Distance of image of $A$ 'is $v_{A}$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v_{A}}+\frac{1}{-60}=\frac{1}{60} \Rightarrow v_{A}=-30 \mathrm{cm}$
Distance of image of $B^{\prime}$ is $v_{B}$
$\frac{1}{v_{B}}+\frac{1}{-90}=\frac{1}{60}$
$\frac{1}{v_{B}}=\frac{1}{60}+\frac{1}{90} \Rightarrow \frac{1}{v_{B}}=\frac{5}{180}$
$v_{B}=36 \mathrm{cm}$
$m=\frac{\text { length of image }}{\text { length of object }}=\frac{6}{30}=\frac{1}{5}$
Distance of image increases with convex mirror.
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| List$-I$ | List$-II$ |
| $UV$ rays | $(i)$ Diagnostic tool in medicine |
| $X-$ rays | $(ii)$ Water purification |
| Microwave | $(iii)$ Communication, Radar |
| Infrared wave | $(iv)$ Improving visibility in foggy days |
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