= Solution
{c}
Interpret the stated detector width $d=20\,\mathrm{\mu m}$ as one square spatial sample. At the point directly below the spacecraft, the <small-angle approximation> gives pixel angular width $d/f\simeq\Delta x/h$. Thus
$$
\boxed{f=\frac{dh}{\Delta x}=\frac{(20\times10^{-6})(1.70087\times10^7)}{500}\,\mathrm m\simeq0.680\,\mathrm m.}
$$
There is a definition error in the source: the usual <focal ratio>, or f-number, divides <focal length> by the entrance-aperture diameter $D$, whereas the printed parenthesis divides by detector width $d$. Reporting both avoids silently changing the question:
$$
\boxed{\left(\frac f d\right)_{\rm printed}\simeq3.40\times10^4,\qquad
\left(\frac f D\right)_{\rm conventional}\simeq\frac{0.680}{0.50}=1.36.}
$$
The actual optical design is therefore approximately f/1.36, a fast <optical telescope>, not an optically slow f/34000 instrument. Such a focal ratio is demanding but possible in principle: <optical aberrations>, obscuration, field coverage, <stray light> and alignment need careful design. The single-sample angular width is $2.94\times10^{-5}\,\mathrm{rad}$. Spatial sampling alone does not guarantee true resolving power; <diffraction>, pointing and <signal-to-noise ratio> also matter. At large viewing angles, the same angular pixel covers a larger, distorted surface footprint. If the detector is literally one element, scanning is additionally needed to form a map.
Back to article page
