$(a)\;15\;cm \\ (b)\;30\;cm \\ (c)\;\frac{100}{15}\;cm \\ (d)\;\frac{10}{3} cm $

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Focal length of lens required for the man is

$u= \infty \qquad v= -30\;cm$

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

$\large\frac{1}{f} =\frac{1}{-30}-\frac{1}{\infty}$

$f= -30 \;cm$

If he is wearing the lens,his near point is

$\large\frac{1}{-30}=\frac{1}{-15} -\frac{1}{u}$

$u= 30 cm$

Hence b is the correct answer.

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