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Doctor has prescribed a lens having power +1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
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Given: Power of lens, = +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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