= Solution
In <Fraunhofer diffraction>, the phase difference from aperture coordinate y to observation angle theta is $qy$, with $q=2\pi\sin\theta/\lambda$. Coherent superposition of the transmitted field elements therefore gives
$$
\boxed{\psi(\theta)\propto\int T(y)e^{iqy}\,dy.}
$$
The common propagation factor is omitted. This is the <Fourier transform> of the complex amplitude transmission, not of its intensity.
For a uniform slit of total width $W=0.2\,\mathrm{mm}$, integration gives $\psi_0=W\operatorname{sinc}(qW/2)$, where $\operatorname{sinc}s=\sin s/s$. Its normalized intensity is $\operatorname{sinc}^2(\pi W\sin\theta/\lambda)$, with a central maximum and first zeros at $\sin\theta=\pm\lambda/W=\pm0.0025$.
For a half-covered slit choose $T=+1$ on the left half and $T=-1$ on the right. Put $h=W/2$ and $s=qh/2$. The two equal rectangles have centres separated by h, so
$$
\psi_\pi=h\operatorname{sinc}s\,(e^{-is}-e^{is})=-2ih\operatorname{sinc}s\sin s,
\qquad
\boxed{\frac{I_\pi}{I_{0,\rm peak}}=\operatorname{sinc}^2s\,\sin^2s=\frac{\sin^4s}{s^2}.}
$$
This <half-slit phase-step diffraction factorization> is the phase-reversed two-slit interference factor multiplied by the diffraction envelope of each half-slit. Here the half-slit width and centre separation both equal the earlier0.1mm separation. The central intensity is zero; the largest off-axis pair satisfies $\tan s=2s$, giving $|\sin\theta|\simeq0.00186$. Other zeros are at $s=m\pi$. The plot includes the requested uniform-slit and half-covered patterns.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2010/paper-1-pa2010-phase-apertures.png]
{title=Phase-step diffraction}
{description=Complex field transmissions and their computed intensities; narrow rectangle widths are illustrative.}
{source=paper-1-pa2010-phase-apertures.py}
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