The magnification produced when an object is placed at a distance of $\displaystyle 20$ cm from a spherical mirror is+$\displaystyle 1$/2. Where should the object be placed to reduce the magnification to+$\displaystyle 1$/3.

Given, Distance of object, \(\displaystyle \mathrm{u}=-20 \mathrm{~cm}\) Magnification \(\displaystyle \mathrm{m}=\frac{1}{2}\) \[\therefore \mathrm{m}=\frac{-v}{u} \Rightarrow \mathrm{v}=-\frac{1}{2} \times-20=10 \mathrm{~cm} \] Using mirror formula \[\begin{aligned} & \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ & \frac{1}{f}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20} \end{aligned} \] f = $\displaystyle 20$ cm If \(\displaystyle \mathrm{m}=\frac{1}{3}\) and \(\displaystyle \mathrm{v}^{\prime}\) and u' are the corresponding image distance and object distance. \(\displaystyle \frac{1}{3}=\frac{-v^{\prime}}{u^{\prime}} \Rightarrow v^{\prime}=\frac{-u^{\prime}}{3} \frac{1}{f}=\frac{1}{v^{\prime}}+\frac{1}{u^{\prime}} \frac{1}{20}=\frac{-3}{u^{\prime}}+\frac{1}{u^{\prime}} \Rightarrow \frac{1}{20}=\frac{-2}{u^{\prime}}\) \(\displaystyle \mathrm{u}^{\prime}=-40 \mathrm{~cm}\).