the-cornea-of-the-eye-has-a-radius-of-curvature-of-approximately-050-cm-and-the-aqueous-humor-behind

Question

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?

Bruce Edelman · Accepted Answer

Okay, so we&#x27;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&#x27;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&#x27;s keep it in centimeters, though. So let&#x27;s D&#x27;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&#x27;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&#x27;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&#x27;t know the image distance here, so we&#x27;re going to solve for that. So when the U. S prime, this equals 1.35 minus one over&#x27;s Europol. Six for eight. Awesome. Let&#x27;s rearrange. You should have s prime equals on 0.93 centimeters. Sorry. That was completely not the right number, so let&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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.

Farhanul Hasan · Answer

No, I said throughout this problem will be using this formula. Ah, but refractive index of medium a over object distance from refracting object plus for fact of index about meeting be over image distance from refracting object is equal to ah, the difference over the radius of curvature radius off the res of curvature off the refracting object. So in part and we know the object is at a very large distance. It&#x27;s a distant object and distant object means we can make this approximation that distance from object to reflecting surfaces very large. So one of a very large number is approximately zero. So this first term drops out okay. Ah, so we&#x27;re left with this proactive index off media be over s prime equals and B minus and a over R and R is what we want. So medium be would be their refractive index specified in the problem 1.35 image distance is 2.5 centimeters on DH. That&#x27;s equal to 1.35 minus 11 being refracted the necks of air over our and so this gives us our equals 0.648 centimeters or 6.48 millimeters. Okay, Uh, part B. Wei don&#x27;t make this approximation anymore. So we have all the terms of this equation and part B. We have a fraction Vanek Severa one over the distance of the text from the cornea. And so that would be 25 centimeters plus 1.35 over as prime, which is where the image will form. And that&#x27;s what we don&#x27;t know here, a sequel to 1.35 My Swan over R and R. We just found the last problem are in the last part, and so are is 0.64 eight centimeters, right? So this works out to 1.35 over us. Prime equals 0.5, so you can see that as prime is 22.7 centimeters, which is equal to 27 millimeters. We were given in the problem that the depth of the average human eyes about 25 millimeters and so this image is formed deeper than the than the eye. So it&#x27;s it&#x27;s form. So if we consider the retina as those farthest deepest point of the eye, this image it&#x27;s formed behind the retina and finally heart Syrians again we go back to the states. An object approximation so that this term doesn&#x27;t factory. So again, you have this equation here, SL for part B part. See, you have 1.35 over s prime. That&#x27;s what you don&#x27;t know. Is it cool? 2.35 over the actual radius of curvature, which is 0.5 centimeters. And so, as prime comes out to be 1.93 centimeters or 19.3 millimeters and so this is less than s O. This is within the death of the eye. And so this ah tells us that the image is in front his form in front of the retina, which means that you need additional lens to focus the image that the mix

Bruce Edelman · Answer

Okay, so we&#x27;re doing Chapter 34. Problem 35 here. So this one talks about the cornea is a simple linds, and it says it behaves about as a thin linds of approximate focal length of 1.8 centimeters. Even though this varies a bit, we can assume that the materials made out of a zip index of refraction of 1.38 here in his front surface is convex with a radius of curvature of five millimeters. Okay, so for part A, it asks if the boat this focal length is an air. What is the radius of curvature of backside of the cornea. So we&#x27;re gonna use the lens maker equation for this where we know that one over equals in minus one. Turns one over R one minus one over R two. Cool. So it&#x27;s gonna notice Plug stuff in here are really simplify things. Curse. We have won over F times in minus one plus one over r two equals one over R one. So now we can say orn over our two equals one over r on minus one over tough and months one or are two equals one are one minus one over F times minus one. I heard it. Cool. So it&#x27;s bugging our numbers for that. And we should have one over 0.5. That&#x27;s half centimeter. Minus one over 1.8 times 1.38 minus one. Cool. That should be everything now. So let&#x27;s plug that in a calculator, and we should see us come out to the 1.86 centimeters. This is art too cool. So now Part B asks the closest distance at which a typical person can focus on an object is about 20 force in the readings. Where were the cornea focused? The image of an eight millimeter tall object at that point. Okay, so if we rearrange the thin Lindsay equation, we should see. We should know that s crime. His s finest troops. It s F over Essman, Ms F Free plug back in and we should see Espen 24 football limping. 1.8. You shouldn&#x27;t see. This comes up to me. 1.9 centimeters go. So is when we&#x27;re looking at an eight millimeter tall object at the close. In the near point, the image receives about 1.9 centimeters away. from our cornea. So what&#x27;s, um that&#x27;s what we focus it on. Do we want to know what is the height of the image? And if it&#x27;s really or Jule and direct or inverted? Cool South Park. Si. So for part C, here we can figure out what the magnification is, and we know that that&#x27;s gonna be negative. S prime over s that equals Why prime over? Why we re arrange that. We see that. Why Crime? Yeah, Listen, like trying that someone was trying figure out is equal. Thio negative. 1.9 centimeters over 24 centimeters. Times eight millimeters. That means the centimeters cancel out. We&#x27;ll be left with millimeters. And we should see this because negative 0.61 about centimeters millimeters and its negative. So that means we know that the image is inverted. And we also know that the image is real because it has a positive image distance. So s s prime being positive means we&#x27;re interviewing so inverted and a really image

Farhanul Hasan · Answer

All right. So the cornea is treated as a thin lens here, and so that then leads equation applies. One of her focal length is equal to refractive index minus one times one of radius of curvature of fun and minus one. Overuse. Um, coach of the back it And it is this our two that we want here in part eso let&#x27;s we&#x27;re angels equation and write one of our one writes or one of our two is equal to one over f times one over and minus one. That&#x27;s just, um, pulled to the left hand side on dso One of our two is just one of her. Are one minus this whole thing of one of our aft times one over and minus one. Okay, on DSO. One of our one is five millimeters less of directing staff. Trip to millimeters. Ah f is 1.8 centimeters. Ah, so that will be 1.8 centimeters are about 18 millimetres. 18 millimeters. Oops. Okay, so that&#x27;s 18 millimeters, and that&#x27;s and er multiplied by one over one point 38 minus 11 country. It was a given refractive index. That&#x27;s one of our two and So this implies that are too Take a reciprocal. What you&#x27;ve got get is 18.6 millimeters or 1.86 centimeters. Part B wants the image distance and so he is one over f equals one of us, plus one over X prime and one of us plus one arrest prime So hopes Assassins as prime there on dso one arrest as primarily once So could be right. This when I were f minus one over R s and half again is eyes one little strain two centimeters from 20.8 centimeters. That&#x27;s F and s image distances 20 Object distance 25 And so we get that as prime is 1.9 centimeters. Finally for part. See Ah, we first configure out magnification shows negative image distance over object distance. So that&#x27;s negative. 1.9 over 25. That&#x27;s negative. 250.76 This means that the images, um inverted lips on inverted, um, and then because it&#x27;s negative and then em is also God, em is also Ah, negative. Whoops. Positive. Why prime over why My prime is what we want. The image height. So that&#x27;s m times. Why. So that&#x27;s negative 0.76 times eight millimeters. And so this comes out to be negative 80.61 millimeters. So it&#x27;s a D magnified image and then ask prima&#x27;s positive. Let&#x27;s be found in part B. So this is ahh, real image.

Donald Albin · Answer

in part a were using two millimeters as the diameter of the retina of an eye and a wavelength of 550 nano meters. So we&#x27;re used also using a distance to the object of 0.25 meters, um, and were asked if we can resolve a 50 micrometer tall object. So, um, the sign of Fada is 1.22 Lambda over D. And the tangent of Fada is X over L where X is the resolution. So X is L Tangent of Fatal, which is the inverse sign of 1.22 Lambda over D. So I&#x27;m going to put that into a calculator. L is 0.25 kinds Tangent of inverse sign of 1.22 times Lambda, which is 150 times 10 to the negative ninth power over DE, which is 0.2 That gives me X of 123456 um, 84 micrometers. I just want to make sure I didn&#x27;t make any mistakes. 0.25 10 Jin inverse sign 1.22 Lambda, uh, over 0.2 Okay, so can we resolve a 50 micron tall object. The answer&#x27;s no. You cannot, because 85 is the minimum. Be the shortest object we could resolve that looks like it&#x27;s the same question again. It&#x27;s just the 84 micrometers. See what bank? Okay, what angle would it sub tend The I? Well, the tangent of data is X over. L so Theda would be the inverse tangent of X over L. So I&#x27;m going to get that in degrees. Inverse tangent Set my calculator two degrees x over l l is 0.25 That gives me 0.19 degrees, but one degree is 60 minutes. So if I multiply that by 60 could give me minutes 1.2 minutes. OK, moving on Teoh Part de least. I think there&#x27;s a part D Let&#x27;s see here. So which is more limiting diffraction or the retinal cells? Well, the retinal cells have a real resident resolution of 50 micrometers and diffraction had a resolution of 84 micrometers so diffraction ISMM or limiting good 84 versus 50

Dading Chen · Answer

All right. So the question was asking us, um what should be the radius off the coverage? Er okay, so it&#x27;s looking for the value off the are. So we know for a satirical reflecting surface, we have such formula, which is an over ass plus and be over as prime. It was M V minus and a over art are sti radius of curvature. Uh, you see index of refraction from the object in these cases Is the air OK and be is the index off a refresh in off? Um, the actress and retrace humors. And lance okay, was given as 1.40 Okay, as is the distance from the object to a cornea, which is the object distance as prime is the distance between cornea to the Latina was just the image distance. Okay, so if we do samaranch me here, we&#x27;ll get it eventually that R is equal to MB minus and eight tons ask times as prime over a test as prime. Plus Embiid has asked. Okay, and we know that C m P is equal to 1.40 and a is equal to while going although, and as is equal to Was this The option business was given his 40.7 meters. And the image this is Oh, this is from Cornetto Latina scheme is true 0.6 old centimeters. Okay, now we know that our is equal to I remember I was Eagle Teoh empty my unsung eight times s Francis prying over on May as pie last MBS So we get that ours equal to, um b minus some languages. One point full, minus one point. Oh, times s which is 40.0, centimeters centimetres. Times 2.60 some years over. Look eight times as prime, which is 1.0, times 2.6 old, same years plus and b times asked which is one point for all times 40 only always centimeters and this will give us our which is the radius of curvature is equal to zero horn 710 sending year. Okay, so this is a radius of curvature. So as art is positive, Therefore the cornea presents a convex service 40 objects. Okay, so that&#x27;s the solution will question Thank you

Problem

The radii of curvature of the surfaces of a thin …

Problem 83 Hard Difficulty

Answer

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

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Hugh D. Young

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Wave Optics - Intro

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

University Physics

Liev B.

Farnaz M.

Zachary M.

Aspen F.

Wave Optics - Intro

Multiple Aperture Interference - Overview

Hugh D. YoungUniversity Physics

Liev B.

Farnaz M.

Zachary M.

Aspen F.

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

Hugh D. Young

University Physics