In a compound microscope, the magnified virtual image is formed at a distance of \(25\, cm\) from the eye-piece. The focal length of its objective lens is \(1\, cm .\) If the magnification is \(100\) and the tube length of the microscope is \(20\, cm ,\) then the focal length of the eye-piece lens (in \(cm\) ) is

also magnification \(\left| m _{1}\right|=\left|\frac{ v _{1}}{ u _{1}}\right|=\frac{1}{ x -1}\)

for \(2^{nd}\) lens this is acting as object

so \(u _{2}=-\left(20- v _{1}\right)=-\left(20-\frac{ x }{ x -1}\right)\)

and \(v _{2}=-25 cm\)

angular magnification \(\left| m _{ A }\right|=\left|\frac{ D }{ u _{2}}\right|=\frac{25}{\left| u _{2}\right|}\)

Total magnification \(m = m _{1} m _{ A }=100\)

\(\left(\frac{1}{x-1}\right)\left(\frac{25}{20-\frac{x}{x-1}}\right)=100\)

\(\frac{25}{20(x-1)-x}=100 \Rightarrow 1=80(x-1)-4 x\)

\(\Rightarrow 76 x=81 \Rightarrow x=\frac{81}{76}\)

\(\Rightarrow u _{2}=-\left(20-\frac{81 / 76}{81 / 76-1}\right)=\frac{-19}{5}\)

now by lens formula

\(\frac{1}{-25}-\frac{1}{-19 / 5}=\frac{1}{f_{e}} \Rightarrow f_{e}=\frac{25 \times 19}{106} \approx 4.48 cm\)