= Solution
With $D=5\ \mathrm{mm}$, the circular-aperture <diffraction limit> is $\theta_R=244\lambda$ when $\lambda$ is in metres. No <wavelength> is specified, so a unique numerical diffraction angle cannot be deduced. Taking representative visible light at $\lambda=550\ \mathrm{nm}$ gives
$$
\theta_R=1.34\times10^{-4}\ \mathrm{rad}\simeq0.46\ \mathrm{arcmin}.
$$
For a roughly square receptor arrangement with surface density $n=10^5\ \mathrm{mm}^{-2}$, the characteristic receptor pitch is $s\simeq n^{-1/2}=3.16\ \mathrm{micrometres}$. With the retina $f_{\rm eye}=17\ \mathrm{mm}$ behind the lens, that pitch subtends
$$
\boxed{\theta_{\rm pitch}\simeq\frac{s}{f_{\rm eye}}=1.86\times10^{-4}\ \mathrm{rad}\simeq0.64\ \mathrm{arcmin}}.
$$
This is an order-of-magnitude estimate of the point-separation scale from retinal sampling. A periodic intensity pattern needs at least two samples per period under the <Nyquist–Shannon sampling theorem>, giving a corresponding period angle of about $3.72\times10^{-4}\ \mathrm{rad}$ or $1.28\ \mathrm{arcmin}$. A different receptor packing changes the pitch by an order-one geometric factor; density alone does not specify a full sampling lattice or a contrast threshold.
At a $5\ \mathrm{mm}$ pupil and the chosen visible <wavelength>, the optical <diffraction limit> is somewhat finer than the receptor pitch. The supplied model is therefore not purely diffraction limited: retinal sampling is comparable to, and slightly more restrictive than, ideal aperture diffraction. Real <optical aberrations> and neural processing can further reduce resolution. Different <wavelengths> or a smaller pupil can change which effect dominates.
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