In a YDSE with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen in moved by  $5\times {10}^{-2}m$  towards the slits, the change in fringe width is $3\times {10}^{-5}m$ . If the distance between the slits is  ${10}^{-3}m$ , then the wavelength of light used is

Distance between the slits $d={10}^{-3}m$

Screen distance  $\Delta D=5\times {10}^{-2}m$

Change in fringe width $\Delta \beta =3\times {10}^{-5}m$

Wavelength $\lambda =?$

Fringe width $\beta =\frac{D\lambda }{d}$

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

                  $\Rightarrow \lambda =\frac{d.\Delta \beta }{\Delta D}$

                           $=\frac{\left({10}^{-3}\right)\left(3\times {10}^{-5}\right)}{5\times {10}^{-2}}$

                             $=6000\times {10}^{-10}m$

                      $\therefore \lambda =6000A^0$