the same angle., The rays that strike an atom in the second layer are diffracted at/the same angle `theta` .but because the second layer of atoms away from the X - ray .source, the distance that the X - rays has to traver is more to reach the second Iayer. The.path difference is indicated by AB in the figure. AB is equal to the distance between atomic layers d(=zB) times the sine Of the angle `theta`

The X -rays have to travel a total extra distance AB + BC
AB + BC = 2d sin `theta` ( AB = BC) The different X - rays striking the two layers of atoms are in -phase initially and also they will be in - phase after reflection only the extra distance AB + BC is equal to a whole number of wavelengths n`lambda` where n is an integer (1, 2, 3,4....) .
AB + BC = 2d sin `theta = nlambda`
. This is known as Bragg.s equation, By knowing the value of `phi,lambda` the value of `theta` can be measured, and the value of n is a small integer, usually 1. Thus trie-distance d between layers of atoms in a crystal can be calculated.