The near point of a hypermetropic eye is 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?

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2.66D

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3.8D

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Solution

The correct option is C
2.66D

Focal length of convex lens to be used can be found by
$(\frac{1}{v}) - (\frac{1}{u}) = \frac{1}{f} H e r e u = - 25 c m = - \frac{1}{4} m v = - 75 c m = \frac{- 3}{4} \frac{1}{f} = p o w e r = \frac{1}{\frac{- 3}{4}} - \frac{1}{\frac{- 1}{4}} P o w e r = \frac{1}{f} = \frac{8}{3} = 2.66 D$
So, the person needs a convex lens of power 2.66D.

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The near point of a hypermetropic eye is 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?

The correct option is C 2.66D Focal length of convex lens to be used can be found by (1v)−(1u)=1fHere u=−25 cm=−14mv=−75 cm=−341f=power=1−34−1−14Power=1f=83=2.66D So, the person needs a convex lens of power 2.66D.