= Solution
Represent equal narrow slits at $x=\pm D/2$ by two contributions with equal <complex amplitudes>. In a direction $\theta$ the path difference is $D\sin\theta$, so
$$
\psi(\theta)\propto e^{iqD/2}+e^{-iqD/2}=2\cos(qD/2),\qquad q=\frac{2\pi}{\lambda}\sin\theta.
$$
Therefore <double-slit interference> gives
$$
\boxed{I(\theta)\propto\cos^2\!\left(\frac{\pi D\sin\theta}{\lambda}\right)}.
$$
The bright <interference fringes> satisfy $D\sin\theta=m\lambda$, with a central maximum. In the small-angle region,
$$
\boxed{\Delta\theta\simeq\lambda/D=4.00\times10^{-3}\,\mathrm{rad}=4.00\,\mathrm{mrad}}.
$$
Dark fringes lie halfway between bright ones. Exactly, spacing is uniform in $\sin\theta$, not in $\theta$ at large angles. The first panel of the pattern figure below shows the narrow-slit result.
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