In a compound microscope the focal length of objective lens is \(1.2\, cm\) and focal length of eye piece is \(3.0\, cm\). When object is kept at \(1.25\, cm\) in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be

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

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 \(200\)