Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Diagram:
Diagram A
Diagram B
In the above diagram, image is formed on the left side so the both image and the object should be taken in the minus (-).
Image distance (v), given = -1m = -100cm
Object distance (u), given = -25 cm
Formula of focal length on the basis of image and the object distance,
1/v – 1/ u = 1/ f
- 1/ 100 – 1/ -25 = 1/ f
-1/ 100 + 1/ 25 = 1/f
- 1 + 4 / 100 = 1/ f
3/ 100 = 1/f
F = 100/ 3 = 33.3 cm. = 0.33m.
And, power of the lens (P) = 1/f
Put the value of f in this formula,
P = 1/ 0.33
= + 3D
Hypermetropia is the eye defect also called far sightedness. A person who is having this defect can see the distant objects clearly but cannot see the near objects very clearly. This is because, light rays from the object focussed at the point behind the retina.
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