Question

The near point of eye is 75 cm. Calculate the power of lens required to correct this defect.

Answer 1

Answer:

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Answer 2

Explanation:

v= -75 cm

u =-25cm

f= (u-v)/vu = (50)/75×25 cm = 2/75 cm

P = 100×75/2= 50×75D = 3750D