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Question
Q. The near point of a hypermetropic eye is 1m. Find the power of the lens required to correct this defect. Assume that near point of the normal eye is 25cm.
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Solution
For Hypermetropic eye,
v=-1m=-100cm
u=-25cm
Using lens formula,
1/f=1/v -1/u
=1/(-100) - 1/(-25)
=3/100
or f=100 /3 cm=1/3 m
Now,
Power,P=1/f=3 D
For Hypermetropic eye,
v=-1m=-100cm
u=-25cm
Using lens formula,
1/f=1/v -1/u
=1/(-100) - 1/(-25)
=3/100
or f=100 /3 cm=1/3 m
Now,
Power,P=1/f=3 D
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