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A person wears eye glasses with a power of `-5.5 D` for distance viewing. His doctor prescribes a correction of `+ 1.5 D` for his near vision. What is the focal length of his distance viewing part of the lens and also for near vision section of the lens ? |
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Answer» (i) For distance viewing, `f = ?, P = -5.5 D` Clearly, `f = (100)/(P) = (100)/(-5.5) = -18.2 cm` (ii) For near vision correction `P = + 1.5 D` Therefore, `f = (100)/(P) = (100)/(1.5) = 66.7 cm`. |
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