The near point and the far point of a child are at 10 cm and 100 cm , respectively. If the retina is located 2.0 cm behind the eye-lens, determine the range of the power of the eye-lens.

Step-by-step solution

Using the lens formula, the power of the eye-lens is given by P = 1 f = 1 v - 1 u , where v and u are in meters. The image distance is fixed at the retina, so v = +2.0 cm = +0.02 m . For the far point, the object distance is u = -100 cm = -1.0 m . The minimum power of the eye-lens is: P_ min = 1 0.02 - 1 -1.0 = 50 + 1 = +51 D For the near point, the object distance is u = -10 cm = -0.1 m . The maximum power of the eye-lens is: P_ max = 1 0.02 - 1 -0.1 = 50 + 10 = +60 D The range of the power of the eye-lens is +51