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The far point of a myopia eye is at `40 cm`. For removing this defect, the power of lens required will beA. `40D`B. `-4D`C. `-2.5 D`D. `0.25D` |
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Answer» Correct Answer - C For myopic eye `f=-`(defected far point ) `impliesf=-40cmimpliesP=(100)/(-40)=-2.5D` |
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