The near point of a Hypermetropic eye is $50cm$, What is the power of the lens required to correct this defect? Assume near point of normal eye is $25cm$.

Step 1: Given data

Near point of Hypermetropia's eye, $v=-50cm$.

Near point of a normal eye, $u=-25cm$.

Step 2: Formula & solution

Expression for the focal length $\left(f\right)$ is shown below

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

Here $v$ is Near point of Hypermetropia or image distance from the lens, $u$ is object distance or near point of normal eye.

Step 3: The calculation for focal length

Substitute $-50cm$ for $v$ and $-25cm$ for $u$ in the above expression of the power.

$$\begin{gathered} \frac{1}{f}=\frac{1}{-50cm}-\frac{1}{-25cm} \\ =-\frac{1}{50cm}+\frac{1}{25cm} \\ =\frac{-1+2}{50cm} \\ =\frac{1}{50cm} \\ f=50cm \end{gathered}$$

Step 4 Expression for the power of the lens

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

Here $f$ is the focal length in the meter.

Calculation for power

Substitute $50cm$ for $f$ in above expression for power.

$$\begin{gathered} P=\frac{1}{50cm} \\ =\frac{100}{50}D \\ =2D \end{gathered}$$

Hence required power of the convex lens is $+2D$.