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# Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Long-sightedness, or hypermetropia, is a defect of vision because of which a person cannot see nearby objects clearly but has normal distant vision.

The two causes of long-sightedness (or hypermetropia) are as follows:

1. The low converging power of the eye lens (because of its large focal length), which causes the formation of the images of objects behind the retina, or

2. The eye ball is too short, which causes light to focus beyond the retina, instead of directly on the retina.

(i) The ray diagram showing the eye-defect long-sightedness.

(ii) The ray diagram showing the correction of long-sightedness by using a lens.

The near point of the hypermetropic person is 1 m (the person can see objects placed at the normal near point of 25 cm in front of the eye if the images of the objects are formed at the person\'s own near point of 1 m from the eye).

Object distance, $u=-25\ cm$

Image distance, $v=-100\ cm$

From lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{-100}-\frac{1}{-25}=\frac{1}{f}$

Or $\frac{3}{100}=\frac{1}{f}$

Or $f=\frac{100}{3}\ cm$

Or $f=\frac{1}{3}\ m$

Power, $P=\frac{1}{focal-length(f)}$

$=\frac{1}{\frac{1}{3}}$

$=+3\ D$

Thus, the power of the convex lens required to rectify the defect is +3 dioptre.

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