Physics, asked by nehaputhan, 10 months ago

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 of the normal eye is 25 cm.

Plz give the answer with explanation

Answers

Answered by SANJAY01010

0

4-1/Nearpoint = 1/f= power

4-1/1= +3 dioptre

Answered by Anonymous

1

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