= 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:
$$
\boxed{I(y)=4J_1\cos^2\left(\frac{\pi\Delta}{\lambda_1}\right)+4J_2\cos^2\left(\frac{\pi\Delta}{\lambda_2}\right).}
$$
For equal single-slit intensities $J_1=J_2=J$, define $\bar\sigma=(1/\lambda_1+1/\lambda_2)/2$ and $\Delta\sigma=1/\lambda_2-1/\lambda_1$. Then
$$
I=4J\left[1+\cos(\pi\Delta\sigma\Delta)\cos(2\pi\bar\sigma\Delta)\right].
$$
The 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.
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