fitting-contact-lenses-with-a-keratometer-when-a-patient-is-being-fitted-with-contact-lenses-the-cur

Question

Fitting Contact Lenses with a Keratometer When a patient is
being fitted with contact lenses, the curvature of the patient's
cornea is measured with an instrument known as a keratometer.
A lighted object is held near the eye, and the keratometer measures the magnification of the image formed by reflection from the front of the cornea. If an object is held 10.0 $\mathrm{cm}$ in front of a
patient's eye, and the reflected image is magnified by a factor of
$0.035,$ what is the radius of curvature of the patient's cornea?

Nathan Silvano · Accepted 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.

Matthew Kegley · Answer

So in order to find the curvature of the cornea in order to use contact lenses and get the right lenses for your eyes, there&#x27;s an object that&#x27;s placed in front of your cornea at a certain distance, and the image has formed. Um, by this object is measured, and it uses, um, this this calibrated calibration to find out how your what&#x27;s the right curvature for your contact lenses. When this problem, we have an object that&#x27;s a distance away from a cornea, and it&#x27;s given that this distance here is, uh, P equals to 30 centimeters. And it&#x27;s known that the magnification of the image, uh, as measured by the equipment, is zero point zero 130 These are the two given quantities that we have in order to solve this problem in the problems asking us to, um, find the radius of curvature of this person&#x27;s cornea. So what we do is we&#x27;ll just used the equations that we know from this chapter, like the mirrors equation equation for magnification, um, in order to solve through the radius of curvature. And we&#x27;re going to treat the cornea as a curved mirror in this case. So we&#x27;re we&#x27;re gonna use the fact that for curved mirrors, the focal length is equal to our over to where r is the radius of curvature of the mirror of the curved mirror. So if we wanted to solve for the curved mirror, we would know that are equal to two times the foreplay. So given this information, let&#x27;s go ahead and start to solve this problem here. The first thing that we need to do is to find the object, are started the image length, and we can use that we can find out using the magnification for me lower and is equal to minus que overpay. Where Q is the image, length and P is our object length, which is given to us, uh, as 30 centimeters. So cue is equal to minus M p. Those two quantities we know so well, just substitute them in 0.130 for the magnification. And we have 30 centimeters for our object length, and the answer we get here is minus zero point 39 centimeters. Now, when we do problems with, um in with optics or any kind of optical equipment, we have to figure out where is our positive lengths, and where are negative legs? So P was our objects and object was to the left of the mirror. So to the or the ruler cornea, which retreating as a curved mirror So the object senses to the left of the curved mirror. Here we&#x27;re treating this as a positive distance. So p he is positive, and I&#x27;ll just straw a line to connect to between the object of the cornea. It&#x27;s positive peas. He is greater than zero here. So to the left of the curved mirror, we get a positive length to the right of the mirror. We&#x27;re getting negative legs. So the negative, um, image distance that we have here, minus 0.39 centimeters, it means that the image is being formed behind the eye. It&#x27;s being performed behind where you have, you know, your retina and going. It&#x27;s being basically going toward the rain. The images formed behind this mirrors curtain here. So I&#x27;ll just put image distance up here que which is, uh, less than zero in our image. Our our image is formed behind the mirror. Our object is in front, so the mirror itself is like a zero on a number line. So that&#x27;s how we treat our optics for this problem here. So that&#x27;s when we have in minus, um a minus image distance for this problem. It&#x27;s because the images being formed behind I what whereas the object is in front of the I, that&#x27;s my ass. A positive. So now we have this answer. We can use the mirror equation, the mirror equation, which allows us to solve for the focal length and from the focal in if we confined our radius of curvature. So that Mary equation is one overpay. Let&#x27;s one over. Q equals one over half, Um, where f is our focal lights people from before. So what we do, we can substitute the equation that we have for the focal length, which is a physical toe over to, and that would give us two over our okay. So in this case, if we were to write and substitute everything in, we can say we have to up to over our equals to one over P and P is 30 centimeters positive 30 centimeters. When Cuba&#x27;s negatives, we actually have to do a minus 1/0 300.39 centimeters because the sign changes because we have a negative, uh, negatives image. Distance. Okay, so in this case, what we can do is we can divide both sides by two. And that would give us one over r equals 1/60 minus one over 70.78 Now, if we choose our common denominator to combine these two fractions as 60 then that would mean this would equal to one minus 76.9 over 60. How did I get that? I kept the one from this first fraction since it has the common denominator that we want. And then we would say that if we want to make this 60 then we have to We have to put this into equation of the same. Um, sir, 0.78 Times X is equal to 60. Ex, um will be multiplied by 0.78 We have to find what that is when to find what exes, which is 60/1 78. That&#x27;s the number that you won&#x27;t apply on the numerator and the denominator. And it turns out that this number is 76 point nine. Okay, so that&#x27;s how you how you do that when you have a denominator, that&#x27;s a little bit complicated and you want to be this want to be another denominator. So you just find what multiple all multiplier you need, and that will be the number here. So you have one month of 76.9. This is going to give you minus 75 point 9/60. And so now you get are equals two minus 60/17 5.9 that it gives you zero minus strike minus 0.79 centimeters. Okay, that is the answer that we have for the radius of curvature of our corn. Now, in the minus sign indicates that the radius is behind the mirror, cause everything behind the mirror gets a negative value. So if we were to say, our radius of curvature is negative, it means it&#x27;s like right here. We&#x27;ll just say this is our radius of curvature just behind the mirror. And as we know, we can see that this is how this is a curved, um, mirror with a negative radius of curvature, which means that any light that comes in will be diverged from the surface and it will appear as if it will appear a ziff. The light source is coming. Uh, from this point right from this point, but coming from behind, uh, coming from behind this a cornea, that&#x27;s what. That&#x27;s why the images being formed behind all it appears as if the light is coming from behind the cornea. And what what this means is that this is cornea that&#x27;s being it&#x27;s acting like a con decks. Let&#x27;s are sorry. Convex mirror. Exacting is a convex mirror and convex mirrors having negative radius of curvature. They have a negative raised of curvature, and because of that, they also have a negative focal length. So as like, comes in like it&#x27;s diverged. And, um, the light that intersects with each other in a sex behind the come convex mirror and forms an image. So that&#x27;s what this problem was treating the the, uh, the cornea, as in this case. So, um, but just to recap who it is, we took our given way. We got our object length, which was positive 30 centimeters in our magnification. We knew what our relationship between focal length and radius of curvature was. So we wrote that down. We use the magnification for me. Like to get the image length, which was negative, which is means that it formed behind the mirror. And this from mayor. We got to use the Mary equation, which allowed us to find the radius of curvature for our cornea, which was a minus 0.79 centimeter length. So that&#x27;s how you find the answer to this question.

Keshav Singh · Answer

we know that the magnification em is equal to minus que overpay Where Q. The image, distance and peace object is still. So if you re arrange this, we could find the image distance Cute. Thank you about his minus him. Tang&#x27;s peak. So we have a magnification off minus 0.0 one. Okay, what image? Locate that centimeters ends to get Q two B minus. You know, point B nine centimeters. And we also know that one over P that&#x27;s one of Q is equal to one over yes, which is the same as two over our and so we can calculate the reduce of perfect job cornea. So one over 30 centimeters us one o minus. You know, going nine centimeters is equal to two. Well, I and if you re arrange the above, you can find the radius of coat coverage over the cornea. Made this a curvature are is you go to minus zero 0.79 centimeters. So what? This means that the cornea is convicts with the radius of curvature off 0.79 centimeters

Zachary Warner · Answer

our question wants us to determine the radius of curvature of the cornea when an image is 30 centimeters away and the magnification of the cornea m 0.13 Okay, so, uh, there are two equations we&#x27;re gonna use to do this. So the 1st 1 is the equation for the magnification, which relates the, uh, object distance to the image distance where the image distances negative. So it&#x27;s negative. Que overpay. Okay, so, solving for Q, this is going to be equal to negative magnification, times the image or the object distance. Okay. And, ah, now that we have the image distance in terms of the object distance than identification which we know we can use the equation which relates the radius of curvature of the lens to over r to the distance the object in the distance of the image that says one over P plus one over Q. So rearranging this to solve for the radius of curvature which we want radius is equal to two times the object distance times the image distance divided by the object distance plus the image distance. We cannot replace the image distance, uh, with the first equation that we found. So this is too times p times negative NP divided by P minus MP. Okay, so, plugging all those values in we find that the radius of curvature is negative zero 0.790 centimeters. This could be boxed in. That&#x27;s the solution to the question.

Salamat Ali · Answer

but a magnification is given by height of the image. Right by height, off the object. I definitely have the support. 167 centimeter divided by I don&#x27;t object. 1.5 sent me to this gives us on a fishing trip when 11 one year would be the eyes given by minus times magnification all times to you. So expression We just found these little 0.111 charms. Do you know we have a three centimeter. So excuse us a minus 0.32 entries in it minus sign indicates that image is behind the mirror. See we confined. Ready something region from the human equation That is one over. If when Woody I plus one more deal. So what from here? If can beard Amazing one would be I plus inward D&#x27;oh! More in this one we know our is equal to two times off f more off the f. Then we can substitute there any off f here. So our our will become here two one word D I plus moon war. Do you hope our minus one and substance saurian the Mord substituting the values two into one or zero point artisan 0.0.33 centimeter Waas one war three centimeter. My This one So so I&#x27;m trying this This part defying this part gives us here are to be the record 75 full centimeter.

Mayukh Banik · Answer

and object. Age not equals 1.5 centimeter high is held three centimeter from the corona. So the object distances three centimeter. It&#x27;s reflected images measure. Toby H I equals 0.167 What is the magnification? Magnification is H I over h hold on. That will come out to be 0.1113 Now magnification also equals negative. D I over dio so immense Distance happens to be negative magnification percio equals negative reported 113 times three equals negative. 0.339 Sandy, we don&#x27;t focal end is given by one of our deal. That&#x27;s one of a D. I in verse equals one of what three to us one of what negative 0.339 inverse equals negative 0.376 centimeter, which is equal to our over to. So we have our equals zero corned 752 centimeter with a negative sign. But we don&#x27;t use that

Problem

Pricey Stamp $A$ rare 1918 "Jenny" stamp, depicti…

Problem 88 Hard Difficulty

Answer

7.3 $\mathrm{mm}$

Related Courses

Physics

Related Topics

Discussion

Top Physics 103 Educators

Elyse G.

Andy C.

Farnaz M.

Aspen F.

Physics 103 Courses

Wave Optics - Intro

Multiple Aperture Interference - Overview

Recommended Videos

Watch More Solved Questions in Chapter 27

Video Transcript

James S. Walker

Physics

Elyse G.

Andy C.

Farnaz M.

Aspen F.

Wave Optics - Intro

Multiple Aperture Interference - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

7.3 $\mathrm{mm}$

Physics

Elyse G.

Andy C.

Farnaz M.

Aspen F.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Elyse G.

Andy C.

Farnaz M.

Aspen F.

James S. Walker

Physics