Question

The near-point of a person suffering from hypermetropia is at 50 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is 25 cm).

Answer 1

Any object placed at $25\text{cm}$ should be viewed clearly by the person.
u = $-25\text{cm}$

The lens will form a virtual image at $50\text{cm}.$ (the distance at which the person can see)
v = $-50\text{cm}$

Using the lens' equation,

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

$\frac{1}{f}=\frac{1}{-50}-\frac{1}{25}$

$\frac{1}{f}=\frac{-1}{50}-\frac{1}{25}$

$\frac{1}{f}=\frac{1}{50}$

f = $50\text{cm}=0.5\text{m}$
Power of the lens, $\text{P}=\frac{1}{f}=\frac{1}{f\left(inm\right)}=\frac{1}{0.5}=+2\text{D}$

So, the power of the lens needed should be $2$ dioptre.