A person uses a lens of power $+ 3 D$ to normalize vision. Find the near point of the hypermetropic eye (in m).

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Solution

Suppose near point of hypermetropic eye is $v .$

Then $u$ = minimum distance of distinct

Vision $= - D = - 25 c m$

Focal length of the lens is $f = \frac{100}{3} c m$ .
From lens formula

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

$\Rightarrow \frac{3}{100} = \frac{1}{v} - \frac{1}{- 25}$

$\Rightarrow v = - 100 c m$

Near point of the hypermetropic eye is 1 m.

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A person uses a lens of power +3D to normalize vision. Find the near point of the hypermetropic eye (in m).

Suppose near point of hypermetropic eye is v. Then u = minimum distance of distinct Vision =−D=−25cm Focal length of the lens is f=1003cm. From lens formula 1f=1v−1u ⇒3100=1v−1−25 ⇒v=−100cm Near point of the hypermetropic eye is 1 m.

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