Two light waves of wavelengths 800 and 600nm are used in Young's...

Fringe width for both cases can be written as.${\omega}_{1} = \dfrac{{\lambda}_{1}D}{d} \text{\text{ and } } {\omega}_{2} = \dfrac{{\lambda}_{2}D}{d}$.

Using values of wavelength, we get ${\omega}_{1} = 16mm \text{\text{ and } } {\omega}_{2} = 12mm$.

Let y be the common distance of the bright fringes by the both wavelength, then

$y = {n}_{1}{\omega}_{1} = {n}_{2}{\omega}_{2}$.

As LCM \left({\omega}_{1}, {\omega}_{2}\right) = 48mm$, therefore at$y = 48mm distance both bright fringes will be found.