A normal eye has its retina positioned 2 cm behind the eye-lens. Determine the power of the eye-lens when the eye is most strained.

Step-by-step solution

The image distance for the eye-lens is the distance to the retina, so v = 2 cm = +0.02 m . When the eye is most strained, it focuses on an object placed at the near point. For a normal eye, the near point is at 25 cm . The object distance is u = -25 cm = -0.25 m . Using the lens formula, the power P of the eye-lens is given by: P = 1 f = 1 v - 1 u P = 1 0.02 - 1 -0.25 P = 50 + 4 = 54 D