(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?

(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.

The two causes of long-sightedness (or hypermetropia) are as follows:

1. The low converging power of the eye lens (because of its large focal length), which causes the formation of the images of objects behind the retina, or

2. The eye ball is too short, which causes light to focus beyond the retina, instead of directly on the retina. 

(i) The ray diagram showing the eye-defect long-sightedness.

(ii) The ray diagram showing the correction of long-sightedness by using a lens.

(b) An eye has a near point distance of 0.75 m (or, 75cm), which means it is a hypermetropic eye because a hypermetropic eye has a near point more than 0.25 m (or, 25cm). 

To correct the defect of hypermetropia, spectacles containing convex lenses should be used. 

In order to find the power of the convex lens required, first, we have to calculate its focal length.

Given:

The near point of the hypermetropic person is 0.75 m (the person can see objects placed 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).

Object distance, $u$ = $−$0.25 m  (the distance of the object at the normal near point)

Image distance, $v$ = $−$0.75 m (the near point of the defective eye in front of the lens)

To find: Focal length, $f$, and power of the lens, $P$.

Solution:

From the lens formula we know that-

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

Substituting the given values we get-

$\frac {1}{(-0.75)}-\frac {1}{(-0.25)}=\frac {1}{f}$

$-\frac {100}{75}+\frac {100}{25}=\frac {1}{f}$

$-\frac {4}{3}+\frac {4}{1}=\frac {1}{f}$

$\frac {-4+12}{3}=\frac {1}{f}$

$\frac {8}{3}=\frac {1}{f}$

$f=\frac {3}{8}$

$f=+0.375m=+37.5cm$

Thus, the focal length, $f$ is 37.5 cm.

Now, 

We know that power of the lens is given as-

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

Putting the value of $f$ we get-

$P=\frac {1}{0.375}$

$P=\frac {1000}{375}$

$P=+2.67D$

Thus, the power of the convex lens required to rectify the defect is +2.67 dioptre.

This eye is long-sighted, as the power of the lens has plus sign $(+)$, which implies that it is a convex lens, and we know that a convex lens is used in correcting a long-sighted or hypermetropia.

(c) An eye has a far point distance of 2 m (or, 200cm), which means it is a myopic eye because a myopic eye has a far point less than infinity. 

To correct the defect of myopia, spectacles containing concve lenses should be used. 

In order to find the power of the concave lens required, first, we have to calculate its focal length.

Given:

The far point of the myopic person is 2 m (the person can see objects kept at the normal near point of 0.25 m if the images of the objects are formed at the person's own near point of 2 m from the eye).

Object distance, $u$ = $\infty$  

Image distance, $v$ = $−$2 m (the far point of the defective eye in front of the lens)

To find: Focal length, $f$, and power of the lens, $P$.

Solution:

From the lens formula we know that-

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

Substituting the given values we get-

$\frac {1}{(-2)}-\frac {1}{\infty}=\frac {1}{f}$

$-\frac {1}{2}-0=\frac {1}{f}$      $(\because Any\ number\ divided\ by\ infinity\ is\ equal\ to\ 0)$

$-\frac {1}{2}=\frac {1}{f}$

$f=-2m$ 

Thus, the foccal length $f$ is -2 m.

Now, 

We know that power of the lens is given as-

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

Putting the value of $f$ we get-

$P=\frac {1}{-2}$

$P=-0.5D$

Thus, the power of the concave lens required to rectify the defect is -0.5 dioptre.

This eye is short-sighted, as the power of the lens has plus sign $(-)$, which implies that it is a concave lens, and we know that a concave lens is used in correcting a short-sighted or myopia.

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Updated on: 10-Oct-2022

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