= Solution
For a rectangular slit centred at $x_c$, the one-dimensional <Fraunhofer diffraction> integral is
$$
\int_{x_c-a/2}^{x_c+a/2}e^{-iqx}\,dx
=a\operatorname{sinc}(qa/2)e^{-iqx_c},\qquad \operatorname{sinc}z=\frac{\sin z}{z}.
$$
The two equal <apertures> have centres at $\pm D/2$; the glass subtracts their fields. Therefore
$$
\psi\propto a\operatorname{sinc}(qa/2)\left(e^{iqD/2}-e^{-iqD/2}\right),\qquad
\boxed{I\propto\sin^2(qD/2)\operatorname{sinc}^2(qa/2)}.
$$
This is <phase-reversed double-slit diffraction>: the narrow-slit <interference> factor is multiplied by the <single-slit diffraction> envelope. In the limit $a\to0$, the normalized envelope tends to one, recovering part (d).
The <interference> minima and envelope minima obey, respectively,
$$
\boxed{\sin\theta_m=m\lambda/D=0.004m},\quad m\in\mathbb Z;\qquad
\boxed{\sin\theta_n=n\lambda/a=0.020n},\quad n\in\mathbb Z\setminus\{0\}.
$$
Near the axis these are $\theta_m\simeq4m\,\mathrm{mrad}$ and envelope zeros $\theta_n\simeq20n\,\mathrm{mrad}$. Here $D/a=5$, so every envelope zero already coincides with an <interference> minimum. The envelope strongly reduces the outer fringes; its central value is nonzero, but the total intensity still vanishes at $\theta=0$ because the two fields cancel. The sketch uses the small-angle coordinate; the equations above specify the exact angular zeros.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2-b7-phase-reversed-slits.png]
{title=Unshifted and phase-reversed narrow-slit fringes, and the phase-reversed finite-width pattern with its envelope}
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