JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 3)
A convex mirror of radius of curvature $$30 \mathrm{~cm}$$ forms an image that is half the size of the object. The object distance is :
$$-$$45 cm
$$-$$15 cm
45 cm
15 cm
Explanation
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Given $$\mathrm{R}=30 \mathrm{~cm}$$
$$\mathrm{f}=\mathrm{R} / 2=+15 \mathrm{~cm}$$
Magnification $$(\mathrm{m})= \pm \frac{1}{2}$$
For convex mirror, virtual image is formed for real object.
Therefore, $$\mathrm{m}$$ is +ve
$$\begin{aligned} & \frac{1}{2}=\frac{f}{f-u} \\ & u=-15 \mathrm{~cm} \end{aligned}$$
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