In Young's experiment the distance of the screen from the two slits is  $4m$ . When light of wavelength $\lambda$  is allowed to fall on the slits, the width of the fringes obtained on the screen is $4mm$ . The distance between the two slits is $0.06cm$ . The wavelength $\lambda$  is ....... (nm)

Distance of the screen $D=4m$

Fringe width $\beta =4mm=4\times {10}^{-3}m$

Distance between the slits $d=0.06cm=0.06\times {10}^{-2}m$

Wavelength $\lambda =?$

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

Wavelength $\lambda =\frac{\beta d}{D}$

                        =$\frac{\cancel{4}\times {10}^{-3}\times 6\times {10}^{-4}}{\cancel{4}}$

                        $=6\times {10}^{-7}$

                       $\therefore \lambda =600nm$