Question

The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Answer 1

1. The person is suffering from an eye defect called myopia. In this defect, the image is formed in front of the retina. Hence, a concave lens is used to correct this defect of vision.

2. Given:
far point of a normal eye = ∞
but this person can only seee upto 80cm from his eyes, not beyond that. So to be able to see objects beyond 80cm, let at ∞, the image needs to be formed at 80cm so that the person can see them distinctly
Therefore
Object distance, $u=-\infty$
Image distance, $v=-80cm$
Focal length = f

According to the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$-\frac{1}{80}-\frac{1}{-\infty }=\frac{1}{f}$
$f=-80cm=-0.8m$

3. We know, Power,
$P=\frac{1}{f}$(in meters)
$P=\frac{1}{-0.8m}=-1.25D$

Negative sign shows that the lens required is concave.