= Solution
A relative <phase> $\pi$ changes the sign of one slit field. Their sum becomes
$$
\psi(\theta)\propto e^{iqD/2}-e^{-iqD/2}=2i\sin(qD/2),\qquad
\boxed{I(\theta)\propto\sin^2\!\left(\frac{\pi D\sin\theta}{\lambda}\right)}.
$$
The central <interference fringe> is now dark. The maxima move to $D\sin\theta=(m+1/2)\lambda$, while the minima occur at $D\sin\theta=m\lambda$. Thus the pattern shifts by half a fringe, $2.00\,\mathrm{mrad}$ near the axis, but \b[its angular fringe separation remains] $\boxed{4.00\,\mathrm{mrad}}$. This assumes the glass changes <phase> uniformly and introduces negligible absorption or beam deflection.
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