1.

For an eye, the defective near point is 150 cm. Calculate power of correcting convex lens to correct this vision defect.

Answer»

Solution :The FOLLOWING values are GIVEN
Hypermetropic Near point=150cm
THEREFORE image distance=v=-150
Object distance=normal near point=-25
(1/v)+(1/u)=1/f
(1/v)+(1/25)=1/f
(1/-150)+(1/u)=1/f
f=150/5cm=0.3
Power=1/f=3.3D (dioptre)
It means convex lens of power 3.3 DIOPTRES is required to correcct the VISION.


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