A person with defective eyes can see the objects distinctly up to the distance 20 cm . What type of lens should be used and of what power?

Explanation

The far point of the defective eyes is 20 cm i.e under this condition the focal length of the convex lens of the eye, `f_1=20cm` . An external lens of focal length `f_2` is to be used with the eye lens to shift the far point to infinity. The meaning of the far point at infinity is that the focal length of the combination of eye lens and external lens, F `=oo` we have,
`1/F=1/(f_1)+1/(f_2) or, 1/(oo)=1/(20)+1/(f_2) or, f_2=-20 cm `
Since `f_2` is negative, so the external lens is a concave one. Its power , `P_2=(100)/(f_2) =-(100)/(20)=-5m ^(-1)`