The near-point of a person suffering from hypermetropia is at 50 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is 25 cm).

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Question

The near-point of a person suffering from hypermetropia is at 50 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is 25 cm).

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The person needs a convex lens to rectify this defect.Calculation of power of the lens:
This hypermetropic eye can see the nearby object kept at 25 cm clearly if the image is formed at its own near point i.e. 50 cm.
Object distance, u = –25 cm
Image distance, v = –50 cm
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{- 50} - \frac{1}{- 25} = \frac{1}{f}$

f = + 50cm

P = \frac{100}{f (in cm)} = \frac{100}{50} = + 2 D