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What Type of the Defect of an Eye is Myopia? What do you Mean by a Bifocal Lens?
Defects Of Vision And Their Remedies
We know that the range of vision of a normal eye extends from 25 cm from the eye to infinity i.e., if an object is situated any-where within this long range, its image is formed on the retina. If the range of vision of an eye is less than this normal range the eye is said to be defective. The different defects of vision are stated below:
- Long sight or hypermetropia
- Short sight or myopia
- Presbyopia
- Astigmatism
The nature of these defects and the processes of their removal are discussed below.
Long sight or hypermetropia: it is the defect of the eye due to which we cannot see the near objects distinctly. The defect is attributed to either of the two causes:
- The eyeball has become too short.
- The focal length of the eye lens has become too long.
The power of accommodation of an eye has a certain limit. For a normal eye the near point is 25 cm. But for a long-sighted eye it is greater than 25 cm i.e., a long-sighted person can see the objects situated more than 25 cm away, distinctly.
Suppose, the point N is the near point of a normal eye [Fig.], For the defect of long sight the image of an object situated at N will be formed behind the retina instead of being formed on the retina causing blurring of the image. For him the near point is situated at N’. So, for this defect the least distance of distinct vision is more than 25 cm.
Remedy: This defect can be removed by placing a convex lens of suitable focal length in front of the eye. The focal length of the convex lens is such that the light rays starting from the object at N appears to come from N’ after refraction in the lens. So N’ is the virtual image of the object at N [Fig.],
If D be the least distance of distinct vision for the normal eye and d be the corresponding distance for the long-sighted eye, then the focal length f of the convex lens placed in front of the eye is given by (from the equation of lens)
\(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) [Here, u = -D, v = -d]
\(\frac{1}{-d}-\frac{1}{-D}\) = \(\frac{1}{f}\) or, \(\frac{1}{f}\) = \(\frac{1}{25}-\frac{1}{d}\)
or, f = \(\frac{25 d}{d-25}\)cm [∵ d > 25 ∴ P is positive]
The power of that convex lens,
P = \(\frac{100}{f}\) = \(\frac{100}{25 d}\)(d – 25)
or P = 4 – \(\frac{100}{d}\) [∵ d > 25 ∴ P is positive]
So long sight can be corrected by using spectacles having convex lens. As the focal length of convex lens is positive its power is also positive that is why the power of the spectacles is obviously positive.
Short Sight or Myopia: It is the defect of the eye which can see near objects but not the distant objects distinctly. The cause of this defect may be either
- The eyeball has become a little bit elongated or
- The focal length of the eye lens has become too short.
For these reasons the image of a distant object is formed not exactly on the retina, but at a point C in front of it [Fig.]. Applying the power of accommodation the image cannot be focussed on the retina. So the far point of this short-sighted eye does not remain at infinity but comes nearer to eye i.e., the distance of the far point of this eye is shorter than that of the normal eye.
Suppose, F is the far point of a short-sighted eye [Fig.]; i.e., if an object is situated at F the eye lens forms its image on the ret-ina. Of course the near point of this eye remains unchanged, i.e., the least distance of distinct vision of a short-sighted eye remains 25 cm.
So a short-sighted eye does not see distinctly the objects from infinity up to F but can see nearer objects distinctly.
Remedy: In order to correct the defect, a concave lens of suitable focal length should be used. From the Fig. it is seen that the parallel rays from distant objects are focussed at C in front of the retina. So to focus the rays on the retina, the point of convergence should be shifted to a further point such that the virtual image of the distant object is brought to the far point F, Fig. This is possible when the focal length of the lens is equal to the distance of the far point F, as shown below.
Suppose, the distance of the far point is d and focal length of the lens L is f. In this case, object distance u = ∞ and image distance v = d.
So from the equation \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) we have,
\(\frac{1}{-d}-\frac{1}{\infty}\) = \(\frac{1}{f}\) or, \(\frac{1}{f}\) = \(\frac{1}{-d}\) or, f = -d
Power of the concave lens,
P = \(\frac{100}{f}\) or, P = \(\frac{-100}{d}\) dioptre
So, for remedy of short sight, the spectacles to be used should have concave lens of focal length equal to the distance of the far point. Focal length of a concave lens is negative, so its power is negative. Hence, a person having the defect of short sight should use spectacles having negative power.
Sometimes, it is difficult to use high power concave lens in spec-tacles. In this case, far point is brought from infinity to a rela-tively nearer point of distance d’. If the focal length of lens is f in this case, then
\(\frac{1}{f^{\prime}}\) = \(\frac{1}{-d}-\frac{1}{-d^{\prime}}\) = \(\left(-\frac{1}{d}+\frac{1}{d^{\prime}}\right)\)
Power of this lens,
P = +\(\frac{100}{f^{\prime}}\) dioptre
= 100\(\left(-\frac{1}{d}+\frac{1}{d^{\prime}}\right)\) dioptre = \(\left(-\frac{100}{d}+\frac{100}{d^{\prime}}\right)\) dioptre
Presbyopia: With the advancement of age the muscles of eye lose their elasticity. Hence, the normal power of accom-modation of eye decreases. As a result the least distance of distinct vision increases. So nearer objects are not visible distinctly. This defect is called presbyopia. For remedy of this defect con-vex lens of suitable focal length is to be used. However, far point of the eye having presbyopia is normal i.e., at infinity.
Again sometimes, the far point of the defective eye comes nearer from infinity. To see distant objects concave lens is to be used. To remove both the defects, therefore a convex and a concave lens are used together, in a circular framing. Distant objects are seen by the concave lens and nearer objects by the convex lens [Fig.]. This type of lens is called bifocal lens.
Astigmatism: It is an optical defect in which vision becomes blurred along different axes. It is due to the inability of the eye to focus a point object lying in different directions into a sharp focussed image on the retina. While a normal eye can see equally distinctly vertical and horizontal lines drawn in a plane an eye having astigmatism does not see all the lines with the same distinctness.
In Fig.(a) a few sets of three parallel lines are drawn inclined at different angles. An eye with astigmatism cannot isolate all of them with the same distinctness. This defect is due to the unequal curvature of the vertical and horizontal sections of the cornea, and is remedied by using cylindrical or spherocylindrical lenses in order to vary the focal length in one plane. The lenses so used are often called toric lenses [Fig.(b)].
Numerical Examples
Example 1.
A person having long sight cannot see things distinctly at a distance less than 40 cm. If he wants to see things situated at 25 cm from him, what should be the power of his spectacles?
Solution:
Let the focal length of the lens of the spectacles = f.
Here the power of the spectacles will be such that if an object is situated at a distance of 25 cm its virtual image will be formed at a distance of 40 cm from the eye i.e., u = 25 cm; v = 40 cm ; both u and v are negative.
From the equation \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) we have,
\(\frac{1}{-40}-\frac{1}{-25}\) = \(\frac{1}{f}\) or, \(\frac{1}{f}\frac{8-5}{200}\) = \(\frac{3}{200}\) or, f = \(\frac{200}{3}\) cm
∴ Power of the lens of the spectacles,
P = \(\frac{100}{f}\) = 100 × \(\frac{3}{200}\) = 1.5 m-1 = 1.5 D
[Short solution: distance of near point, d = 40cm
∴ Power of the lens, P = 4 – \(\frac{100}{40}\) = 4 – 2.5 = 1.5 m-1
As power is positive, the lens is a convex lens]
Example 2.
A short-sighted person can see distinctly the object sit-uated at a distance of 20 cm from him. What type of lens will he use to see the objects situated at a distance of 100 cm from him? What will be the power of the lens?
Solution:
In this case a lens is to be used to form the image of the objects situated at a distance of 100 cm at 20 cm from the eyei.e., u = -100 cm; D = -20 cm; u and v are both negative.
From the equation \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) we have,
\(\frac{1}{-20}\) – \(\frac{1}{-100}\) = \(\frac{1}{f}\) or, \(\frac{1}{f}\) = \(\frac{-5+1}{100}\) = –\(\frac{1}{25}\)
or, f = -25 cm
So, a concave lens of focal length 25 cm is to be used.
Power of the lens, P = \(\frac{100}{f}\) = \(\frac{100}{-25}\) = -4 D
[Short solution: distance of far point, d = 20 cm
After using lens distance of far point, d’ = 100 cm
∴ Power of lens,
P = \(\frac{100}{d^{\prime}}\) – \(\frac{100}{d}\) = \(\frac{100}{100}\) – \(\frac{100}{20}\) = -4D
Since power is negative, the lens is a concave lens]
Example 3.
A short-sighted person can read a book only up to 15 cm from his eyes. To read a book placed at a distance of 25 cm from him what type of spectacles should he use? What will be the power of the spectacles?
Solution:
In this case a lens is to be used which will form the image of the objects situated at a distance of 25 cm at 15 cm from the man. i.e., u = 25 cm, v = 15 cm, u and v are both positive.
From the equation \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) we have,
\(\frac{1}{15}\) – \(\frac{1}{25}\) = \(\frac{1}{f}\) or, \(\frac{1}{f}\) = \(\frac{5-3}{75}\) = \(\frac{2}{75}\) or, f = \(\frac{75}{2}\) = 37.5 cm
So the lens to be used is a concave lens of focal length 37.5 cm.
Power of the lens,
P = \(\frac{-100}{f}\) = \(\frac{-100}{75}\) × 2 = \(\frac{-8}{3}\) = -2.67 dioptre
Example 4.
A person can see distinctly up to a distance of 2 m and no further. To see distinctly up to a long distance what type of spectacles should he use? What will be the power of the lens of the spectacles?
Solution:
As the person can see distinctly up to a distance of 2 m, the defect of the eye is short sight, so to see distinctly up to a long distance he will have to use concave lens. The focal length of the lens will be such that the image of the objects at infinity will be formed at a distance of 2 m from the eye i.e., u = ∞, v = – 2 m = – 200 cm (negative).
Suppose the focal length of the lens = f.
From the equation \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) we have,
\(\frac{1}{-200}\) – \(\frac{1}{\infty}\) = \(\frac{1}{f}\) or, f = -200 cm
∴ Power of the lens,
P = \(\frac{100}{f}\) = \(\frac{100}{-200}\) = -0.5 m-1
Example 5.
A person with spectacles of power 3 m-1 can see dis-tinctly the letters of newspaper placed at a distance of 25 cm from the eye. At what distance should the news-paper be kept to be able to read it without spectacles?
Solution:
The power of the lens of the spectacles = 3 m-1. If f be the focal length of the lens, then,
3 = \(\frac{100}{f}\) or, f = \(\frac{100}{3}\)cm
Again u = -25 cm and f = \(\frac{100}{3}\)cm
So from the equation \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) we have, \(\frac{1}{v}\) + \(\frac{1}{25}\) = \(\frac{3}{100}\) or, \(\frac{1}{v}\) = \(\frac{3}{100}\) – \(\frac{1}{25}\) or, v = -100 cm
So to read the newspaper without spectacles the person will have to place it at a distance of 100 cm from the eye.
Example 6.
A person can see distinctly any object situated in between the distance 50 cm and 300 cm. What type of spectacles are to be used
(i) to extend the far point up to infinity and
(ii) to bring the least distance of distinct vision at 25 cm ? What will be the range of distinct vision in each pair of spectacles?
Solution:
i) To extend the far point from 300 cm to infinity spectacles should have a concave lens. Since the far point of the defective eye is at 300 cm, so the focal length of the concave lens of the spectacles will be 300 cm. Because the parallel rays coming from infinity after refraction by the concave lens appear to diverge from a point at a distance 300 cm from the lens.
With these spectacles, the near point of the range of vision will be that object distance for which the image distance is 50 cm.
Here, v = -50 cm and f = -300 cm.
From the equation \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) we have
\(-\frac{1}{50}\) – \(\frac{1}{u}\) = \(-\frac{1}{300}\) or, \(-\frac{1}{u}\) = \(\frac{1}{50}\) – \(\frac{1}{300}\) = \(\frac{1}{60}\)
So with this pair of spectacles, the range of vision will be from 60 cm up to infinity.
ii) To bring the near point to 25 cm from 50 cm spectacles with convex lens are to be used. Here, u = -25 cm and v = -50 cm .
So from the equation \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) we have,
\(\frac{1}{f}\) = –\(\frac{1}{50}\) – \(\frac{1}{-25}\) = \(\frac{1}{50}\) or, f = 50 cm
So a convex lens of focal length 50 cm is to be used.
To find out the far point v = -300 cm, f = 50 cm, u = ?
∴ \(\frac{1}{-300}\) – \(\frac{1}{u}\) = \(\frac{1}{50}\) or, \(\frac{1}{u}\) = \(\frac{1}{-300}\) – \(\frac{1}{50}\) = –\(\frac{7}{300}\)
In this case the range of vision of the man extends from 25 cm to 42.86 cm .
Example 7.
A person using spectacles having power + 2.5 m-1 can see the objects distinctly at a distance of 25 cm. What is the near point for the person? What type of defect of vision does the eye have?
Solution:
Power of spectacles, P = + 2.5 m-1. If f be the focal length of the lens of the spectacles, then
P = \(\frac{100}{f}\) or, 2.5 = \(\frac{100}{f}\) or, f = \(\frac{100}{2.5}\) = 40cm
Suppose, the distance of the near point of the defective eye = x cm
The lens will form the image of the object situated at a distance of 25 cm at a distance x cm.
Here, u = -25 cm; v = -x cm and f = 40 cm.
From the equation \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) we have,
\(\frac{1}{-x}\) – \(\frac{1}{-25}\) = \(\frac{1}{40}\) or, \(\frac{1}{x}\) = \(\frac{1}{25}\) – \(\frac{1}{40}\) = \(\frac{3}{200}\)
This is the distance of the near point for the person i.e., the near point has been shifted away from normal distance (25 cm). So the defect of the eye is long sight.
Example 8.
A person with defective eyes can see the objects dis-tinctly up to the distance 20 cm. What type of lens should be used and of what power?
Solution:
The far point of the defective eyes is 20 cm i.e., under this condition the focal length of the convex lens of the eye, f1 = 20 cm . An external lens of focal length f2 is to be used with the eye lens to shift the far point to infinity. The meaning of the far point at infinity is that the focal length of the combination of eye lens and external lens, F = ∞ we have, Since f2 is negative, so the external lens is a concave one. Its power,
P2 = \(\frac{100}{f_2}\) = –\(\frac{100}{20}\) = -5 m-1
Example 9.
A boy can clearly see objects between distance 15 cm to 200 cm from his eye. To clearly see an object situated at infinity, what will be the power of the lens that should he use? If he wears that lens then what will be the least distance of distinct vision in that case?
Solution:
To see an object placed at infinity, focal length of the lens should be such that parallel rays coming from infinity seem to be coming from a point at a distance 200 cm.
∴ \(\frac{1}{-200}\) – \(\frac{1}{\infty}\) = \(\frac{1}{f}\) or, f = -200 cm
As f is negative, the lens is concave.
∴ Power of the lens, P = \(\frac{100}{f}\) = \(\frac{100}{-200}\) = -o.5 m-1
Let, while he is wearing spectacles,, the least distance of distinct vision = u.
From the equation \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\), we get v u f
\(\frac{1}{-15}\) – \(\frac{1}{u}\) = \(\frac{1}{-200}\)
or \(\frac{1}{u}\) = –\(\frac{1}{15}\) + \(\frac{1}{200}\) = \(\frac{-40+3}{600}\) = –\(\frac{37}{600}\)
or, u = –\(\frac{600}{37}\) = -16.22 cm
Hence, the required distance is 16.22cm.
Example 10.
A long-sighted man can clearly see at any distance beyond 2.5 m. What kind of lens in his spectacles does require to read books placed 25 cm from his eyes? [HS (XI) ’10]
Solution:
Let the focal length of the spectacles be f.
Now, if an object is placed at 25 cm from the eye then virtual image will be created at a distance 250 cm from the eye.
Hence, u = -25 cm, v = -2.5 m = -250 cm
From the equation \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\), we get
\(\frac{1}{-250}\) – \(\frac{1}{-25}\) = \(\frac{1}{f}\)
or, f = \(\frac{250}{9}\) = 27.78 cm
So that person should use convex lens of focal length 27.78 cm.