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Exercise · Q23

Q.A person suffering from hypermetropia cannot focus on objects closer than 1 m1\ \text{m} (his near point has shifted to 1 m1\ \text{m}). Find the power of the spectacle lens that lets him read comfortably at the normal near point of 25 cm25\ \text{cm}.

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To let the hypermetropic person read comfortably at the NORMAL near point of 25 cm25\ \text{cm}, the corrective lens must take an object actually held at u=−0.25 mu=-0.25\ \text{m} and form a virtual image of it at the person's own (farther, defective) near point, v=−1 mv=-1\ \text{m} — the closest distance that person's own eye can still focus on unaided. Using the lens formula, 1f=1v−1u=1−1−1−0.25=−1−(−4)=−1+4=3 m−1\dfrac1f=\dfrac1v-\dfrac1u=\dfrac1{-1}-\dfrac1{-0.25}=-1-(-4)=-1+4=3\ \text{m}^{-1}, so $f=1/3\ \text{m}\approx0.333\ \text …

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