A man can see only between 75 , cm and 100 , | vedclass

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(A) To correct the near point, we need to place an object at the standard near point $(u = -25 \, cm)$ such that its virtual image is formed at the ma

Vedclass · Accepted Answer

$+ \frac{8}{3} \, D$

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(A) To correct the near point, we need to place an object at the standard near point $(u = -25 \, cm)$ such that its virtual image is formed at the man's actual near point $(v = -75 \, cm)$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} = \frac{-1 + 3}{75} = \frac{2}{75} \, cm^{-1}$.Thus, $f = \frac{75}{2} \, cm = 0.375 \, m$.The power of the lens is given by $P = \frac{1}{f(	ext{in meters})} = \frac{1}{0.375} = \frac{100}{37.5} = +\frac{8}{3} \, D$.

$A$ man can see only between $75 \, cm$ and $100 \, cm$. The power of the lens required to correct the near point will be:

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