The near point and far point of a person are | vedclass

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(C) The person is suffering from hypermetropia (farsightedness) because the near point has shifted from the normal $25\,cm$ to $40\,cm$.To read a book

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$1.5$

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(C) The person is suffering from hypermetropia (farsightedness) because the near point has shifted from the normal $25\,cm$ to $40\,cm$.To read a book at $u = -25\,cm$,the lens must form a virtual image at the person's near point,$v = -40\,cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $P = \frac{1}{-0.4} - \frac{1}{-0.25}$.$P = -2.5 + 4.0 = 1.5\,D$.Thus,the required power of the lens is $1.5\,D$.

The near point and far point of a person are $40\,cm$ and $250\,cm$ respectively. Determine the power of the lens he/she should use while reading a book kept at a distance of $25\,cm$ from the eye $..........D$

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