(a) Suppose Willebrord Snell's near point of vision is 55cm from his unaided eye. What power, inDiopters, is needed to correct Willebrord's vision to the normal near point of 25cm if contactlenses are used?(b) If eyeglasses are used instead, and the distance from Willebrord's eye to the lens of theglasses is 2.1cm, what power in Diopters is needed for the glasses?(c) Willebrord also has difficulties focusing clearly on objects far away. If his unaided far point ofvision is 3.52m, what power corrective lens (in glasses) would help his vision for very distantobjects?(d) Describe a solution to his overall vision problem.

We will use the thin lens formula, 1v-1u = 1fa) given that:- willebrord snells' near point vision is 55 cm normal near point is 25 cm so, v = -55cmu= -25 cm using the thin lens formula 1v-1u = 1f 1f=1-55 - 1-251f =-25+551375 1f =6275 cm-1 now the power of the lens is , P = 100fP = 600275P = 2.18D b) if lens is placed at a distance 2.1 cm away from the eye then, the new image will formed at a distance, 55-2.1 = 52.9 cm and the normal vision will be at, 25 - 2..1 = 22.9 cm using these values in thin lens formula 1v-1u = 1f 1f=-152.9 + 122.9 as P = 100fP = 2.5 D c) using the formula, 1v-1u = 1f here v = -352 cm u = -∞ 1f = 1-352- 1-∞1f = -1352power of lens required P = -100352 = -0.28 Dif glass is placed at 2.1 cm away then P = -100352-2.1 = -0.28 Dd) if the person is struggling with his/her near vision and far vision then the person may suffering form presbyopia.

Answered: (a) Suppose Willebrord Snell's near…