1.

A person hasfarsightedness with the minimum distance he couldsee early is 75 cm. Calculate the poer of the spectacts necessary to recify the defect.

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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