Question

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

• A. +3 D
• B. + 4 D
• C. +5 D
• D. +6 D

Answer 1

The correct option is A

+3 D

The near point of eye is 1 metre or 100 cm. It means the eye cannot see objects placed closer than 1 metre.
If an object is placed at 25 cm , then to be visible its image has to be formed at 100 cm on the same side as the object, which acts as the object for eye, which forms the final image at retina.

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

v = -100 cm or -1 m, u = -25 cm or 1/4 m

$(-\frac{1}{100})-(-\frac{1}{25})=\frac{1}{f}$
$(-\frac{1}{100})+\frac{1}{25}=\frac{1}{f}$
$\frac{+3}{100}=\frac{1}{f}$
f = +33.3 cm
$P=\frac{1}{f\left(inmetres\right)}or\frac{100}{f\left(incms\right)}$
$P=\frac{100}{33.3cms}$
P = +3D

Thus power is +3D