Question

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

Answer 1

Hint: For a myopic eye, a diverging lens has to be used since the rays from the object converge due to the eye lens before reaching the retina of the eye. The power of the lens should be such that if an object is placed at infinity, it must form an image at the far point of the eye, so that the eye lens can converge it properly on the retina.

Formula used:
The power $P$ of a lens is given by
$P=\dfrac{1}{f\left( \text{in metre} \right)}$

Complete step by step answer:
For a myopic eye, the eye lens shape is changed in such a way that it cannot converge the light rays coming from an object at infinity upon the retina of the eye but does so before it. Hence, for the light rays to converge at the retina, the object must be brought closer and hence, the greatest distance that the eye can clearly see an object (the far point of the eye) decreases.
To correct this, a diverging concave lens is used such that it diverges the light rays coming from the object at infinity, before they reach the eye lens and hence, the eye lens can converge the slightly diverged rays perfectly onto the retina. This can also be imagined in the sense that the concave lens forms a virtual image of the object at infinity at the far point of the eye.
Hence, let us analyze the question.
The diverging lens must be such that the image of the object at infinity must be formed at the far point of the eye. Since, for a lens, the image of an object at infinity is formed at its focus, the focus of the lens is equal to the far point of the eye.
Hence, by the question, the focus of the lens will be $f=-80cm=-0.8m$ --(1) $\left( 1cm=0.01m \right)$.
The focus is negative since the focus and the object are on the same side of the lens and according to the sign convention, the focal length is measured from the centre of the lens in a direction opposite to the light rays emanating from the object and hence, the length will be negative.
Also, the power $P$ of a lens is given by
$P=\dfrac{1}{f\left( \text{in metre} \right)}$ --(2)
Putting (1) in (2), we get,
$P=\dfrac{1}{-0.8}=-1.25D$
Where $D=1{{m}^{-1}}$ stands for dioptre, the unit of lens power.
Hence, the nature of the lens is a concave lens and its power is $-1.25D$.

Note: Students must properly know the nature of the lenses that are used to correct the various eye defects. They need not be memorized but by properly understanding the defects, and the inherent properties of the corrective lenses due to which they are employed to correct the defects, it becomes very clear and intuitive about the nature of corrective lenses that must be used to correct an eye defect. Looking at diagrams of these eye defects and the effects of these corrective lenses makes it even clearer.
Students must also remember to properly use the sign conventions, since they hold significant meaning in optics. They tell us about the nature of lenses, magnifications of images and their properties. For example, if we had not written the proper sign convention in the problem above, we would have got a positive value for the power, which would mean that we required a convex lens (convex lenses have positive powers), which would have been absolutely wrong.