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Q.

The near point of hypermetropic person is 50 cm from the eye. The power of the lens required to enable the person to read clearly a book held at 25 cm from the eye is

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a

2 D

b

8 D

c

4 D

d

1 D

answer is A.

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Detailed Solution

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \begin{array}{l} = \frac{1}{- 50} - \frac{1}{(- 25)} \\ = \frac{- 1}{50} + \frac{1}{25} \\ = \frac{- 1 + 2}{50} = \frac{1}{50} \\ f = 50 cm \end{array} \begin{array}{l} P = \frac{100}{f} \\ = \frac{100}{50} \\ P = 2 D \end{array}$

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