Answer
(a) $0.91arc\space seconds$
(b) $1.7Km$
Work Step by Step
(a) We can find the required minimum angular separation as follows:
$D=6.0\space inches=6.0\times 0.0254m=0.1524m$
Now $\theta_{min}=1.22\frac{\lambda}{D}$
We plug in the known values to obtain:
$\theta_{min}=1.22(\frac{550\times 10^{-9}m}{0.1524m})=440.88\times 10^{-9}$
$\theta_{min}=(440.88\times 10^{-9})(\frac{360^{\circ}}{2\pi \space rad})(\frac{1}{\frac{1}{3600^{\circ}}})=0.91arc \space seconds$
(b) The required minimum distance can be determined as follows:
$y=Ltan\theta_{min}$
We plug in the known values to obtain:
$y=(384,400Km)tan[(1.22)(\frac{550\times 10^{-9}m}{0.1524m})]$
We know that if $\theta$ is small then $tan\theta \approx\theta$
$\implies y=(384,400Km)(4.4028\times 10^{-6})$
$y=1.7Km$
