Question

A person with a myopic eye cannot see objects beyond a distance of $1.5m$. What would be the power of the corrective lens used to restore proper vision?

Answer 1

Far point of myopic eye is $1.5m=v$ to change far point to infinity $=u$
focal length of power $\to P=?,F=?$
$\begin{array}{cc}\therefore \frac{1}{f}& =\frac{1}{\tau }-\frac{1}{u}\text{(Lens formula)}\\ \frac{1}{f}& =\frac{1}{-1.5}-\frac{1}{-\infty }=-\frac{1}{1.5m}\\ f& =-1.5m\\ \therefore & =\frac{1}{f}=\frac{1}{-1.5}=-0.67\text{dioptre.}\end{array}$