make a diagram to show how hypermetropia is corrected. the near piont of a hypermetropic eye is 1 m.what is the power of Lena required to correct this defect .assume that the near point of the normal eye is 25 cm?
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Hypermetropia is a condition in which a person cannot see a nearby object but can clearly see an object at a distance.
It is referred to as far-sightedness.
A convex lens is used to overcome the problem and allow a person to look appropriately.
Object distance, u = -25 cm
Virtual image distance, v = -100 cm
Using lens formula;
1/f = 1/v – 1/u = 1/(-100) – 1/(-25)
1/f = -1/100 + 1/25 = 3/100
Hence, f = 100/3 cm = 1/3 m
Therefore, Power of lens,
P = 1/f = 3/1 = +3D
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