 Solution

Let $J_1,J_2$ be the single-slit intensities at the two wavelengths. At a screen position $y$, the path difference is approximately $\Delta =d\sin \theta$, with $\sin \theta =y/\sqrt{D^2+y^2}$. Fields from the two slits interfere at each wavelength, while a detector averaging over the optical beat period adds the intensities of the two distinct wavelengths:For equal single-slit intensities $J_1=J_2=J$, define $\bar{\sigma }=(1/{\lambda }_1+1/{\lambda }_2)/2$ and $\Delta \sigma =1/{\lambda }_2-1/{\lambda }_1$. ThenThe first cosine gives the interference visibility envelope. Near the centre one may further replace $\Delta$ by $dy/D$. The formula neglects unspecified single-slit envelopes and the slow change of single-slit intensity across the screen.