Answer to Question 11

(a) Power of eye at near point = 1/3 + 1/0.02 = 0.33 + 50 = 50.33 D

(b) Power needed = 1/0.25 + 1/0.02 = 4 + 50 = 54 D

(c) Power of the corrective lens = Power of the corrected eye - power of the uncorrected eye

Power = 54 - 50.33 = 3.67 D  It's a convex lens as the result is positive.

(d) Focal length = 1/3.67 = 0.27 m