Thank you for reporting, we will resolve it shortly
Q.
The least distance of vision of a longsighted person is $60\, cm$. By using a spectacle lens, this distance is reduced to $12\, cm$. The power of the lens is
Here, $v=-60\, cm , u=-12\, cm$
$\therefore $ By using the relation, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
We have, $\frac{1}{f}=\frac{1}{-60}-\frac{1}{-12}$
$\Rightarrow \frac{1}{f}=\frac{1}{15} \,cm$ or $\frac{100}{15} m$
So, the power of lens, $P=\frac{100}{15}=+\frac{20}{3} D$