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A far sighted person has his near point $50 \mathrm{~cm}$, find the power of lens he should use to see at $25 \mathrm{~cm}$, clearly.
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1649 Upvotes
Verified Answer
The correct answer is:
$+2 \mathrm{D}$
Here $u=-25 \mathrm{~cm}, v=-50 \mathrm{~cm}$
We have $\frac{1}{f}=\frac{1}{-u}+\frac{1}{v}$
i.e, $\frac{1}{f}=\frac{1}{25}-\frac{1}{50} \quad$ or $f=50 \mathrm{~cm}$
Power of lens he should use,
$$
P=\frac{100}{f}=\frac{100}{50}=+2 \mathrm{D}
$$
We have $\frac{1}{f}=\frac{1}{-u}+\frac{1}{v}$
i.e, $\frac{1}{f}=\frac{1}{25}-\frac{1}{50} \quad$ or $f=50 \mathrm{~cm}$
Power of lens he should use,
$$
P=\frac{100}{f}=\frac{100}{50}=+2 \mathrm{D}
$$
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