Answer
$27cm$
Work Step by Step
We were given that patches are $60\mu m$ across. Using a wavelength of $550*10^{-3}m$ and $3.0mm$ as the diameter of our pupil in equation (36-17), we can find L, the maximum viewing distance in question.
$L=\frac{Dd}{1.22 \lambda}$
$L=\frac{60*10^{-6}m*3.0*10^{-3}m}{1.22* 550*10^{-9}m}$
$L=0.27m$
$L=27cm$
