 Solution

At focal length $f=80\mathrm{m}$, the stroke maps to a focal-plane width $w_i=fs/H=12\mu \mathrm{m}$. A practical estimate with two detector samples across that stroke isThe required sensor extent in that direction is $f(500\mathrm{m})/H=0.20\mathrm{m}$. One $12\mu \mathrm{m}$ pixel per stroke would give about $1.67\times 10^4$ pixels across, but cannot reliably encode a stroke's position and contrast as the image moves relative to the pixel grid.

For a precise interpretation of “fully resolve” as sampling the entire diffraction-limited incoherent image, use the optical intensity-transfer cutoff ${\nu }_c=D/(\lambda f)$. The Nyquist–Shannon sampling theorem requires $1/(2p)\ge {\nu }_c$. With the circular mirror chosen in 10d,This stricter value retains the full circular-aperture spatial bandwidth; the $6\mu \mathrm{m}$ and $3.3\times 10^4$ estimates use two samples per specified feature instead. An engineering design would provide further sampling and contrast margin. The distinction states the sampling convention rather than treating the order-of-magnitude estimates as contradictory.