(a) The instrument is called compound microscope because the focal length of objective lens is smaller than the focal length of eyepiece.
(b) Power of objective $={\mathit{P}}_{\mathit{o}}=100\mathrm{D}$
$\therefore {\mathit{f}}_{\mathrm{o}}=\frac{1}{100}\mathrm{m}=1\mathrm{cm}$

$\therefore {\mathit{f}}_{\mathrm{E}}=\frac{1}{40}\mathrm{m}=2.5\mathrm{cm}$
Power of eyepiece $={\mathit{P}}_{\mathit{E}}=40\mathrm{D}$
$\therefore {\mathit{f}}_{\mathrm{E}}=\frac{1}{40}\mathrm{m}=2.5\mathrm{cm}$
Length of tube $=\mathrm{L}=20\mathrm{cm}$
$\mathrm{D}=$ least distance of distinct vision $=25\mathrm{cm}$
Angular Magnification $=\frac{\mathit{L}}{{\mathit{f}}_{\mathit{o}}}\times \frac{\mathit{D}}{{\mathit{f}}_{\mathit{R}}}$
$\therefore$ Angular Magnification $=\frac{20}{1}\times \frac{25}{2.5}=200$