The far point of a myopia eye is 10 cm from the eye. The focal length of a lens for reading at normal distance

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-8.35cm

-16.7cm

-32.7cm

-35.4cm

answer is B.

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Detailed Solution

$\frac{1}{f} = \frac{1}{v} = \frac{1}{u} \begin{array}{l} = \frac{1}{- 10} - \frac{1}{(- 25)} \\ = \frac{1}{- 10} + \frac{1}{25} \\ \frac{1}{f} = \frac{- 5 + 2}{50} = \frac{- 3}{50} \\ f = \frac{- 50}{3} = - 16 .7 cm \end{array}$

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