In Fresnel’s biprism experiment the width of  $10$  fringes is $2cm$  which are formed at a distance of two $2$  meter from the slit. If the wavelength of light is  $5100A^0$ then the distance between two coherent sources will be

Wavelength $\lambda =5100A^0=5100\times {10}^{-10}m$

Fringe width $\beta =\frac{2}{10}cm=\frac{2}{10}\times {10}^{-2}m$

Distance between slit and screen $D=2m$

Distance between two coherent sources $d=?$

$\Rightarrow d=\frac{D\lambda }{\beta }$   $\left(\because \beta =\frac{\lambda D}{d}\right)$                            

           $=\frac{2\times 51\times {10}^{-8}}{2\times {10}^{-3}}=5.1\times {10}^{-4}m$

$\therefore d=5.1\times {10}^{-4}m$