Question

The nearer point of hypermetropic eye is $40\, cm$. The lens to be used for its correction should have the power

• A. $+ 1.5 \,D$
• B. $- 1.5 \,D$
• C. $+ 2.5 \,D$
• D. $+ 0.5 \,D$

Answer 1

Answer: (C)

$+ 2.5 \,D$

Answer 2

Hypermetropia is corrected by using convex lens. Focal length of lens used $f = +$(defected near point) $f = +d = +40\,cm$ $\therefore$ Power of lens $= \frac{100}{f(cm)}$ $= \frac{100}{+40} = +2.5\,D$