 Solution

The etendue is the area–solid-angle acceptance of a ray bundle, more generally $\mathcal{G}=n_{mopt}^2\int \cos \theta dAd\Omega$. In an ideal lossless optical system it is conserved. Increasing pupil area increases accepted power only together with the appropriate field and detector cone; passive optics cannot arbitrarily increase radiance. For a small nadir pixel here, $A\Omega$ denotes entrance-pupil area times the solid angle of its surface footprint.

Choose $1.00$–$1.03\mu \mathrm{m}$ within the surface-sensitive window. A representative band-mean total radiance read from the original blue curve is ${\bar{L}}_{\lambda }\simeq 4\mu \mathrm{W}\mathrm{c}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mu {\mathrm{m}}^{-1}$, uncertain at roughly the tens-of-per-cent level. Its band integral isThe provided SNR formula depends on $L\Delta \nu$ only through this band integral. If $\nu$ is wavenumber, the selected band spans about $291\mathrm{c}{\mathrm{m}}^{-1}$ and $L_{\nu }d\nu =L_{\lambda }|d\lambda |$; numerical radiance densities in the two coordinates must not be mixed. Convert the specific detectivity to SI: $D^{\ast }=5\times 10^9\mathrm{m}\mathrm{H}{\mathrm{z}}^{1/2}{\mathrm{W}}^{-1}$ and $\sqrt{a}=2\times 10^{-4}\mathrm{m}$. Using the footprint dwell time from part (c),A square $200\mathrm{m}$ footprint at altitude $259\mathrm{km}$ subtends $\Omega \simeq (200/259000{)}^2=5.96\times 10^{-7}\mathrm{sr}$. Thus the collecting area and circular aperture diameter areThe Rayleigh criterion independently requires only $D\gtrsim 1.22\lambda /(200/h)\simeq 1.6\mathrm{mm}$ at $1.015\mu \mathrm{m}$, so the noise requirement dominates diffraction. The $200\mu \mathrm{m}$ detector size corresponds to focal length about $0.259\mathrm{m}$ for this pixel field. This assumes one detector footprint per stated feature size; stricter spatial sampling would need smaller footprints and a revised photon budget.

The estimate assumes a clear, non-scattering atmosphere as in the graph, a night-side or otherwise isolated thermal signal, negligible extra stray-light and read-noise penalties beyond the supplied detectivity, and transmission $0.7$ for the instrument alone. Atmospheric attenuation is already included in the plotted incoming radiance. The selected band has little atmospheric-only signal, but actual hot-spot detection concerns contrast relative to the background: SNR $50$ on total radiance is not a guarantee for an arbitrarily small temperature anomaly. An illustrative radiance range $3$–$5$ in the plotted units would give diameters about $0.20$–$0.15\mathrm{m}$.