Question

A person needs a lens of power, −5.5 D for correcting his distant vision. For correcting his near vision, he needs a lens of power, + 1.5 D. What is the focal length of the lens required for correcting:
(a) distant vision?
(b) near vision?
​What type of lens is being used for correcting (i) distant vision, and (ii) near vision?

Answer 1

(a) According to relation, $\mathrm{Power}=\frac{1}{f(\mathrm{in}\mathrm{meter})}$

For distant vision, P = $-$5.5 D

Focal length, $f_{\mathrm{distant}}=\frac{1}{-5.5}\mathrm{m}=-\frac{100}{5.5}\mathrm{cm}=18.18\mathrm{cm}$

(b) According to relation, $\mathrm{Power}=\frac{1}{f(\mathrm{in}\mathrm{meter})}$

For near vision, P = +1.5 D

Focal length, $f_{\mathrm{near}}=\frac{1}{+1.5}\mathrm{m}=\frac{100}{1.5}\mathrm{cm}=66.66\mathrm{cm}$

For distant vision, the person is using a lens of negative power, so it is a concave lens. For near vision, the person is using a lens with positive power, so it is a convex lens.