= Solution
At <focal length> $f=80\,\mathrm m$, the stroke maps to a focal-plane width $w_i=fs/H=12\,\mathrm{\mu m}$. A practical estimate with two detector samples across that stroke is
$$
\boxed{p\lesssim\frac{fs}{2H}=6.0\,\mathrm{\mu m},\qquad
N\gtrsim\frac{f(500\,\mathrm m)/H}{p}\simeq3.33\times10^4\ \text{pixels across}.}
$$
The required sensor extent in that direction is $f(500\,\mathrm m)/H=0.20\,\mathrm m$. One $12\,\mathrm{\mu m}$ pixel per stroke would give about $1.67\times10^4$ 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,
$$
\boxed{p\le\frac{\lambda f}{2D}=\frac{fs}{2(1.22)H}=4.92\,\mathrm{\mu m},\qquad
N\ge4.07\times10^4.}
$$
This stricter value retains the full circular-aperture spatial <bandwidth>; the $6\,\mathrm{\mu m}$ and $3.3\times10^4$ 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.
Back to article page
