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?

The individual is suffering from myopia. In this defect, the image is formed in front of the retina. Therefore, a concave lens is used to correct this defect of vision.

Object distance (u) = infinity = ∞

Image distance (v) = - 80 cm

Focal length = f

According to the lens formula,

\( \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\( -\frac{1}{80}-\frac{1}{∞}=\frac{1}{f}\)

\( \frac{1}{f}=-\frac{1}{80}\)

\( f=-80\,cm=-0.8\,m\)

We know,

\( Power,P=\frac{1}{f(in \,meters)}\)

\( P=\frac{1}{-0.8}=-1.25\,D\)

A concave lens of power - 1.25 D is required by the individual to correct his defect.