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The least distance of vision of a longsighted 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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$+(20 / 3) \mathrm{D}$
Using, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Here, $v=-60 \mathrm{~cm}, u=-12 \mathrm{~cm}$.
$\begin{gathered}\therefore \quad \frac{1}{f}=\frac{1}{(-60)}-\frac{1}{(-12)}=\frac{-1}{60}+\frac{1}{12}=\frac{1}{15} \\ \frac{1}{f}=\frac{1}{15} \Rightarrow f=15 \mathrm{~cm}\end{gathered}$
Power, $P=\frac{100}{15}=\frac{20}{3} \mathrm{D}$
Here, $v=-60 \mathrm{~cm}, u=-12 \mathrm{~cm}$.
$\begin{gathered}\therefore \quad \frac{1}{f}=\frac{1}{(-60)}-\frac{1}{(-12)}=\frac{-1}{60}+\frac{1}{12}=\frac{1}{15} \\ \frac{1}{f}=\frac{1}{15} \Rightarrow f=15 \mathrm{~cm}\end{gathered}$
Power, $P=\frac{100}{15}=\frac{20}{3} \mathrm{D}$
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