Change the chapter
Question
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
$2.00\textrm{ m}$
Solution Video

# OpenStax College Physics for AP® Courses Solution, Chapter 26, Problem 6 (Problems & Exercises) (1:15)

Rating

No votes have been submitted yet.

Quiz Mode

Why is this button here? Quiz Mode is a chance to try solving the problem first on your own before viewing the solution. One of the following will probably happen:

1. You get the answer. Congratulations! It feels good! There might still be more to learn, and you might enjoy comparing your problem solving approach to the best practices demonstrated in the solution video.
2. You don't get the answer. This is OK! In fact it's awesome, despite the difficult feelings you might have about it. When you don't get the answer, your mind is ready for learning. Think about how much you really want the solution! Your mind will gobble it up when it sees it. Attempting the problem is like trying to assemble the pieces of a puzzle. If you don't get the answer, the gaps in the puzzle are questions that are ready and searching to be filled. This is an active process, where your mind is turned on - learning will happen!
If you wish to show the answer immediately without having to click "Reveal Answer", you may . Quiz Mode is disabled by default, but you can check the Enable Quiz Mode checkbox when editing your profile to re-enable it any time you want. College Physics Answers cares a lot about academic integrity. Quiz Mode is encouragement to use the solutions in a way that is most beneficial for your learning.

## Calculator Screenshots

Video Transcript
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.