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The lens-to-retina distance of a patient is 2.03 cm and the totally relaxed strength of his...

The lens-to-retina distance of a patient is 2.03 cm and the totally relaxed strength of his eye is 50.0 D. A)What is the most distant object he can see clearly? B)What spectacle lens strength will correct his distant vision? Please provide formula and step by step solution.

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Answer #1

1/f = 1 / di     + 1/do  

to find whether he is far sighted or near sighted,

1/f = 1/0.0203 + 0

f = 49.26 D which is < 50D, so the patient is near sighted

50 = 1/0.0203 + 1/do

do = 50- 49.26 = 74.9 cm

diverging lens are to be used

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