The least distance of distinct vision of a lo | vedclass

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(B) For a long-sighted person (hypermetropia),the person cannot see objects clearly at the normal near point $(u = -25 \ cm)$. However,the problem sta

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$+20/3 \ D$

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(B) For a long-sighted person (hypermetropia),the person cannot see objects clearly at the normal near point $(u = -25 \ cm)$. However,the problem states the person's current near point is $v = -60 \ cm$. We want to bring the near point to $u = -12 \ cm$ using a lens.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$Here,$v = -60 \ cm$ (image formed at the person's actual near point) and $u = -12 \ cm$ (object placed at the desired near point).$\frac{1}{f} = \frac{1}{-60} - \frac{1}{-12} = \frac{-1 + 5}{60} = \frac{4}{60} = \frac{1}{15} \ cm^{-1}$.Since $f$ is in $cm$,$f = 15 \ cm = 0.15 \ m$.The power of the lens $P = \frac{1}{f(m)} = \frac{1}{0.15} = \frac{100}{15} = +\frac{20}{3} \ D$.

The least distance of distinct vision of a long-sighted person is $60 \ cm$. By using a spectacle lens,this distance is reduced to $12 \ cm$. The power of the lens is

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