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The maximum distance a person having defect of short right can see clearly is 4 meters. The power of lens required to correct is

$(a)\;4 \; dioptre \\ (b)\;-4\;dioptre \\ (c)\;+0.25\; dioptre \\ (d)\;-0.25\; dioptre $

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An object at infinity must form a virtual image at $4 m$ due to a lens of focal length f.
$\therefore u=\infty \qquad v= -4 m$
Power $= \large\frac{1}{f} =\frac{1}{\infty}+\frac{1}{-4}$
Power $=-0.25\; diopter$
Hence d is the correct answer.
answered Jan 23, 2014 by meena.p
 

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