Question

The near point of a hypermetropic eye is 1m. Assume that near point of the normal eye is $25$ cm. The power of the lens required to correct this defect will be .........$(+3,-3)$ dioptre.

Answer 1

Object distance, $u=-25cm$

Image distance, $v$ = $-1m=-100cm$

Focal length, $f$

Using the lens formula,

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

here $u=-25cm$

$v=-100cm$

substituting & calculating for $f$, we get

$f=+33.3cm=+0.33m$

Power $P=\frac{1}{f}=+3D$