Power of the corrective lens that should be used to account for a hypermetropic eye having a near point of 40 cm is:

Options

A. `+1.5 D`
B. `-1.5 D`
C. `+2.5 D`
D. `+0.5 D`

Answer

C

Explanation

Hypermetropia is corrected by using convex lens. Focal length of lens used f = + (defected near point)
f = +d= +40 cm
`therefore` Power of lens `100/(f(cm)) = 100/(+40) = +2.5 D`