Physics, asked by devula9595, 11 months ago

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).

Answers

Answered by Anonymous
8

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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