OpenStax College Physics, Chapter 26, Problem 6 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. When a person's eyes are relaxed then they are viewing the far point of their eye and the far point is the object distance that is the maximum possible. So we know that the power when viewing the far point is 50.5 diopters and the image distance is 2.00 centimeters for the typical eye— that's the distance from the lens to the retina— and power is 1 over object distance plus 1 over image distance and our job is to solve for d o. So we'll subtract 1 over d i from both sides and we have 1 over d o then is equal to the power minus 1 over d i and then we raise both sides to the exponent negative 1 to solve for d o. So the object distance then which is the far point in this case because it's the power when the eye is relaxed is power minus 1 over d i all to the negative 1. So that's 50.5 diopters minus 1 over 2.00 times 10 to the minus 2 meters and then take that difference and then take the result to the exponent negative 1 and you get 2.00 meters. So this person is very near-sighted so they can't see very far only 2.00 meters at most, without glasses.