resolution-of-the-eye-the-maximum-resolution-of-the-eye-depends-on-the-diameter-of-the-opening-of-th

Question

Resolution of the Eye. The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0$\mu \mathrm{m}$ in diameter) limits the size of an object at the near point $(25 \mathrm{cm})$ of the eye to a height of about 50$\mu \mathrm{m} .$ To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50 -\mum-tall object at 25 $\mathrm{cm}$ from the eye with light of wavelength 550 $\mathrm{nm}$ ? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the $25-\mathrm{cm}$ near point with light of wavelength 550 $\mathrm{nm} ?(\mathrm{c})$ What angle would the object in part (b) subtend at the eye? Express your answer in minutes $\left(60 \min =1^{\circ}\right),$ and compare it with the experimental value of about 1 min. (d) Which effect is
more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

Donald Albin · Accepted 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

Ryan Hood · Answer

Let&#x27;s first figure out the angular size of the object. And so the angler size is going to be the height of the object. Yeah, which is that over the distance to the object, which is 25 centimeters or that many meters. And so this expresses the actual angular size from your perspective, and this is equal to 2.0 times 10 to the last four. That&#x27;s in radiance. Now that we have that, let&#x27;s figure out what the angular resolution of your eyes and so we can do this. Using release Criterion 1.22 times claimed over D flinging in Lambda, which is 5 50 times 10 to the minus nine Andy for your eye, which is two times 10 to the minus 30 meters. This gives 3.4 times 10. Tim lies for radiance. And so since data, which is the actual angular size of the object, is less than your resolution ability, you cannot resolve the object. You cannot resolve the object for part B. We want to figure out what size object we can resolve. And so to do this, we&#x27;re going to take our resolution angle here. We&#x27;re going to equate it to the height of the object or the distance of the object. And so why is equal to the resolution angle Times s and we know what s is s is still 25 centimeters and we know what the resolution. Inglis. We found it here. And so it&#x27;s equal to 3.4 times 10 to the minus four radiance. And so this equals 8.5 times 10 to the minus third centimeters, which is equal to 85 microns. And so this is the size of the object that you can resolve with this resolving power here. And if you&#x27;re 25 centimetres away from it now, Marcie, we want to figure out what the angle is of the resulting power. So in this case, data is equal to the resolving eagle because we&#x27;re on the limit of what we can resolve here and again. This is 3.4 times 10 to the minus for and that&#x27;s in radiance. But if you want degrees, you Khun, convert easily two degrees to get 20.19 degrees. And then the problem asked for this in minutes and 60 minutes. Aaron, one degree. So if you convert this two minutes you get, we&#x27;ll play one Our commitments. I&#x27;ll put arc minutes. The problem just puts minutes, but stark minutes, and this is very close toe 1.0, which is the experimental value. So that&#x27;s the experimental value, and it&#x27;s very close to it. Party asked. What&#x27;s more important, the diffraction or the other effect of your eye and answers that the diffraction is more important. But in order to really see this, you have to investigate the problem that they referred to in this question. The problem is 51 and Chapter 25 which tells you a little bit more about the other effect. But if you investigate this a little more, you&#x27;ll find that diffraction. The fraction is is more important. But again, it&#x27;s hard to answer this question with just this question along. You have to really go back to Chapter 25. Problem 51. The diffraction is is more important, and that completes the problem

Kelley Commeford · Answer

in this problem. We are looking at the resolution angle of the eye and seeing if diffraction please Ah, significant role in how is see objects. So assuming that our field of vision was a distance of 25 centimeters away on the diameter of our people, it was two millimeters with a wavelength of light at 550 nanometers. What is the maximum? Various? The minimum size that we can see. So are really angle given by 1.22 Lambda over D And as with most of our really problems, we can assume that our angle is approximately. Why, over s because the size of the object is going to be smaller than the distance that we&#x27;re observing it up. So if we put these two things together, me at 1.2 to Lambda over D is equal to why Over S and we want to solve for y because that&#x27;s gonna be the size of the object that we&#x27;re looking at. So 1.2 to Landover de multiplying by s to the other side saying at 1.22 Landa was 550 nano tend to the negative nine meters D is two times 10 to the negative three Millie meters and us is 25 senti so 10 to the negative, two liters. So if I played that into my calculator, I get 83.9 micro meters, which is really small.

Ryan Kutayiah · Answer

problem. 90 for, uh, a Jew. A diagram to indicate this sort of information were given. So we&#x27;re told that we have a spherical I, which has a diameter of 2.5 centimeters. So what this means for us is that, uh, the distance from the cornea to the retina, where the image so this the corny will act like your lens and the retina will be where the image is formed said that sets are as prime distance. So the diameter of the eye is the distance from the cornea to the retina or the ass prime distance because the eyes sparkle. And we&#x27;re also told that an object is placed at the near point. And we know that from the textbook that that&#x27;s about 25 centimeters, so that gives us our S. So already we know are asked, That&#x27;s prime. And we&#x27;re told that in the retina charities receptor cells, which are about five micro meters and I converted two centimeters and what this means is that we know that essentially the height of of these of these receptor cells, uh, is five microns, because that&#x27;s a diameter right? So anywhere you measure it from one into the other. You&#x27;ll get five my crowds. So that gives us our Y prime, right? Since the images form at the retina, then they it could only basically spent for you to see yourselves need to receive the image of the image can cannot really be bigger than then the receptor cells. So this limits the size of your your y prime, right, your image height. So we will set the image hide to the diameter of the receptor cells. Now, we&#x27;re asked in the first part in part a of the question to find the smallest object that&#x27;s at the near point that you&#x27;re I can resolve, um, that are that you can. Then you can see clearly eso what we&#x27;ll do is use equation. So let me a little box here, and we&#x27;re going to solve our problem inside this little box. So we have we will use equation 17 which tells us that you&#x27;re lateral magnification is minus s prime over as, uh and we know as private s. And the reason why we want to use this equation is because we have a If you read the paragraph, maybe just before this equation, you&#x27;ll see Also, that M is equal to why Primed over why? So what we&#x27;ll do is rearranged this equation. And I&#x27;m just gonna put some absolute values here because you only care about the size and at the sign. So I&#x27;ll have wise too. Why primed over em? We know why. Prime we know why we don&#x27;t know him. But we know Ehsan as primary s. So then recon unsolved reminisce equation. And once we find them, we just plug it in here, and then we can find why. And this will be the smallest object that we can resolve at the near point. So mm is equal to now s prime. Over s, which is so s prime we have is 25 our 2.5 centimeters divided by s, which is 25 centimeters, and that gives me a 0.1. Okay. And now, to find why so now why is equal to why prime, which is five times 10 to the minus four centimeters, divided by 0.1? I could write this sensitive minus one, and that gives me a why primed that right down here. Uh, sorry. Why? Of five times 10 to the minus three Senate leaders. Okay, so that is the smallest. How great that, uh, we can resolved. And now, in part B, you&#x27;re asked to find the angle. That&#x27;s some tended by the ice. So let&#x27;s do that on another page. So let me draw on another diagrams and begin set some things up here. So suppose we have an object over here at the near point, and this object hasn&#x27;t height. Why? No, this just here is why? And and this is the optic axis. So let&#x27;s say that we we have the optic asked axis bisect are objects of. Then this is why over too. So from here to here is why over too. Okay, uh, and now let&#x27;s try and maybe some it&#x27;s a raise. Better go like this. The end. So they intersect at this point, which will be, uh, the cornea, right? And then and then drama, my diagram. So well, but this is the corn in, and this is supposed to be the white prime. So, uh, endure it, obviously not to scale. But imagine that this was what we had before this little green line here. So we know that the Y prime should be, uh, five times 10 to the minus three and our Why? Sorry. But times 10 to the minus four and the her Why? Height is five times 10 to the minus three. This isn&#x27;t some years. Okay, so this is what we we found before we sell for why? And then we also know we also know that this distance this is s on. That&#x27;s 25 centimeters, and this is a s prime. That&#x27;s 2.5 senators. Okay? And now we wish to find this angle here. So the way through this image, I mean, my my lines, they&#x27;re not straighten someone, but, um, this let&#x27;s take this. This angle with the double line is the same as this angle here. So his big angle. So if I call this data and that&#x27;s called a state O s, then the S is, uh, data. Plus, they rank because this is data and this is also fada, and data ask is twice data. Okay, so that&#x27;s something we&#x27;re going to keep in mind now and what we have had, What we have here is a right triangle, right? So I said the object is bisected by the optic access so and enjoy the right angle. So then we have this triangle here. Uh, let me judge this. Imagine that&#x27;s a straight line. Although my life is not very straight, I&#x27;m not going to drive to scale. So put in, we have their cakes, it connects. And this is data, all right? And we know this is why over too. And we know that this is s okay. And I wish I want to find data so I can say OK from trigonometry. You know that the tangent of data is the opposite over adjacent. So that is why over too divided by yes, which is why over to s I just took this too down there. And if I want to find data, then I take the inverse tangent Nerdy r tendon of who I over to s okay, but my goal is to find Data s so that&#x27;s twice stay down. So we know the data, and then they&#x27;d I ask is to date, huh? So then they don&#x27;t ask is twice tangent. Inverse of why Over to s. Okay, so now we have a formula for findings. The angle that&#x27;s attended, buddy I and what we do, you know, is Larry. Just plug in the numbers so you can do this near scientific calculator. Why we know is five times 10 to the minus three. The sulfur is in part a centimeters. Forget to over there, and then that&#x27;s two times s, which is 25 centimeters. And when you play this in, uh, you get 0.2 radiance. So now they were asked to find this in in minutes on. You were given that, uh So you want to find it in minutes, right? You won&#x27;t Beta ass in minutes. But you have data s in radiance. Uh, but you&#x27;re given that one degree is equal to 60 minutes. So what you need to do first is take birth date s two degrees and then two minutes. So they asked and agrees to remember that at 0.2 radiance to convert, uh, into degrees, multiplied by 180 over pi, and that gives me 0.11 degrees. Now, using this relationship here, I have that one degree is equal to 60 minutes. So gonna have data acid sequences. Your point, Teo 112 threes. Uh, times 60 minutes per I want to create. And that gives me roughly 0.7 minutes and warehouse to compare this to the standard. Uh, so the usually we&#x27;re told the standard is measured by experiment. Uh, is that a degree in minutes of one minute? So we have an answer that&#x27;s a little bit less than the mattered one minute. And the textbook says that in this particular approximation of this very cool guy, we ignore the bending of the light as it comes in at the cornea. So we&#x27;re essentially ignoring the fact of the neck. The I said that could probably be why we have something that&#x27;s a little less in this approximation.

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 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.

Problem

A glass sheet is covered by a very thin opaque co…

Problem 70 Hard Difficulty

Answer

a. 84
b. 84$\mu \mathrm{m}$
c. 1.155 $\mathrm{min}$
d. that the effect of diffraction is more important in limiting the resolution of our eyes

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Hugh D. Young

University Physics with Modern Physics

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Liev B.

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a. 84b. 84$\mu \mathrm{m}$c. 1.155 $\mathrm{min}$d. that the effect of diffraction is more important in limiting the resolution of our eyes

University Physics with Modern Physics

Christina K.

Liev B.

Farnaz M.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

Hugh D. YoungUniversity Physics with Modern Physics

Christina K.

Liev B.

Farnaz M.

Jared E.

a. 84
b. 84$\mu \mathrm{m}$
c. 1.155 $\mathrm{min}$
d. that the effect of diffraction is more important in limiting the resolution of our eyes

Hugh D. Young

University Physics with Modern Physics