Answer
a) $0.00126^{\circ}$
b) $1.8\times10^{-5}mm$
Work Step by Step
a) $D=2.5cm$
$f=0.8mm$
$lambda=450nm$
Minimum angular separation so that the specimen are just resolved :
$\theta_{min}=\frac{1.22\times wavelength}{D}=\frac{1.22\times 450\times 10^{-9}}{2.5\times10^{-2}}=0.000022rad=0.00126^{\circ}$
b) Resolving power of the lens =$ s = \frac{1.22\times lambda\times f}{D}=\frac{1.22\times450\times10^{-9}\times 0.8\times10^{-3}}{2.5\times10^{-2}}=1.8\times10^{-5}mm$
