= Solution
For an ideal circular pupil, the <point spread function> along a central line is
$$
\boxed{I(x)=I(0)\left[\frac{2J_1(q)}q\right]^2,\qquad q=\frac{2\pi\mathrm{NA}}{\lambda}\frac{x}{M},\quad M=100,\ \mathrm{NA}=1.3.}
$$
Here $x$ is image-plane position and the central ratio is interpreted by its limit at $q=0$. The <Airy disk> has a bright central maximum, a first zero at object-space radius $0.61\lambda/\mathrm{NA}$ and weaker side lobes. The <Rayleigh criterion> gives two-source <optical resolution> $d\simeq0.61\lambda/\mathrm{NA}$. At an illustrative $\lambda=550\,\mathrm{nm}$ this is $258\,\mathrm{nm}$ in object space or $25.8\,\mu\mathrm m$ in the image. No unique dimensional width follows without a wavelength. The right panel of the figure in <past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s12/2/c/solution> uses wavelength-normalized coordinates.
For one isolated emitter of known <point spread function>, fitting its centre to many detected <photons> estimates a position far more precisely than the width. For approximately Gaussian width $s$ and $N$ independent signal <photons>, the ideal localization <standard deviation> scales as $s/\sqrt N$. Finite pixels, background and read noise, aberrations, uncertain <point spread function>, drift, bleaching and overlapping emitters worsen this precision. Single-source localization precision is a different task from resolving two unknown nearby sources.
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