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A student’s eyes, while reading the blackboard, have a power of 51.0 D. How far is the board from his eyes?
Question by OpenStax is licensed under CC BY 4.0.
$1.00\textrm{ m}$
Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 26, Problem 12 (Problems & Exercises) (0:56)

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Video Transcript
This is College Physics Answers with Shaun Dychko. When viewing a blackboard, the student has a power of 51.0 diopters in their eye and the image distance is the distance between the lens and the retina, which is typically 2.00 centimeters, or 2.00 times 10 to the minus 2 meters and the question is how far is the blackboard from the student and so that's the object distance in this case. So power is 1 over object distance plus 1 over image distance and we'll solve for 1 over d o by subtracting 1 over d i from both sides and we get 1 over object distance is power minus 1 over d i and then we raise both sides to the exponent negative 1, which on the left flips this fraction to solve for d o and on the right side, we write P minus 1 over d i all to the negative 1. So that's 51.0 diopters minus 1 over 2.00 times 10 to the minus 2 meters all to the negative 1 and that's 1.00 meters. So the distance between the student and the blackboard is 1.00 meter.