Change the chapter
Question
In a LASIK vision correction, the power of a patient’s eye is increased by 3.00 D. Assuming this produces normal close vision, what was the patient’s near point before the procedure?
Question by OpenStax is licensed under CC BY 4.0.
1.0 m
Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 26, Problem 9 (Problems & Exercises) (1:52)

Sign up to view this solution video!

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

OpenStax College Physics, Chapter 26, Problem 9 (PE) calculator screenshot 1
Video Transcript
This is College Physics Answers with Shaun Dychko. This LASIK patient's eye initially had some power that was then increased by three diopters to create normal power. So, I have the subscript N for normal and a subscript I for initial. And, we want to know what is the initial near point of this person. So, we'll substitute this formula in place of the power. So, we have one over the initial near point for object distance plus one over the image distance, which is the distance between the lens and the retina. And, that's being substituted for the initial power. And then, we add three diopters to that, and it equals one over the normal object distance, which we know is 25 centimeters plus one over the image distance, which again is the lens retina distance. It doesn't change. So, this does not get a subscript initial or normal because it's the same value in each case. The lens retina distance does not change and so, this can be subtracted from both sides and so that disappears. And then, we solve for one over initial object distance by subtracting three diopters from both sides. And, we end up with one over initial object distance is one over the normal object distance minus three diopters. Then, we raise both sides to the exponent negative one, and we take the reciprocal of the left to solve for the initial object distance and that equals one over the normal object distance minus three diopters all to the power of negative one. So, that's one over 25 centimeters written as times ten to the minus two meters minus three diopters to the negative one is 1.0 meters. And, so this is their initial near point before the LASIK surgery.