(13%) Problem 5: A person has a near point of NP-55 cm and a far point of FP = 1

Question

(13%) Problem 5: A person has a near point of NP-55 cm and a far point of FP = 195 cm. The person wishes to obtain a pair of bifocal eyeglasses to correct these vision problems. The glasses will sit a distance d= 1.9 crn 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 person to see distant objects clearly Grade Summary Pi=11 Attempts remaining: 5 FP 4 5 6 (4% per attempt) detailed view 0 NP Feedback: 0% deduction per feedback. the power, Pi, 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. t (d) Calculate the power, P2, of the lower portion of the bifocals in units of diopters. Use N 25 em, which is for normal human vision

Explanation / Answer

(a)For far point

p=-(195-1.9)=-193.1cm

q=infinity

1/p +1/q=1/f

power =100/f=100/p where p is in cm

(b) P=-0.52D

(c) for near point

p=25 cm

q=55-1.9=53.1 cm

Using

1/p +1/q=1/f

f=pq/(p+q)

P=100/f=100(p+q)/pq where p,q are in cm

(d) P=5.9D

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