The near point of hypermetropic eye is 75 cm from the eye. What is the power of the lens required to enable him to read a book held at 25 cm from the eye?

Step 1: Given that:

Near point of hypermetropic eye= $75cm$

Distance of the book from eye= Object distance($u$) = $25cm$

Step 2: Calculation of the focal length of the required lens:

1. A hypermetropic person can not see the objects placed at normal near point that is $25cm$ .
2. The near point of the hypermetropic person is increased from $25cm$ .
3. In order to make him see the objects at the normal near point, specs are provided with the convex lens that makes the image of the object placed at $25cm$ at his near point.

Thus,

Object distance($\text{u}$)= $-25cm$

The image distance(v) = The near point of hypermetropic eye

$v=-75cm$

Using lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ , we get

$\frac{1}{f}=\frac{1}{-75cm}-\frac{1}{-25cm}$

$\frac{1}{f}=-\frac{1}{75}+\frac{1}{25}$

$\frac{1}{f}=\frac{-1+3}{75}$

$\frac{1}{f}=\frac{2}{75}$

$f=\frac{75}{2}cm=\frac{75}{2}\times \frac{1}{100}m=\frac{3}{8}m$

Step 3: Calculation of power of the lens:

Power of a lens= $\frac{1}{Focallength\left(inm\right)}$

Thus,

The power of the lens required for the hypermetropic eye will be;

$P=\frac{1}{\frac{3}{8m}}$

$P=\frac{8}{3}$

$P=2.66D$

Thus,

Option D) $2.66D$ is the correct option.