Answer
$D\approx 50m$
Work Step by Step
We are given all of the information needed to find thesmallest linear width on Earth’s surface that the astronaut can resolve, D, using equation (36-17).
$L=\frac{Dd}{1.22 \lambda}$ (36-17)
Solving for D, we obtain:
$D=\frac{L*1.22\lambda}{ d}$
$D=\frac{1.22*400*10^{3}m*550*10^{-9}m}{0.005m}$
$D\approx 50m$
