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
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