Change the chapter
Question
A student’s eyes, while reading the blackboard, have a power of 51.0 D. How far is the board from his eyes?
$1.00\textrm{ m}$
Solution Video

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

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.

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.