predictexplain-intracorneal-ring-an-intracorneal-ring-is-a-small-plastic-device-implanted-in-a-perso

Question

Predict/Explain Intracorneal Ring An intracorneal ring is
a small plastic device implanted in a person's cornea to change its
curvature. By changing the shape of the cornea, the intracorneal 
ring can correct a person's vision. (a) If a person is nearsighted,
should the ring increase or decrease the cornea's curvature?
(b) Choose the best explanation from among the following:
I. The intracorneal ring should increase the curvature of the
cornea so that it bends light more. This will allow it to focus
on light coming from far away.
II. The intracorneal ring should decrease the curvature of the
cornea so it's flatter and bends light less. This will allow parallel rays from far away to be focused properly on the retina.

Nathan Silvano · Accepted Answer

Okay, so imagine this problem. We have the corn, the shape of the cornea changing and inter cornea rink can correct a person&#x27;s vision. Uh, we know that we need to consider the correction required for a near sightedness type of person. So you just need to remember to answer this question that ah, near sighted person converge too much of the light. So in order to correct the vision of a near sighted person, we need a diverging type of lands because the diverging type of lens diverged, delight, race and correct the image off the object that is being produced. So you the answer to the first I thin is the inter cornering should decrease the need to put again should decrease, increase the corneas curvature. Because if we decrease the core of the corneas curvature, we&#x27;re not bending that much of the light, and then we can see properly. Based on that, we can say that the best explanation is the explanation two, because this position to say that the inter cornering should decrease the curvature of the cornea for this flatter and bends light less Israel a battle Oh, race from far away to be focused properly on director, and that&#x27;s exactly what we should think based on the correction of the near sighted person, So that&#x27;s all things.

Steven Emmel · Answer

Hello, everyone, this is problem 36 from Chapter 28. It&#x27;s a problem about Newton&#x27;s rings says monochromatic Light with a given wavelength shines down on a plane. Oh, convex lens lying on a piece of plate glass has shown the figure when viewing from above one season set of concentric dark and bright fringes for it to his Newton&#x27;s rings party says if the radius of the 12th darkroom ring from the center is measured to be 1.56 centimeters, what is the radius of curvature? Big are of the lens. All right, so we have this Ah, big lens sort of sitting on a piece of plate glass and we&#x27;re getting some dark fringes. And so we can use our equation for, ah, wedge of plate glass because that&#x27;s really Ah, a wedge of glass here, which is really what we have, um, in order to solve for this this height D. Okay, so how does that help us solve for the radius of curvature? Ah, well, if we have this race of d, we can actually relate it in some way to ah, the distance. We are away from the, um from a line down the center here and the radius of curvature. So it&#x27;s actually draw these in. I know that both these air big r and then if we have d So this is D um Then we can use the Pythagorean theorem because we have little are this is given to us in the problem And so we know that our squared is equal to little are square, large squared plus big arm ice t squared And then using this equation rearranging a bit we can solve for big R in terms of these things. Okay, so now we know that we can actually find big are if we have D and so we can use this equation Tau solve for D. Okay, so this is the equation for dark fringes on here. I&#x27;m just going to cancel the 1/2 on either side. And so we know that D is equal to em. Lambda over two were given Linda in the problem and we know the M is equal to 12 live leaves. All right, it&#x27;s the 12th dark fringe. There should be 12 and from this we obtained the D. A. Is equal to 3.89 times 10 to the negative six meters or 3.9 micro meters. So you want to make sure that you carry carry along this times 10 to the negative sixth here, Um, so that you get the right answer out on our big r. And so remember, you&#x27;re using for a little r. You&#x27;re using 1.56 centimeters and using this value we just found for D, we obtain 31.3 meters. All right, then, Prop Part B says if light with a longer wavelength is used with the system, will the radius of the 12th dark ring be greater than or less than 1.56 centimeters? Explain. Okay, So if we use a longer wavelength So this was a party party. If we use a lot larger wavelength wavelength is going up. This means that Di is going up. If G goes up. This means we must be moving farther out here, right? Because if if d a smaller removing further in, they&#x27;re smaller. D&#x27;s in here and bigger D&#x27;s indicate going outward. So that means that little are we will also increase. And so uh huh it will ask, Will the radius be larger radius will be, in fact larger. And then what? We were given off 1.56 centimeters.

Nathan Silvano · Answer

Okay, so in this problem, we have intra ocular replacement. Lens is adjusted for viewing of distance objects and corrective glasses are warm when viewing nearby. Object. So again we have the lens. Let&#x27;s put the lense is for distance objects with nearby valleys. Distance this tent objects but the glasses, the glasses, our foreign year bye objects. Okay, so the solution to the first i 10 of the problem to view close objects D I must refract light by a greater amount then and when viewing distance objects afford the corrective glasses The Gless It should be a converging type of glass to produce this additional Convergencia incoming light that so knowing that we need to convert to the light, we know that the glasses must have a positive a positive refraction refractive our mystery meat Positive positive refractive power. Okay. And what is the answer to the second letter? Second item of the problem. The second item The best explanation should be explanation one because the refractive power should be positive converging because danger or garlands will make the person far sighted. It&#x27;s what we just need to understand that the glasses for nearby therefore we need to coverage the light race. So that&#x27;s the answer. Thanks for

Steven Emmel · Answer

Hello, everyone, this is problem. 84 from Chapter 28 says sodium light shines downward onto the system shown in the figure. When viewed from above, you see a series of concentric circles known It&#x27;s Newton drinks Party says, Do you expect a bright or dark spot at the center of the pattern? Explain. Okay, so let me try to sort of draw out this configuration. We have basically a lens here, not necessarily a semi circle here, as I&#x27;ve kind of drawn it. But there&#x27;s lens like this. And so if first we want to figure out whether there&#x27;s Ah, dark patch or a bright patch near the center and consider rays reflecting here, going out, reflecting off the bottom of this lends and also those that, um, spend a small, just barely exit villains and then hit the glass. What happens to those race? So raise their reflected from the bottom from the bottom portion of the last. Let me just kind of blow this up. So this is just, uh, enhanced view of what&#x27;s going on down here. So raise that go here. Ah, and come back up, of course, are having new phase change because they&#x27;re reflecting off the bottom of the lens. But raise that, uh, go out slightly out of the lens and then back out are going to have a 180 degree face change. And so, as this gap goes to zero, um, the difference, Um, in phase of these two, these two reflected rays is 108 degrees, so there will be a dark French will be a dark spot in the middle. Okay, So then part B asks if the radius of curvature of the plane a convex lens is Article 26.1 meters was the radius of the 10th largest dark ring. Um, and so it also gives us the wavelength of sodium like earlier in the problem. So we should be able to figure the sound. So we have an equation, Um, for whenever we have Ah, a situation like this where we have ah, piece of glass a distance D above. Ah, um, above where? We&#x27;re viewing the fringes. Um, we have the equation. 1/2 plus two D over Lambda is equal to m plus 1/2. And this is for the dark fringes. Okay? And the dark fringe is what we want. Um Okay, so we confined D and member D is just the height here. That&#x27;s the height D So, um, if we also know so So from this, we can get d because we know him. We know Lambda. And, um, what do we do we actually want here. We want this this linear distance from the center to the point in which we&#x27;re seeing that that fringe, um, so called this little are It&#x27;s a normal thing to call it, um and so we want little are and were given. We know, actually, the reduce of curvature. So imagine that this is so the way I&#x27;ve drawn it kind of looks like the centers here, But imagine that this is maybe not so not so curved here. The edges. Um, imagine that this is the center Ah, of the circle. Um, that this Ah, that this the radius of curvature describes on the lens describes. Okay, so this is big are then and then. So this length from here to here, we&#x27;ll use a different color here. So we realized I&#x27;m not talking about this whole thing. Just this red part, big R minus D. Okay, So since we can get d from this equation and were given big are in the problem. We could just use the Pythagorean theorem to find little arm. So you know, the little or just equal to the square root off R squared minus quality or mice d squared. Or it could also be written as a squared of two r d s d squared. All right, so we can go forward with our calculation here, and we see from this equation cancel out the one half&#x27;s on either side and bring the lambda on the two over. We get that d equals m limbed over two, and here M is 10. And so if we we use that and the value of lambda they give, we obtain fractionally. Um, sorry. I know that we obtain approximately two point look, 2.945 micro meters and then plugging in here, obtain 1.24 centimeters of their final answer

Nathan Silvano · Answer

Okay, So in this problem, we have a light foreign object, reflects off the front of the cornea, as if were a convex mirror and produce a reduce retour image. And we want to calculate the radius of curvature off the corn ear. So what we have we must know First of all, that radios. Of course. Richard defined two times the focal lens off the Cory. Therefore, if we discover the focal Lynch, we discovered the reaches a curvature. So what? We know off this problem, we know that the distance off the object is just let&#x27;s see. 10 same team enters. We know the magnification, which is miners, you know, 0.0 35. And that&#x27;s it is the only thing we know about this problem. So knowing that we have the magnification in the distance of the object, we know that to discover, in order to discover the focal length, gonna have to use the fuel ins equation, which is one divided by zero plus one divided by the I all to the bar of miners one. But the problem here is we don&#x27;t have a distance of the image. So let&#x27;s find out first the distance of the image using the magnification because we know the distance of the images. Just the magnification, times the distance of the object. So this is going to be a loss. Actually, magnification is positive in here, so position gonna be negative. Negative. Do you know 0.0 35 time, Stan. So this is just miners, you know, point 35 ST. Team enters. Okay, Well, now that we have the distance of the image, let&#x27;s calculate the focal length, which should be one divided by 10 plus one divided by minors. You know, 10.35 minus one. So this is your smile is you know, point 36 tree. Same team enters. So now that we have a focal lens, the curvature radius of curvature, Cher&#x27;s going to be That&#x27;s pointing here. Just going to be two times zero point 363 So this is just 7.3 me limiters, and that&#x27;s a final instructor. Just problem. Thanks.

Nathan Silvano · Answer

Okay, So in this problem, we have the relax it cormorant. I focus on distance objects. And we know that in order to focus objects underwater, the cormorant is able to change the shape of the lens and change the reflective power by 45 die officers. Okay, so the first item of the problem we need to explain if this change of 45 doctors is going to be an increase or decrease in the refractive power. Okay, so if remember and chapter 26 we know that the refraction index off the water needs to be closer through the reflection index off the lens, then to the refraction index off the air. So to compensate this, the refractive power off the lens must increase. That&#x27;s the answer to the first item. The cormorant must increase the reflective power. So let&#x27;s see the 2nd 1 We need to calculate the focal and underwater. So we know that the focal and is just the inverse of the reflective power. So, first of all, we need to calculate what is the reflective power off the cormorant. I relax it. So the reflective power, it&#x27;s just going to be Ah, let&#x27;s see what is difficult and the problem gives us is 4.2 millimeters before entered the water. So it&#x27;s just going to be one divided by you&#x27;re appoints you does, you know, for true. That&#x27;s equal to your 138. Hey officers. So after entering the water, the calmer and eyes must have a focal of focal and off, one divided by 238 plus 45. That is the change. This is going to be go to 3.5 limiters and that&#x27;s the final answer to this problem. Thanks for watching.

Problem

The lens in a normal human eye, with aqueous humo…

Problem 81 Medium Difficulty

Answer

a) decrease
b) II

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Multiple Aperture Interference - Overview

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