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

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

Tardigrade Admin · Accepted Answer

Hypermetropia can be removed by using a convex lens Focal length of used lens f=+d=+40 m = + (defected near point) ∴ Power of lens =(100/f(cm)) =(100/40) =2.5 D

Q. The nearer point of hypermetropic eye is $40 c m$ . The lens to be used for its correction should have the power

+ 1.5 D

-1.5 D

+2.5 D

+0.5 D

Hypermetropia can be removed by using a convex lens Focal length of used lens
$f = + d = + 40 m$ = + (defected near point)
$∴$ Power of lens $= \frac{100}{f ( c m )}$
$= \frac{100}{40} = 2.5 D$

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