Physics - Geometrical Optics Question with Solution | TestHub

Let as shown, 1 and 2 are positions of objects and images in two different situations. It is given$\left|\frac{v_1}{u_1}\right|=2\left|\frac{v_2}{u_2}\right|$ Here, $u_1=-15\mathrm{cm},u_2=-20\mathrm{cm}$ $\therefore v_1=2v_2\times \frac{u_1}{u_2}$ $=2v_2\times \frac{15}{20}=\frac{3}{2}{v}_2$ Now, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $\therefore \frac{1}{f}=\frac{1}{v_1}-\frac{1}{u_1}$ and $\frac{1}{f}=\frac{1}{v_2}-\frac{1}{u_2}$ So, $\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{v_2}-\frac{1}{u_2}$ $\Rightarrow \frac{2}{3v_2}+\frac{1}{15}=\frac{1}{v_2}+\frac{1}{20}$ $\Rightarrow v=20\mathrm{cm}$