Physics - Geometrical Optics Question with Solution | TestHub

Here $v_O=L-f_e=21.7\mathrm{cm}-2.5\mathrm{cm}=+19.2\mathrm{cm}$ $+ve$ sign shows that image is real. For a objective lens $\frac{1}{u_O}+\frac{1}{v_O}=\frac{1}{f_O}$ [using real-virtual sign convention] $\begin{array}{rl}& \therefore \frac{1}{u_O}=\frac{1}{f_O}-\frac{1}{v_O}\therefore \frac{1}{u_O}=\frac{1}{(+1.6)}-\frac{1}{(+19.2)}\\ & \frac{1}{u_O}=\frac{19.2-1.6}{(1.6)(19.2)}\text{ or }{u}_O=+1.745\mathrm{cm}\end{array}$ +ve sign shows that object is real. When the image is formed at infinity, the magnification is given by $m=\left(\frac{v_O}{u_O}\right)\left(\frac{D}{f_e}\right)$ where $D$ is the distance of distinct vision. $\therefore m=\left(\frac{19.2\mathrm{cm}}{1.745\mathrm{cm}}\right)\left(\frac{25\mathrm{cm}}{2.5\mathrm{cm}}\right)=110.$