Question

(a) What is long-sightedness? State the two causes of long-sightedness (or hypermetropia). With the help of ray diagrams, show:
(i) the eye-defect long-sightedness.
(ii) correction of long-sightedness by using a lens.

(b) An eye has a near point distance of 0.75 m. What sort of lens in spectacles would be needed to reduce the near point distance to 0.25 m? Also calculate the power of lens required. Is this eye long-sighted or short-sighted?
(c) An eye has a far point of 2 m. What type of lens in spectacles would be needed to increase the far point to infinity? Also calculate the power of lens required. Is this eye long-sighted or short-sighted?

Answer 1

(a) Long-sightedness, or hypermetropia, is a defect of vision because of which a person cannot see nearby objects clearly but has normal distant vision. Hypermetropia occurs because

1. the eye lens has low converging power, which causes the formation of the images of objects behind the retina, or

2. the eye ball is too short, which causes the formation of the images of objects behind the retina

The ray diagram shows the hypermetropic eye and how the defect is corrected by using a convex lens, which converges the light rays from objects and forms clear images on the retina.

(b) A person suffering from hypermetropia can correct the defect by wearing spectacles containing convex lenses. In order to find the power of the convex lens required, we have to first calculate its focal length.
Given that the near point of the hypermetropic person is 0.75 m from the eye (the person can see objects kept at the normal near point of 0.25 m in front of the eye if the images of the objects are formed at the person's own near point of 0.75 m from the eye).
$u=-0.25m$ (the distance of the object at the normal near point)
$v=-0.75m$ (the near point of the defective eye in front of the lens)
f = ? (focal length)
The focal length can be calculated using the formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Substituting the values in the formula, we get
$$\begin{gathered} \frac{1}{f}=\frac{1}{-0.75}-\frac{1}{-0.25} \\ \frac{1}{f}=\left(\frac{-1+3}{0.75}\right) \end{gathered}$$.
∴$f=2.67m$

Now that we know the focal length of the convex lens, its power can be calculated.
Power $P=\frac{1}{f}(finmeters)$
∴$P=\frac{1}{2.67}=0.37\mathrm{D}$
Hence, the power of the convex lens required to rectify the defect is +0.37 D.
This eye is long-sighted.

(c) The far point of an eye can be increased by wearing spectacles with concave lens. In order to find the power of the concave lens required, we have to first calculate its focal length.

Given that the far point of the short-sighted person is 2 m in front of the eye (the person can see objects kept at infinity if the images of the objects are formed at the person's own far point of 2 m from the eye).
$u=\infty$(the distance of the object)
$v=-2m$ (the far point of the defective eye in front of the lens)
$f=?$ (focal length)
The focal length can be calculated using the formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Substituting the values in the formula, we get $\frac{1}{f}=\frac{1}{-2}-\frac{1}{\infty }=\frac{-1}{2}(S\mathrm{inc}e,\frac{1}{\infty }=0)$.
∴ $f=-2m$

Now that we know the focal length of the concave lens required, its power can be calculated.
Power $P=\frac{1}{f}(forfinmeters)$
∴ P = $\frac{1}{-2}=-0.5\mathrm{D}$
Hence, the power of the concave lens required to rectify the defect is -0.5 D.
This eye is short-sighted.