Fringe width for both cases can be written as.{\omega }_{1}=\frac{{\lambda }_{1}D}{d}&{\omega }_{2}=\frac{{\lambda }_{2}D}{d}.
Using values of wavelength, we get {\omega }_{1}=16\mathrm{mm}&{\omega }_{2}=12\mathrm{mm}.
Let y be the common distance of the bright fringes by the both wavelength, then
y=n1ω1=n2ω2.
As LCM (ω1,ω2)=48mm, therefore at y=48mm distance both bright fringes will be found.