Practice question

**Human eye** least distance D=25 cm, near point, far point infinity for normal eye, accommodation by ciliary muscles changing lens curvature, power ≈60 D total. Defects: myopia far point <∞ corrected by diverging lens, hypermetropia near point >25 cm corrected by converging lens, astigmatism cylindrical lens. Focal length: f = (R/2) = (24/2) = 12 cm . Object distance: u = -8 cm . Mirror equation: (1/v) + (1/-8) = (1/12) ⇒ (1/v) = (1/12) + (1/8) = (2 + 3/24) = (5/24) . v = (24/5) = 4.8 cm (virtual image). Substituting values gives 4.8 cm, which matches expected image position and magnification from