The defective eye of a person has near point 0.5 m and far point 3m .The power for corrective lens required for
i . reading purpose
ii. seeing distant objects ,respectively are

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$+ 2 D and - \frac{1}{3} D$

$0.5 D and - 3.0 D$

$0.5 D and + 3 D$

$- 2 D and + \frac{1}{3} D$

answer is B.

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Detailed Solution

For reading purpose :
$u = - 25 cm, v = 0.5 m = - 50 cm$
$f = ?, p = ?$
$f = \frac{1}{v} - \frac{1}{u} = \frac{1}{- 50} - \frac{1}{- 25} = \frac{1}{50}$
$∴ p = \frac{100}{f (cm)} = 100 \times \frac{1}{50} = + 2 D$
For distant objects
$u = α, v = - 3 m$
$f = ?, p = ?$
$f = \frac{1}{v} - \frac{1}{u} = \frac{1}{- 3} - \frac{1}{- α} = \frac{- 1}{3}$
$∴ p = \frac{1}{f (m)} = \frac{- 1}{3} D$