Physics - Geometrical Optics Question with Solution | TestHub

Given : $f_0=1.2\mathrm{cm};f_e=3.0\mathrm{cm}$ $u_0=1.25\mathrm{cm};M_{\infty }=$ ? From $\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$ $$\Rightarrow \frac{1}{1.2}=\frac{1}{v_0}-\frac{1}{(-1.25)}$$ $$\begin{array}{rl}& \Rightarrow \frac{1}{v_0}=\frac{1}{1.2}-\frac{1}{1.25}\\ & \Rightarrow v_0=30\mathrm{cm}\end{array}$$ Magnification at infinity, $$\begin{array}{rl}& M_{\infty }=-\frac{v_0}{u_0}\times \frac{D}{f_e}\\ & =\frac{30}{1.25}\times \frac{25}{3}(\because D=25\mathrm{cm}\text{ least distance }\end{array}$$ of distinct vision) $$=200$$ Hence the magnifying power of the compound microscope is 200