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The least distance of vision of a long sighted person is $60 \mathrm{cm}$. By using a spectacle lens, this distance is reduced to $12 \mathrm{cm}$. The power of the lens is
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$(20 / 3) \mathrm{D}$
Here, $v=-60 \mathrm{cm}, u=-12 \mathrm{cm}$
$\therefore$ By using the relation, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
We have. $\frac{1}{f}=\frac{1}{-60}-\frac{1}{-12}$
$\Rightarrow \quad \frac{1}{f}=\frac{1}{15} \mathrm{cm}$ or $\frac{100}{15} \mathrm{m}$
So, the power of lens, $P=\frac{100}{15}=+\frac{20}{3} \mathrm{D}$
$\therefore$ By using the relation, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
We have. $\frac{1}{f}=\frac{1}{-60}-\frac{1}{-12}$
$\Rightarrow \quad \frac{1}{f}=\frac{1}{15} \mathrm{cm}$ or $\frac{100}{15} \mathrm{m}$
So, the power of lens, $P=\frac{100}{15}=+\frac{20}{3} \mathrm{D}$
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