A person has a near point of NP = 85 cm and a far point of FP = 145 cm. The pers
ID: 1605672 • Letter: A
Question
A person has a near point of NP = 85 cm and a far point of FP = 145 cm. The person wishes to obtain a pair of bifocal eyeglasses to correct these vision problems. The glasses will sit a distance d = 2.3 cm from the eyes. Write a formula for the power, P_1, of the upper portion of the bifocals, in terms of the given quantities, that will enable the person to see distant objects clearly. Calculate the power, P_1, of the upper portion of the bifocals in units of diopters. Write a formula for the power, P_2, of the lower portion of the bifocals, in terms of given quantities, so that the person can clearly see objects that are located a distance N from his eyes. P_2 = (1/NP) + (1/(FP - (N - d) (d) Calculate the power, P_2, of the lower portion of the bifocals in units of diopters. Use N = 25 cm, which is for normal human Vision.Explanation / Answer
His near point is 85 cm and far point is 145 cm. We will use these numbers to calculate what the Powers of his bifocal lens would be and then indicate a small correction due to the lenses sitting 2.3 cm from the eye.
(1) 1/f = 1/s + 1/s' Or 1/f = 1/infinity + 1/(-1.45) = -1/1.45 . So f = -1.45 m and Power = -1/ 1.45 = -0.69 D
(2) 1/f = 1/0.25 + 1/(-0.85) = 4 - 1.17 = +2.83. So Power = 1/2.83 = +0.35 D
These will slightly change as shown below:
(1) 1/f will be -(1.45-0.023) =-1.427 and Power will be -1/1.427 = -0.70 D
(2) 1/f will be = 1/(0.25-0.023) + 1/(-0.85-0.023) = 4.40-1.29 and Power will be +1/2.11 = +0.47 D
D is the power in Dioptre.
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