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Given: Power of lens, P = +1.5 D, f=? Now, focal length of lens,\( f=\frac{1}{P} =\frac{1}{(+1.5D)} =\frac{10}{15} m \)=+0.67 m P is positive, This shows the lens is convex. Thus, the defect of vision is farsightedness or hypermetropia.
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