The near point of a hyper metropic person is 80 cm from the eye. The power of lens, required to enable the person to real clearly a book held at 25 cm from the eye is:

5.25 D

The correct answer is Option (2) → 5.25 D

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ [len's maker formula]

where,

f = focal length of the lens

v = image distance = 80 cm

u = object distance = -25 cm

$\frac{1}{f}=\frac{1}{80}-\frac{1}{-25}$

$=\frac{1}{80}+\frac{1}{25}=\frac{21}{400}$

$⇒f=\frac{400}{21}≃19.05cm$

Power, $P=\frac{1}{f}$ (in meter)

$P=\frac{1}{19.05}cm=5.25D$