The near-point of a person suffering from hypermetropia is at 50 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is 25 cm).
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Answer:
Explanation:
u=-25 cm
v=-50 cm
p=1/f(meter)
therefore,1/f=1/50
therefore,f=50cm
therefore,p=1/0.5
=100/50=2D
therefore,power=2 D(dioptre)
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