Question

A man can see the objects upto a distance of one metre from his eyes. For correcting his eye sight so that he can see an object at infinity, he requires a lens whose power is

or A man can see upto 100 cm of the distant object. The power of the lens required to see far objects will be

• A. +0.5 D
• B. +1.0 D
• C. +2.0 D
• D. –1.0 D

Answer 1

Answer: (D)

–1.0 D

Answer 2

f = –(defected far point) = – 100 cm. So power of the lens $P = \frac { 100 } { f } = \frac { 100 } { - 100 } = - 1 D$