The near point of a hypermetropic eye is 75 cm from the eye. What is the power of the lens required to read 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

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

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Similar questions

The near point of a hypermetropic eye is 1 m, what is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is 25 cm)

Q. The near-point of a hypermetropic eye is

75 cm

from the eye. What is the power of the lens required to enable the person to read a book held at

25 cm

from the eye?

Q. An object is placed at

25 c m

in front of a convex lens. Its image is formed at

75 c m

from the lens on the same side as the object. Find the power of the lens?

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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 read a book held at 25 cm from the eye?

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