the-photosensitive-cells-rods-and-cones-in-the-retina-are-most-densely-packed-in-the-fovea-the-part-

Question

The photosensitive cells (rods and cones) in the retina are most densely packed in the fovea- the part of the retina used to see straight ahead. In the fovea, the cells are all cones spaced about $1 \mu \mathrm{m}$ apart. Would our vision have much better resolution if they were closer together? To answer this question, assume two light sources are just far enough apart to be resolvable according to Rayleigh's criterion. Assume an average pupil diameter of $5 \mathrm{mm}$ and an eye diameter of $25 \mathrm{mm}$. Also assume that the index of refraction of the vitreous fluid in the eye is $1 ;$ in other words, treat the pupil as a circular aperture with air on both sides. What is the spacing of the cones if the centers of the diffraction maxima fall on two nonadjacent cones with a single intervening cone? (There must be an intervening dark cone in order to resolve the two sources; if two adjacent cones are stimulated, the brain assumes a single source.)

Mayukh Banik · Accepted Answer

in this particular problem, we have to use the formula. A sign get the tenor is equal to one point to do them there when the rather criteria gay satisfied on the objects are barely result. So signed ticker or sign of Delta Tater comes out to be one point to do Lambda over a ik was one point to do. We&#x27;re using 400 nanometer light. This is the least visible where event So we will use 400 Nana meter are 400 times change the minus six, maybe a meter and the diameter off the que bill is five millimeter so that comes out to be 1.2 times 400 divided by five it was 97.6 times 10 to the minus six. So using small angle approximation, this is the valley off Delta terror as well This much Radian Oh, 97.6 time stand to their minus three Me the radian which is 0.0 976 May the ready in now because the anger, the small that&#x27;s why we can use jianjun off their dictator is equal to sign off the Arteta. It was do that the data. So Tandon terror is given by X over l because Celta Tetteh so ex Equus and just the tender. Where else is the diameter? 25 millimeter. So 25 damn stand to the minus three times. Get the data 97.6 Time stand to the minus six equals 25 times 97.6 equals 2440 2440 Damn Stand to the minus nine meter which comes out to be 2.44 times 10 to the minus six Major, which is equal to two point for four my cross.

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

Abhishek Jana · Answer

so to see that off text objects separately, we need tohave a minimum separation of the image in the four year off one corn. And we&#x27;re considering the images as point like so, if we use these two informations on DH, then if we try to look at this his own, as you could see that for one corn on the cob, the diameter of one cornice, two micrometers. So we have full micrometers written here. So the catch here is if we try to zoom in this idea, we first know that it has to have ah, one corn separation between the two images. So let&#x27;s say that&#x27;s the one corn which is two micrometer now then we&#x27;LL have the two objects. So these are the two objects and since it&#x27;s point like we can take, we can take the objects at at the center. So that means object started the image of the center. So that means the image will be here and here. And we know that the diamond is two micrometers. So that means this guy is two micrometers and this is half off the diameter, which is one micrometer, and this party is half of the diamond services one micrometer so that all together gives us Formica meters. So that&#x27;s what&#x27;s going on here. Andi, stop! The problem is simple. If you If we look at the geometry here, we see that we have two triangles. And if we take the ratio off base and aunt itude off these triangles, we get a relation like and over twenty five centimeters, which is equal to for micrometer, um, divided by two centimeters. So from there we can find the separation off the object. So l will be twenty five centimeters times for micro reader divided by two centimeters. And that will give us L as fifty. My point. So that&#x27;s the suppression between the That&#x27;s the suppression between the objects. Thank you.

Carlos Henrique De Lima · Answer

I have Ah, to know how far parts two points are that we can resolve it with the I So the schematic is here. So we have the I hear in the internal part of the eye has something to say meters, the image is for me Grow me, grow meters a side and we are 25 sentiments away and the five point of the two objects we call this distance l and we can go basically trigonometry here to find El. So I get that l divided by 25 centimeters. Isn&#x27;t Cousteau for make Ramirez divided by two centimeters? So the relation between the this distance here and this This is here because of these two complimentary trials and you can find that l iss modified by for 25 centimeters. This gives us L&#x27;Echo 6 50 metre mirrors. So this is the they the hell far apart. The objects are so we can resolve them

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.

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

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2.44 x $10^{6}m$

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Alan Giambattista, Betty McCarthy Richardson, Robert C. Richardson

Physics