A person uses +1.5 D glasses to have normal vision from 25 ~cm...

Let $d$ be the distance of near point of the person without glasses. Focal length of glasses, f_1=$\frac{100}{P_1}=\frac{100}{1.5}=\frac{200}{3} \mathrm{~cm}$Now $u=-25 \mathrm{~cm} ; v=$ ? From lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1} or \frac{1}{v}=\frac{1}{f_1}+\frac{1}{u}=\frac{3}{200}-\frac{1}{25}=-\frac{5}{200}$ \therefore v=-$\frac{200}{5}=-$40 $\mathrm{~cm}$Therefore, the near point of the person is $40 \mathrm{~cm}$. Focal length of microscope, $\begin{gathered}f_2=\frac{100}{P_2}=\frac{100}{20}=5 \mathrm{~cm}\\ \therefore M=1+\left|\frac{d}{f_2}\right|=1+\frac{40}{5}=9 \end{gathered}