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 :

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{m}=\frac{1}{2}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}} \ & \frac{1}{2}=\frac{\mathrm{f}}{\mathrm{f}-(-40)} \ & \mathrm{f}+40=2 \mathrm{f} \Rightarrow \mathrm{f}=40 \mathrm{~cm} \ & 	ext { now } \mathrm{m}=\frac{1}{3}=\frac{40}{40-\mathrm{u}} \ & 40-\mathrm{u}=120 \Rightarrow \mathrm{u}=-80\end{aligned}$

When an object is placed 40 cm away from a spherical mirror an image of magnification $\frac{1}{2}$ is…

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