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A person hasfarsightedness with the minimum distance he couldsee early is 75 cm. Calculate the poer of the spectacts necessary to recify the defect. |
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Answer» Solution :Theminimum DISTANCE the person could see elarly is, y = 75 cm The lens should have a focal length of, `f = (y+25cm)/(y-25cm)` `f=(75cmxx25cm)/(75cm-25cm)=37.5cm` it is a CONVEX or CONVERGING lens. The POWER of the lens is, `P = (1)/(0.375m) = 2.67 DIOPTER` |
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