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What is the near point of a person whose eyes have an accommodated power of 53.5 D?
Question by OpenStax is licensed under CC BY 4.0.
28.6 cm
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OpenStax College Physics for AP® Courses Solution, Chapter 26, Problem 7 (Problems & Exercises) (1:08)

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Video Transcript
This is College Physics Answers with Shaun Dychko. The near point means the closest object distance that someone can see something clearly at. And so, the accommodative power of the eye has a formula which is reciprocal of image distance plus reciprocal of object distance. And, it's this object distance we want to solve for. The image distance we know is two centimeters, that's the distance between the lens and the retina. So, we'll solve for one over Do by subtracting one over Di from both sides. And then switch the sides around. And, this gives us one over Do equals power minus one over image distance. And then, raise both sides to the exponent negative one, and then on the left side we flip the fraction and we get object distance. But on the right side, we'll just leave it as bracket to the negative one. So, that's 53 and a half diopters minus one over two centimeters, which is two times ten to the minus two meters, and all to the power of negative one, giving 0.2857 meters, which is 28.6 centimeters.