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A point object $O$ is placed in front of a glass rod having spherical end of radius of curvature $30 \mathrm{~cm}$. The image would be formed at
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$30 \mathrm{~cm}$ left
By using formula $\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$
$\Rightarrow \frac{1.5}{v}-\frac{1}{(-15)}=\frac{(1.5-1)}{+30} \Rightarrow v=-30 \mathrm{~cm}$.
$\Rightarrow \frac{1.5}{v}-\frac{1}{(-15)}=\frac{(1.5-1)}{+30} \Rightarrow v=-30 \mathrm{~cm}$.
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