In an eye where the lens-to-retina distance is fixed at 2.00 cm, an object sits 50.0 cm away. Using P equals 1 over d_i plus 1 over d_o, what power must the eye supply to focus this object onto the retina?

1. Since P = 1/f and the thin-lens equation gives 1/f = 1/d_i + 1/d_o, power is simply P = 1/d_i + 1/d_o. 2. Converting to metres: d_i = 0.0200 m and d_o = 0.500 m. 3. Substituting values: P = 1/0.0200 + 1/0.500 = 50.0 + 2.00. 4. Adding these gives P = 52.0 D. 5. 50.0 D alone only accounts for the retina distance term and leaves out the object distance term entirely. 6. 2.00 D and 0.52 D both undercount the eye's actual power by a wide margin, missing the dominant 1/d_i term. _Source: OpenStax Physics (CC BY 4.0), Ch 16 "Mirrors and Lenses", section 16.3 Lenses_