Question

A person with a hypermetropia eye cannot see objects beyond $1$ metre directly. What should be the type of the corrective lens used? What would be its power?

Answer 1

He should use such a lens which forms the image of far away objects within $1m$ so that he can see clearly.

Hence, for extreme position of vision-

Object distance$=u=-\infty$

Image distance $v=-1m$ . This image formed by lens will be percieved by the eye-lens as the object.

Now,

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

$⟹f=-1m$

Since, $f$ is negative, the lens to be used is $concave$.

Power of lens $P=\frac{1}{f\left(inm\right)}=-1D$