Question

The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is 25 cm)

• A. +3 D
• B. - 3 D
• C. +4 D
• D. -4 D

Answer 1

The correct option is A

+3 D

If the object is placed at $25cm$ $\left(N^{{}^{\prime }}\right)$ , then the image has to be formed at $1m$ $\left(N\right)$ (because $N$ is the far point for the hypermetropic eye) , which act as the object for eye lens, which forms the final image at the retina.

Let $v$ be the image distance, $u$ be the object distance and $f$ be the focal length.
Using lens formula we have,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Now, applying cartesian convention,
$v=-1m,u=-25cm=\frac{-1}{4}m$
Therefore, $\frac{-1}{1}-\frac{-4}{1}=\frac{1}{f}$
$\frac{1}{f}=+3m$
Thus power is $+3D$