A far-sighted person has a near point at 100 cm. What must be the power of the correcting lens? -General Knowledge

A far-sighted person has a near point at 100 cm. What must be the power of the correcting lens? -General Knowledge
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Power of the lens =  To calculate focal length using the near point=100cm f=  Where, u= object size v= image size ∵ image is virtual, it is with a - sign. The near point of a normal eye=u=25 cm Using this, f=  = +33.3 cm=+  Also, Power=P= where f is in meters ⇒P=+ = + 3 D
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Last Answer : +3.0 D

A far-sighted person has a near point at 100 cm -General Knowledge

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Last Answer : a -0.8 D b -3.0 D c +0.8 D d +3.0 D Answer : D Solution : Given near point = 100 cm. ∴v=−100cmandu=−25cm∴v=-100cmandu=-25cm 1f=1v−1u=1−100−1&minus ... =1−100+125=−1+4100=3100cm.1f=1-100+125=-1+4100=3100cm. Power, P=1for,100fD=100×3100D=3D

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