Answer
a) $l=1.8\times10^4m$
b) $\theta=23^o$
This is an overestimate because it does not take into account aberrations in the eye and atmospheric effects.
Work Step by Step
a) $s=l\theta=l\frac{1.22\lambda}{D}$
$l=\frac{Ds}{1.22\lambda}=\frac{(6.0\times10^{-3}m)(2.0m)}{1.22(560\times10^{-9}m)}=1.8\times10^4m$
b) $\theta=\frac{1.22\lambda}{D}=\frac{1.22(560\times10^{-9}m)}{(6.0\times10^{-3}m)}=1.139\times10^{-4}rad\frac{180^o}{\pi rad}\times3600=23^o$
This is an overestimate because it does not take into account aberrations in the eye and atmospheric effects.