= Solution
Let $X$ be horizontal position on the screen, $L=0.50\,\mathrm m$, and use $\sin\theta=X/\sqrt{L^2+X^2}$. The plotted interval is $-50\,\mathrm{mm}\leq X\leq50\,\mathrm{mm}$. With $a=10\,\mu\mathrm m$, $D=20\,\mu\mathrm m$ and $\lambda=0.500\,\mu\mathrm m$, the <four-slit Fraunhofer diffraction> intensity is the previous normalized expression evaluated at this angle. The total aperture width is $3D+a=70\,\mu\mathrm m$, giving a <Fresnel number> of order $0.020$, consistent with the far-field approximation.
Nominal principal array directions satisfy $D\sin\theta_m=m\lambda$. Their screen positions are
$$
X_m=\frac{L(m\lambda/D)}{\sqrt{1-(m\lambda/D)^2}}.
$$
Thus the visible nonzero orders are near $\boxed{X=\pm12.50\,\mathrm{mm}\ (m=\pm1),\quad X=\pm37.61\,\mathrm{mm}\ (m=\pm3)}$, in addition to the central maximum. The even nonzero orders are <missing diffraction orders>: at $m=2$ the first envelope zero occurs at $X=\pm25.03\,\mathrm{mm}$. The next envelope zero and $m=4$ direction lie at about $\pm50.25\,\mathrm{mm}$, just beyond the screen edges.
At the nominal first and third array directions, $I/I(0)=4/\pi^2\simeq0.405$ and $4/(9\pi^2)\simeq0.0450$, respectively. The closest central minima satisfy $D\sin\theta=\pm\lambda/4$, hence $\boxed{X\simeq\pm3.125\,\mathrm{mm}}$. Two small subsidiary array maxima occur between adjacent principal directions. The single-slit envelope slightly shifts the true off-axis maxima relative to the nominal principal array angles.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-2-b8-four-slit-intensity.png]
{title=Calculated four-slit intensity across the specified screen, with principal array directions and missing orders marked}
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