When an object is placed 40 cm away from a spherical mirror an image of magnification \(\frac{1}{2}\) is produced. To obtain an image with magnification of \(\frac{1}{3}\), the object is to be moved :

\(\begin{aligned} & \mathrm{m}=\frac{1}{2}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}} \\ & \frac{1}{2}=\frac{\mathrm{f}}{\mathrm{f}-(-40)} \\ & \mathrm{f}+40=2 \mathrm{f} \Rightarrow \mathrm{f}=40 \mathrm{~cm} \\ & \text { now } \mathrm{m}=\frac{1}{3}=\frac{40}{40-\mathrm{u}} \\ & 40-\mathrm{u}=120 \Rightarrow \mathrm{u}=-80\end{aligned}\)
Thus, the object needs to be placed 80 cm from the mirror, opposite to the initial placement of 40 cm. To achieve this repositioning, the object needs to be moved 40 cm further from its initial 40 cm distance, totaling a move of 40 cm away from the mirror.