Answer
(a) $0.78m$
(b) $1.3m$
Work Step by Step
(a) We know that
$\theta_{min}=sin^{-1}[1.22\frac{\lambda_r}{D}]$
$\implies \theta_{min}=sin^{-1}[1.22(\frac{690\times 10^{-9}m}{12\times 10^{-6}m})]$
$\theta_{min=4.0226}$
Now, we can determine the maximum distance from which the light waves can be resolved as
$L_{max}=\frac{y}{tan(\theta_{min})}$
We plug in the known values to obtain:
$L_{max}=\frac{0.055m}{tan(4.0226)}$
$L_{max}=0.78m$
(b) We know that
$\theta_{min}=sin^{-1}[1.22(\frac{\lambda_r}{D})]$
$\theta_{min}=sin^{-1}[1.22(\frac{420\times 10^{-9}m}{12\times 10^{-6}m})]$
$\theta_{min}=2.447^{\circ}$
Now $L_{max}=\frac{y}{tan(\theta_{min})}$
$\implies L_{max}=\frac{0.055m}{tan(2.447^{\circ})}=1.3m$
