1.

The near point of eye suffering fromhypermetropia is 1 m. Calculate the Power that is needed to correct this eye defect. In this case we are assuming the near point as 25 cm.

Answer»

Solution :Here we KNOW the VALUES of u and v .PUT these values in the fonnula given above and final f i.e FOCAL length
Value of u=-1m =-100cm
Value of u=-1m=-100cm
Value of -25cm
USE the lens formula
(1/v)-(1/u)=1/f
`1/-25-1/-100=1/f
4-1/100=3/100
f=100/3=33cm
=1/.33=3D
Power of lens=1/f=3D


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