Question

If the near point of a far-sighted eye is $1m$, what is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is $25cm$)

• A. +3D
• B. - 3D
• C. +4D
• D. -4D

Answer 1

The correct option is A

+3D

If an object is placed at 25 cm (N'), then the image has to be formed at 100 cm (N) (because N is the far point for the far-sighted eye), which acts as the object for the eye lens, which forms the final image at the retina.

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Now, applying Cartesian convention:
$v=-1m,u=-25cm=-1/4m$
Therefore, $\frac{-1}{1}-\frac{-4}{1}=\frac{1}{f}$
$\frac{1}{f}=+3m$
Thus power is $+3D$