 Solution

For an ideal circular pupil, the point spread function along a central line isHere $x$ 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 $d\simeq 0.61\lambda /\mathrm{NA}$. 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 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.