\(u_{0}=1.25\, \mathrm{cm} ; M_{\infty}=?\)

From \(\frac{1}{f_{0}}=\frac{1}{v_{0}}-\frac{1}{u_{0}}\)

\(\Rightarrow \frac{1}{1.2}=\frac{1}{v_{0}}-\frac{1}{(-1.25)}\)

\(\Rightarrow \quad \frac{1}{v_{0}}=\frac{1}{1.2}-\frac{1}{1.25}\)

\(\Rightarrow \)  \(v_{0}=30\, \mathrm{cm}\)

Magnification at infinity,

\(M_{\infty}=-\frac{v_{0}}{u_{0}} \times \frac{D}{f_{e}}\)

\(=\frac{30}{1.25} \times \frac{25}{3}\)    ( \(\because D=25\, \mathrm{cm}\) least distance of distinct vision) 

\(=200\)

Hence the magnifying power of the compound microscope is \(200\)