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Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropia 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.

Answer :

It is corrected by using convex lens of suitable focal length.

Near point hypermetropia eye v = -1m = -100cm

Object distance u = -25cm
According to formula, \(\frac{1}{v} – \frac{1}{u} = \frac{1}{f} \)

Near point of hypermetropia eye:-
\(\frac{1}{f} = - \frac{1}{100} – \frac{1}{-25} = \frac{-1}{100} + \frac{1}{25} \)
\(\frac{1}{f} = \frac{(-1+4)}{(100)} = \frac{3}{100} \)
f = \(\frac{100}{3} \)

Near point of a hypermetropic eye

Hypermetropia eye:-
f = \(\frac{100}{3} \)
\( P = \frac{1}{f} =\frac{ (3 x 100)}{(100)} = 3D \)

A convex lens of power +3.0 D is required to correct the defect.

Correction of a hypermetropia eye

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