A near-sighted man can see objects clearly up to distance of $1.5m$ . Calculate the power of the lens of the spectacles necessary for the remedy of this defect?

Hint: In this question we have a man having the distinct vision. And we have to find the power of the lens. So by using the formula $P = \dfrac{1}{f}$ , and solving it we will get the solution for the spectacles having the necessary remedy of this defect.

Formula used
Power of the lens, is given by the formula,
$P = \dfrac{1}{f}$
Here, $P$ is the power, $f$ is the focal length.

Complete step by step answer:
So we have a near sighted man which means that the man is a myopic man whose near distance is $1.5m$ .
So we have the object distance which is infinite.
So solving for the power, substituting the values in the formula, we get
$ \Rightarrow P = \dfrac{1}{{1.5}}$
And on solving it we get
$ \Rightarrow P = 0.67D$
Hence, the power of the lens of the spectacles necessary for the remedy of this defect will be $0.67D$.

Note:
This comes under the common eye defects and to remove this we can use the concave lens, as it shoves the rays of light further apart so that it is organized in the proper focus at the back of the eye. And in this way we can get out of it.