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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.
Refer Fig. 10.5
Here u = – 25cm, v = –1m = – 100 cm
Lens formula = 1/v - 1/u
= 1/ -f
1/(-100) - 1/(-25) = 1/f
⇒ 1/f = -1 + 4 / 100
⇒ f = 100/3
f = 100/3 x 1/100
f = 1/3 cm
Power of lens P = 1/f
P = 1/(1/3)
P = +3D
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