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?

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.