Q 12-09-033NEETJEE MainMedium
A person's near point is $50$ cm. The power of the reading glasses needed to read at $25$ cm is
Answer: (D) $+2$ D
Hypermetropia: the lens must image an object at $u = -25$ cm at $v = -50$ cm.
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = -\frac{1}{50} + \frac{1}{25} = \frac{1}{50} \Rightarrow P = +2\ \text{D}$$
Solution by Sreeraj P, M.Sc Physics