Derive Bragg's equation ?

Suppose two X-rays with wavelength $\mathrm{\lambda }$ are incident on two parallel planes of a crystal surface with an angle $\mathrm{\theta }$ , two X-rays are diffracted from lattice points A & B of the crystal.

$\to$ But 2^nd X-ray travelled some extradistance CB + BD, than 1st X-ray. If two X-rays undergo
constructive interference, they will be in same phase.
$\to$Therefore the extradistance travelled by 2nd X-ray must be equal to integralmultiple of wavelength (n$\mathrm{\lambda }$) . 
$\text{ i.e, }\mathrm{CB}+\mathrm{BD}=\mathrm{n\lambda } ---\left(1\right)$
Where n = 1,2.3....... (Order of diffraction)
$\to$From figure,
$\begin{array}{r}\mathrm{△}\mathrm{ABC},\sin \mathrm{\theta }=\frac{\mathrm{CB}}{\mathrm{AB}}=\frac{\mathrm{CB}}{\mathrm{d}}\therefore \mathrm{CB}=\mathrm{dsin}\mathrm{\theta }\\ \mathrm{\Delta ABD},\sin \mathrm{\theta }=\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{\mathrm{BD}}{\mathrm{d}}\therefore \mathrm{BD}=\mathrm{dsin}\mathrm{\theta }\end{array}$
$$\begin{gathered} \therefore \mathrm{From} \mathrm{equation} \left(1\right) \mathrm{dsin}\mathrm{\theta }+\mathrm{dsin}\mathrm{\theta }=\mathrm{n\lambda } \\ \mathrm{n\lambda }=2\mathrm{dsin}\mathrm{\theta } \end{gathered}$$
This is known as Bragg's equation.