Answer
(C)$~11.5ft$
Work Step by Step
We know that
$\theta_{min}=\frac{1.22\lambda}{D n}$
We plug in the known values to obtain:
$\theta=\frac{(1.22)(645\times 10^{-9}m)}{(5.50\times 10^{-3}m)(1.36)}$
$\theta=1.052\times 10^{-4}rad$
We can find the linear separation as
$Linear \space separation=\frac{Height \space of \space screen}{Number \space of \space horizontal \space lines}$
We plug in the known values to obtain:
$Linear\space separation=\frac{15.70}{1080}inch=(\frac{15.70}{1080}inch)(\frac{25.4mm}{1inch})$
$Linear\space separation=0.369mm$
Now $distance_{min}=\frac{Linear\space searation}{\theta_{min}}$
We plug in the known values to obtain:
$distance_{min}=\frac{0.369mm}{1.052\times 10^{-4}rad}$
$distance_{min}=(\frac{0.369mm}{1.052\times 10^{-4}rad})(\frac{1ft}{304.8mm})$
$distance_{min}=11.5ft$
