= Solution
Let $I_s$ be the intensity that either narrow slit would produce alone, and let $y$ measure screen displacement from the perpendicular bisector. The <optical path difference> is approximately $\delta=b\sin\theta\simeq by/D$. Equal coherent field amplitudes then give
$$
I=I_s+I_s+2I_s\cos\left(\frac{2\pi\delta}{\lambda}\right)
=\boxed{4I_s\cos^2\left(\frac{\pi by}{\lambda D}\right)}.
$$
The maxima are at $y_m=m\lambda D/b$; the minima are halfway between them. Here
$$
\boxed{\Delta y=\frac{\lambda D}{b}=0.532\,\mathrm{mm}.}
$$
The expression uses the paraxial approximation and neglects the slowly varying single-slit <diffraction> envelope because the slit widths are unspecified.
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