 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 $-50\mathrm{mm}\le X\le 50\mathrm{mm}$. 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 areThus the visible nonzero orders are near $\boxed{X=\pm 12.50\mathrm{mm}(m=\pm 1),X=\pm 37.61\mathrm{mm}(m=\pm 3)}$, in addition to the central maximum. The even nonzero orders are missing diffraction orders: at $m=2$ the first envelope zero occurs at $X=\pm 25.03\mathrm{mm}$. The next envelope zero and $m=4$ direction lie at about $\pm 50.25\mathrm{mm}$, just beyond the screen edges.

At the nominal first and third array directions, $I/I(0)=4/{\pi }^2\simeq 0.405$ and $4/(9{\pi }^2)\simeq 0.0450$, respectively. The closest central minima satisfy $D\sin \theta =\pm \lambda /4$, hence $\boxed{X\simeq \pm 3.125\mathrm{mm}}$. 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.