In YDSE a parallel beam of incident light consists of two wavelengths ${\lambda }_1=4000 \stackrel{\circ }{\mathrm{A}} and{\lambda }_2=5600 \stackrel{\circ }{\mathrm{A}}$. The minimum distance y on the screen, measured from the central axis, where the bright fringe due to two wavelengths coincide is $\frac{n{\text{λ}}_{1}D}{d}.$ Find $n$.

Hence, the correct answer is 7.