A plane wave of wavelength $6250A^0$  is incident normally on a slit of width $2\times {10}^{-2}cm$ . The width of the principal maximum on a screen distant $50cm$ will be

Wavelength  $\lambda =6250A^0=6250\times {10}^{-10}m$

  Slit width     $a=2\times {10}^{-2}cm=2\times {10}^{-4}m$

  Distance of the screen  $D=50cm=0.5m$

Width of central maximum$=\frac{2\lambda D}{a}=\frac{2\times 6250\times {10}^{-10}\times 0.5}{2\times {10}^{-4}}$

                                            $=312.5\times {10}^{-3}cm.$