(13%) Problem 5: A person has a near point of NP-75 cm and a far point of FP-185

Question

(13%) Problem 5: A person has a near point of NP-75 cm and a far point of FP-185 cm. The person wishes to obtain a pair of bifocal eyeglasses to correct these vision problems. The glasses will sit a distance d = 1.8 cm from the eyes 25% Part (a) Write a formula for the power, P1, of the upper portion of the bifocals, in terms of the given quantities, that will enable the n to see distant objects clearly Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: 5 % per attempt) detailed view 123 0 NP DELCLEAR Submit Hint Hints: 1 % deduction per hint. Hints remaining: Feedback: 0%-deduction per feedback. D 25% Part (b) Calculate the power, P1, of the upper portion of the bifocals in units of diopters - 25% Part (c) Write a formula for the power, P2, 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 D 25% Part (d) Calculate the power, P2, of the lower portion of the bifocals in units of diopters. Use N= 25 cm, which is for normal human vision

Explanation / Answer

given near point = 75 cm

Far point = 185 cm

d = 1.8 cm

a. the upper part makes images of infinity at far point

1/v - 1/u = 1/f

u = inf

v = -(FP - d)

-1/(FP-d) = 1/f = P/100

hence

P = -100/(FP-d)

b. P = -0.54585

c. the lower part makes image of 25 cm at near point of the person

u = -(25 - d)

v = -(NP - d)

hence

1/v - 1/u = 1/f

-1/(NP - d) + 1/(25 - d) = 1/f = P/100

P2 = 100(1/(25 - d) - 1/(NP - d))

P2 = 100(NP - 25)/(25 - d)(NP - d)

d. P2 = 2.944224

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