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The near point of a hypermetropic person...

The near point of a hypermetropic person is at 75 cm from the eye. What is the power of the lens required to enable him to read clearly a book held at 25 cm from the eye ?

Text Solution

Verified by Experts

D = 25 cm
Distance of near point = x= 75 cm
This defect can be corrected by using a concave lens of focal length f which is given by
`f=(xD)/(x-D)`
`f=(75xx25)/(75-25)`
`=(75xx25)/(50)`
`f=37.5 cm = 0.375m`
Power of the corrective lens `=P=(1)/(f("in m"))`
`P= +2.67D`
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