A person cannot see objects clearly beyond $50\text{cm}.$ What should be the power of corrective lens used ?

The correct option is C $-2\text{D}$
The person cannot see objects clearly beyond $50\text{cm}.$
So, his far point$=50\text{cm}$ for the distant object $u=\infty .$

Hence, he should use a lens which forms an virtual image of the distant
object $u=\infty$ at a distance of $50\text{cm}$ in front of him.

Using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Here, $u=-\infty ,v=-50\text{cm}$

$-\frac{1}{50}+\frac{1}{\infty }=\frac{1}{f}$

$f=-50\text{cm}$

$\therefore P=\frac{1}{f}=\frac{1}{-0.5}=-2\text{D}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.