Change the chapter
Question
A boy has a near point of 50 cm and a far point of 500 cm. Will a -4.00 D lens correct his far point to infinity?
1. -0.263 m
2. A negative object distance means the object should be behind the eye, which doesn't make sense.
3. The lens over-corrects. The power should be much less.
Solution Video

# OpenStax College Physics for AP® Courses Solution, Chapter 26, Problem 25 (Problems & Exercises) (2:00) 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. A boy has a far point of 500 centimeters and the question is by reducing the power of their eyes by four diopters, will their far point then be infinity, which would be normal vision. So, let's figure out the power that they have initially. So, that's one over the image distance which is between their eyeball lens and the retina, which is two centimeters, and then add to that one over the initial object distance. And, we'll get an initial power uncorrected, which is one over two centimeters written as times ten to the minus two meters plus one over 500 centimeters, which is 50.2 diopters. Now, their final power then is going to be that initial power minus four diopters. That's 50.2 minus four, which is 46.2 diopters. And, let's figure out what the object distance will be with this power. And so, we have the same one over Di in this case, there's no need for a subscript F for a final here because this is the lens retina distance that doesn't change. But, there is a subscript F here on the object distance and a new final power. So, the object distance is going to be the final power minus one over the image distance all to the power of negative one because we'll subtract one over Di from both sides. And then, raise both sides to the exponent negative one and we get this line here. So, that's 46.2 diopters minus one over two centimeters all to the power of negative one, which is negative 0.263 meters. Now, this is nonsensical because with a negative number there, that means that the object should be behind the eyeball, which doesn't make sense. And so, this lens overcorrects and the power should be less.