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

Explanation

For diagram,
Here, `x' = 1m = 100 cm, d = 25 cm, f = ?`
From `f = (x' d)/(x' - d), f = (100 xx 25)/(100 - 25) = 33.3 cm`
`P = (100)/(f) = (100)/(33.3) = 3 D`.
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