The focal length of the objective and the eyepiece of a telescope are \(50\,cm\) and \(5\,cm\) respectively. If the telescope is focussed for distinct vision on a scale distant \(2\,m\) from its objective, then its magnifying power will be

\(\mathrm{d}=25\, \mathrm{cm}, \mathrm{u}_{0}=-200\, \mathrm{cm}\)

Magnification \(\mathrm{M}=?\)

As \(\frac{1}{\mathrm{v}_{0}}-\frac{1}{\mathrm{u}_{0}}=\frac{1}{\mathrm{f}_{0}}\)

\(\Rightarrow \frac{1}{v_{0}}=\frac{1}{f_{0}}+\frac{1}{u_{0}}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}\)

or \(\quad v_{0}=\frac{200}{3}\, \mathrm{cm}\)

Now \(v_{e}=d=-25\, \mathrm{cm}\)

From. \(\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{c}}\)

\(-\frac{1}{u_{k}}=\frac{1}{f_{e}}-\frac{1}{v_{e}}\)

\(=\frac{1}{5}+\frac{1}{25}=\frac{6}{25}\)

or. \(\quad u_{e}=\frac{-25}{6}\, \mathrm{cm}\)

Magnification \(\mathrm{M}=\mathrm{M}_{0} \times \mathrm{M}_{\mathrm{e}}\)

\(=\frac{v_{0}}{u_{0}} \times \frac{v_{c}}{u_{e}}=\frac{-200 / 3}{200} \times \frac{-25}{-25 / 6}\)

\(=-\frac{1}{3} \times 6=-2\)