Question

A man cannot see objects closer than 1 meter from the eye clearly. What is the power of the corrective lens he should use?

Answer 1

Near point of the person is 1 m , it means that he can see an object kept at 25 cm clearly only if image formed of this object by corrective lens is at 1 m.
Using Sign convention rules we consider

$u=-25cm$ as image is formed real we are using minus sign

$v=-1m=-100cm$

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

$\frac{1}{f}=\frac{1}{-100}-\frac{1}{-25}$

so $f=\frac{100}{3}cm=\frac{1}{3}m$

Power $P=\frac{1}{f}=\frac{1}{1/3}=3D$