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Myopia, Hypermetropia, and Presbyopia
(Study stuff for grade 10 )
Our this post Myopia, Hypermetropia, Presbyopia is dedicated to developing the knowledge base of 10-grade students or other students who have this topic in their science book. Myopia is the defect of the eye due to which a person is unable to see the distant object, it is also called nearsightedness because of the image of the distant objects formed near of the eye lens i.e between the eye lens and retina. Hypermetropia is the defect of the eye due to which a person is unable to see the nearby objects, it is also known as farsightedness because the image of the object is formed far from the eye lens behind the retina. The presbyopia is the defect due to which the person is able to see neither far object nor near object due to aging eyes.
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Myopia
Myopia is also called nearsightedness, it is the defect of the human eye due to which we can’t see the distant object, we can see the nearby object but after a certain distance, we can not see the objects and the so formed image looks blurred,it happens because of elongation of the eye ball which causes the eye lense to decrease focal length due to which image of distant object forms between eye lens and rating(i.e near to eye lens).
Generally, myopia is occurred during childhood due to the growth of eyes within the age of 20 years but it may occur to anybody due to our habit of seeing television by sitting near about it, excess reading or improper way of reading, diabetes, etc. To understand myopia first it is better to understand the functioning of the eyes in seeing the objects, one of the functions of the eyes is the accommodation of the eyes.
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Accommodation of the eyes
The human eye is just like a camera, its lens system forms an image on a light-sensitive screen called the retina. Our eye has the ability to see the near and distant objects due to the flexible capability of the crystalline lens behind the cornea, the crystalline lens is composed of fibrous jelly-like material called ciliary muscles when we see the distant object these muscles relax resulting the eye lens becomes thin means the radius of curvature of the eye lens becomes larger. The larger radius of the curvature results in a larger focal length of the lens focusing the image of the object at the retina.
When we see the nearer object the ciliary muscle contracts reducing the radius of the curvature of lens thereby decreases the focal length of the lens and thus focusing the image of the object at the retina, thus a healthy eye is capable to see near and far objects. The property of the eyes of adjusting its focal length to see the nearby and distant object is known as accommodation of the eye.
Myopia is caused by bulging cornea due to which the radius of the curvature of the whole of the eye decreases and the focal length of the eye is decreased, the eye becomes unable to focus the image of distant objects at the retina and thus image is formed near the eye lens, it is that’s why this defect of the eye is called nearsightedness or myopia, as a result of this defect of eye( myopia) the far objects look blurred. The maximum distance at which an object could be seen by a myopic eye is known as its far point beyond this far point image is formed near the eye lens instead of at the retina.
Treatment of myopic eye- The myopic eye is treated by applying a concave lens of a particular focal length, depends on the far point, as an example shown below.
Example- If the far point of an eye of a person is 100 cm, then find the type of lens and its focal length used for the treatment of clearing his vision.
Answer. The far point of the person’s eye 100 cm reveals that he can not see the things beyond 100 cm, so he is needed a diverging lens(concave ) of focal length f so that the image could be formed at the far point of the eye as seen in the following diagram.
1/f = 1/v – 1/u
f = ?, v = –100 cm, u =
1/f = 1/(-100) – 1/(-∝)
1/f = -1/100 +0
1/f = -1/100
f = –100 cm
Hence the power of the lens will be
P(in diopter) = 1/f(in meter)
Where P is the Power of the lens
P = 1/(-1) = -1D
Power of the lens = –1 D
Its meaning is the lense required for such an eye is –1 D , after applying this lens the person could see beyond the distance of 100 cm.
Hypermetropia
This is the defect of an eye due to which a person is capable to see the distant objects but is unable to see the nearby object. Hypermetropia occurred when the ciliary muscles of the Christlinine lens unable to contract and thus focal length of the eye lens increase that image of the nearby objects is formed beyond the retina that is far from the eye lens that’s why hypermetropia is also called farsightedness.
Example- An eye ‘s near point is 50 cm, find the treatment of this eye to correct vision.
Answer. Near point 50 cm means that the eye is unable to see the object nearer to 50 cm, thus the given eye is suffering from hypermetropia.To overcome this problem a convex lens of focal length f must be used so that if now the distance is 25 from the lens then the image could be formed at the near point of the eye.
u =– 25 cm
v = – 50 cm
1/f = 1/v – 1/u
1/f = 1/(-50) – 1/(-25)
1/f = -1/50 + 1/25
1/f = (-1+2)/50
1/f = 1/50
f = 50
The focal length of the convex lens is 50 cm
50 cm = 0.5 m
P(in diopter) = 1/f(in meter)
Where P is the Power of the lens
P= 1/0.5= +2D
Presbyopia
Presbyopia is caused due to aging, it occurs because the ciliary muscles of the Crestline lens become unable to contract and relax, as a result of this, the human eye is unable to see the object near and far object, this is the defect called presbyopia.
The treatment of such a defective eye is to use a bifocal lens a concave lens and a convex lens such that the lower end of the eyeglass is convex through which the patient of presbyopia can see nearby objects and the upper part of the glass is the concave lens through which he can see the distant object.
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