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The cornea of the eye has a radius of curvature of approximately 0.50 cm, and the aqueous humor behind it has an index of refraction of 1.35. The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 mm. (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were 25 cm in front of the eye? If not, where would it focus that text: in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about 5.0 mm, where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

A. $6.48 \mathrm{mm}$B. $27.0 \mathrm{mm}$C. $19.3 \mathrm{mm}$

Physics 103

Chapter 34

Geometric Optics

Reflection and Refraction of Light

Wave Optics

Simon Fraser University

Hope College

University of Sheffield

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Okay, so we're doing Chapter 34 Problem 83 year. So the corny of the eye has a radius of curvature of approximately the half a centimeter, and this says the awkwardness humor behind it as it index of refraction. I'm gonna call this and be 1.35 And it says the thickness of the cornea is small enough that we can neglect it and the depths of the human eyes around 25 millimeters. Okay, so part ay is the stuffing of this first in part, ay asks for what would have to be the radius of curvature of the cornea, so that it alone would focus the image of a distant mountain on the retina, which is the back of the eye. OK, so first we know that the image formed is gonna be given or follow this equation the over our with in a being air or one, and then be 1.35 as was given. So for this first situation with a map mountain very far away, being ass is infinity, and we want to focus it on the retina, which is we said the depth is 25 millimeters behind the cornea. So that s the the image distances 25. Sending news 25 millimeters. Let's keep it in centimeters, though. So let's D'oh! 2.57 years. Yeah, which is the same as 25 meters. Um, so now that lets us plug in. We say here. So the first term is gonna be zero since as infinity on the bottom. So we're gonna be left with 1.35 over 2.5 centimeters equals 1.35 minus one over r. Awesome will be rearranged for our and we find that our is 0.6487 years. Cool. So just a little bit bigger than what it wa. So what's going to party? Um so part B asks if the cornea focused the mountain correctly on the retina, described in part a. Would it also focused the text from a computer screen on the retina if that scream or 25 centimeters in front of the eye, if not where it goes? Okay, so now we can use the object distance of 25 centimeters and the are of what we found before, and it lets go back to our equation here. So we should have won over 25. 1.35 over 2.5 in this equals. Oh, sorry. We don't know the image distance here, so we're going to solve for that. So when the U. S prime, this equals 1.35 minus one over's Europol. Six for eight. Awesome. Let's rearrange. You should have s prime equals on 0.93 centimeters. Sorry. That was completely not the right number, so let's go ahead in. Ah, write that out. So s prime is gonna be 1.35 minus one over 6 40 minus one over 25. Being inverted times 1.3 sides. Okay, awesome. So now that put that all on your calculator, and that should come out to be 2.7 centimeters. But the back of the eye is 2.57 years, so this actually forms the image behind the retina behind the retina. Cool. Um, what's at any page here is to go under part see so part. See, given that the cornea has a radius of curvature about five millimeters, where does it actually focus the mountain? Okay, so now we have our equation of its kind it out again. Cols and b minus in a over car. So what we're trying to find is s prime. So we have one over infinity. So that zero when you have 1.35 over X prime equals zero court 35 over zero point product. Go. So let's just rearranged that CS crime is your 0.5 over 0.35 times 1.35 And we can usually plug that in our calculator. And this comes out to be in 1.93 Senate leaders. So that means it's in front of the retina, since the retina was two point by seven years back. So this image is in front, um, the, uh right now awesome.

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