In a Young's double slit experiment, slits are separated by $$0.5\,mm,$$  and the screen is placed $$150\,cm$$  away. A beam of light consisting of two wavelengths, $$650\,nm$$  and $$520\,nm,$$  is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

Solution :
For common maxima, $${n_1}{\lambda _1} = {n_2}{\lambda _2}$$
$$\eqalign{ & \Rightarrow \,\,\frac{{{n_1}}}{{{n_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} \cr & = \frac{{520 \times {{10}^{ - 9}}}}{{650 \times {{10}^{ - 9}}}} \cr & = \frac{4}{5} \cr & {\text{For, }}{\lambda _1} \cr & y = \frac{{{n_1}{\lambda _1}D}}{d}, \cr & {\lambda _1} = 650nm \cr & y = \frac{{4 \times 650 \times {{10}^{ - 9}} \times 1.5}}{{0.5 \times {{10}^{ - 3}}}} \cr & {\text{or, }}y = 7.8\,mm \cr} $$