Defects of Vision and Their Correction

1.0Myopia (Near sightedness)

A person suffering from this defect cannot see distant objects clearly. This is because the maximum focal length is less than the distance between the lens and the retina. The parallel rays coming from the distant object focus short of the retina [see figure (a)]. The ciliary muscles are fully relaxed in this case and any strain in it can only further decrease the focal length which is of no help to see distant objects.

Reason

This defect arises because the power of eye lens is too great, due to the decrease in focal length of the eye lens. This may arise due to either excessive curvature of the eye lens or elongation of the eye ball.

2.0Calculations for Corrective Lens for Myopia

As discussed above, myopia is corrected by using a concave lens (diverging lens) which increases the focal length in order to bring the image of the object back on the retina itself. 

Let a person can see clearly to a distance ‘x’ only . If we apply lens equation to the lens, then we have u = – ∞ and v = –x.

By lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{-x}-\frac{1}{-\infty }=\frac{1}{f}$

$f=–x$

And, power is given by,

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

(‘x’ is always taken + ve, ‘x’ and ‘f’ are in metres) 

3.0Hypermetropia (far-sightedness)

A person suffering from Hypermetropia cannot see clearly the objects closer to the eye. The least distance of distinct vision is quite larger than 25 cm for that person and the person has to keep the object inconveniently away from the eye. Thus, the image is not formed at the retina if an object is kept at about 25 cm away from the eye. The rays are focused behind the retina [see figure (b)]. 

Reason

This defect arises because either the focal length of the eye lens is too great or eye ball becomes too short. Due to this, light rays from nearby objects cannot be brought to focus on the retina to give a clear image.

4.0Calculations for Corrective Lens for Hypermetropia

Hypermetropia can be corrected by using a convex lens (converging lens) which decreases the focal length in order to focus the image of a nearby object at the retina. Thus, a person can see the object clearly. Let ‘y’ be the minimum distance at which the person can see the object clearly.

Let the near point of a person having hypermetropia be at y distance (see figure) from the eye (y > 25 cm). If we apply lens equation to the lens, then we have u = – 0.25 m and v = – y.

$\frac{1}{-y}-\frac{1}{-0.25}=\frac{1}{f}$

$\frac{1}{f}=\frac{1}{0.25}-\frac{1}{y}$

Also, power,

$P=\frac{1}{f}=\frac{1}{0.25}-\frac{1}{y}=4-\frac{1}{y}$

(‘y’ is always taken +ve, ‘y’ and ‘f’ are in metres)

Sometimes, a person may suffer from both myopia and hypermetropia. Such people often use bifocal lens. Commonly, the upper portion of bifocal lens is a concave lens (used for distant vision) and the lower portion is a convex lens (used for reading purpose).

5.0Presbyopia 

Rather than resulting from a change in the shape of the eyeball, the inability to see objects that are close to you can also occur because the lens loses its flexibility. This condition is known as presbyopia. As you age, the lens often loses its flexibility and cannot become round enough to create clear images of near objects.

The defect that arises due to ageing in which a person cannot read comfortably and distinctly without corrective eye glasses is called ‘presbyopia’.

● Reason

The power of accommodation of the eye decreases with ageing. For most of the people, the near point recedes, this means, the least distance of distinct vision increases. This phenomenon arises due to the gradual weakening of ciliary muscles and decreasing flexibility of the crystalline eye lens.

Due to ageing, usually a person can see the distant objects clearly. This is because, the rays from a distant object are less diverging as compared to the rays from a near object. Thus, the lens can still focus rays from distant objects on the retina. To cause the more sharply diverging rays from a near object to be focussed on the retina, the lens has to become quite round to shorten its focal length. Since the lens has lost its flexibility (or elasticity), it cannot become quite round to focus the light rays from the near object. Thus, he/she cannot see the near object clearly.

Correction

Presbyopia is literally an “old-age vision” and it is due to a reduction in accommodation ability. The cornea and lens together are not able to bring nearby objects into focus on the retina. The symptoms are the same as with hypermetropia or farsightedness, and the condition can be corrected with converging lens i.e., convex lens.

These days, it is possible to correct the defects of vision by using ‘contact lenses’.

Presbyopia can occur in conjunction with myopia or hypermetropia. If one already wears glasses to correct for myopia, as presbyopia occurs, the bifocal lenses can be used to accommodate both conditions.

6.0Solved Examples

1. The far point of a myopic person is 80 cm in front of the eye. What is the focal length and power of the lens required to enable him to see very distant objects clearly? 

Solution

Here, x = 80 cm = 0.8 m. 

Focal length, f = – x = – 0.8 m (A concave lens).

Power,

$P=\frac{1}{f}=\frac{1}{-0.8}=–1.25\text{d}ioptres.$

2. The near point of a patient’s eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 25.0 cm away? Neglect the eye–lens distance. (b) What is the power of this lens? 

Solution

Here, u = – 25 cm ; v = – 50 cm ; f =?

(a) By lens equation,  $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{f}=\frac{1}{(-50)}-\frac{1}{(-25)}$

$\frac{1}{f}=\frac{-1}{50}+\frac{1}{25}$

$\frac{1}{f}=\frac{-1+2}{50}=\frac{1}{50}$

$f=+50\text{cm}=+0.5\text{m}$

Alter :

Here, y = 50 cm = 0.5 m

$\frac{1}{f}=\frac{1}{0.25}-\frac{1}{y}$

$\frac{1}{f}=\frac{1}{0.25}-\frac{1}{0.5}$

$\frac{1}{f}=\frac{100}{25}-\frac{10}{5}=4-2=+2$

$f=\frac{1}{2}=+0.5\text{m}$

(b) Power, $P=\frac{1}{f}$

$P=\frac{1}{+0.5}=+2\text{D}ioptre$

3. Suppose a lens is placed in a device that determines its power as + 2.75 diopters. Find (a) the focal length of the lens and (b) the minimum distance at which a patient will be able to focus on an object if the patient’s near point is 60.0 cm. Neglect the eye–lens distance.

Solution

Here, P = + 2.75 dioptre ; f =?

(a) $f=\frac{1}{P}=\frac{1}{+2.75}=+0.364\text{m}=+36.4\text{cm}$

(b) Here, v = – 60 cm ; u =?

By lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{-60}-\frac{1}{u}=\frac{1}{+36.4}$

$\frac{1}{u}=-\frac{1}{60}+\frac{1}{36.4}=\frac{-36.4-60}{2184}=\frac{-96.4}{2184}$

$u=\frac{2184}{-96.4}=-22.6\text{cm}$

4. The near point of a hypermetropic person is 75 cm from the eye. What is the focal length and power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?

Solution

Here, y = 75 cm = 0.75 m

$\frac{1}{f}=\frac{1}{0.25}-\frac{1}{y}=\frac{1}{0.25}-\frac{1}{(0.75)}=\frac{100}{25}-\frac{100}{75}=4-\frac{4}{3}=\frac{8}{3}$

$f=+\frac{3}{8}=+0.375\text{m}$

Power, $P=\frac{1}{f}=\frac{8}{3}=+2.67\text{dioptres.}$

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