A hypermetropic person has $75cm$ as his near point. What will be the focal length and power of the convex lens in his spectacles?

Step 1: Given data

1. Near point of the person means image distance, $v=-75cm$
2. Since the near point of the normal eye is $25cm$ means object distance, $u=-25cm$

Step 2: Expression for focal length of the lens is shown below

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

Here $v$ is image distance, $u$ is object distance

Step 3: Calculate the focal length

Substitute $-75cm$ for $v$ and $-25cm$ for $u$ in the above expression of the focal length.

$$\begin{gathered} \frac{1}{f}=\frac{1}{-75cm}-\frac{1}{-25cm} \\ =\frac{-1-(-3)}{65cm} \\ =\frac{-1+3}{75cm} \\ =\frac{2}{75cm} \\ f=\frac{75cm}{2} \\ =37.5cm \end{gathered}$$

Step 4: Expression for the power of the lens is shown below

$P=\frac{1}{f}$

Here $f$ is the focal length and the focal length is in m.

Step 5: Calculate the power of the lens,

Substitute $37.5cm$ for $f$ in above expression of the focal length.

$$\begin{gathered} P=\frac{1}{37.5cm} \\ =\frac{100}{37.5}D \\ =2.67D \end{gathered}$$

Hence required focal length is $37.5cm$ and power is $2.67D$.