Question

A person has near point of his vision shifted to 50 cm. What lens must be used to see an object placed at 25 cm from the eye? What is the power of the lens?

• A. A diverging lens of focal length 50 cm and power + 2 D
• B. A converging lens of focal lens 50 cm and power - 2 D
• C. A diverging lens of focal length 50 cm and power - 2 D
• D. A converging lens of focal length 50 cm and power + 2 D

Answer 1

The correct option is D

A converging lens of focal length 50 cm and power + 2 D

We need a lens to form a virtual image of an object at the near point of the eye. Thus, here u = - 25 cm and v = - 50 cm. Using the lens formula, we get
1/f = 1/v - 1/u
or, 1/f = (-1/50) - (-1/25) = 1/ 50
or, f = 50 cm = 0.5 m
P = 1/f (in m) = 1/0.5 = + 2 D

A converging lens, i.e., a convex lens of focal length 50 cm is required by the person to see an object placed at a distance of 25 cm from his eye. The power of this lens is +2D.