Answer
$1.32\times10^{-7}rad$ ;
$\theta_{min}$ by Hale is 1.6 times as large
Work Step by Step
wavelength $=550nm$
D=$200in=5.08m$
Thus, $\theta_{min}=\frac{1.22\times wavelength}{D}=\frac{1.22\times550\times 10^{-9}}{5.08}=1.32\times10^{-7}rad$
From previous exercise,
$\theta_{min}=8.183\times10^{-8}rad$
Therefore, $\frac{\theta_{minH}}{\theta_{minE}}=\frac{1.32\times10^{-7}rad}{8.183\times10^{-8}rad}=1.6$
