Physics - Geometrical Optics Question with Solution | TestHub

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, $$\begin{array}{c}{f}_2=\frac{100}{P_2}=\frac{100}{20}=5\mathrm{cm}\\ \therefore M=1+\left|\frac{d}{f_2}\right|=1+\frac{40}{5}=9\end{array}$$