predictcalculate-refering-to-example-27-6-suppose-a-persons-near-point-distance-is-670-mathrmcm-a-is

Question

Predict/Calculate REfERING TO EXAMPLE $27-6$ Suppose a person's
near-point distance is 67.0 $\mathrm{cm} .$ (a) Is the refractive power of the
eyeglasses that allow this person to focus on an object just 25.0 $\mathrm{cm}$
from the eye greater than or less than 2.53 diopters, which is the
refractive power when the near-point distance is 57.0 $\mathrm{cm} ?$ The
glasses are worn 2.00 $\mathrm{cm}$ in front of the eyes. (b) Find the required
refractive power for this person's eyeglasses.

Nathan Silvano · Accepted Answer

so similar to the problem. 106 We have a person wearing glasses at a distance off, too. 70 meters from hair eyes and we have, ah, the distance of the object this person can see off 25 centimeters from hair ice. We have to suppose that the near point distance near point distance is going to be 67 70 meters and again referring to the problem 27. For example 27.6. We need to explain if the refractive power off the eyeglass they&#x27;re allowed this person to focus this object is going to be greater than or less than two points. 53 Diop doctors. Okay, so again, look into the problem. 27 6 Since we know that because the near point is farther than off the example the refractive our off, this lance&#x27;s should be greater then two points 50 tree Diop tres Because we need to look, the objects are a long distance. So if the distance is farther, the reflective power also going to need to be greater and we can&#x27;t win now In the second item, we need actually to calculate this value. So how are we going to calculate the refractive power. We know that the refractive hour is just one divided by F. But using the finger ends equation, we know that one divided by F. It&#x27;s just one divided by the eye, one divided by zero. So we just need to calculate the distance of the image and the distance off the object. But remember, this is the Finland&#x27;s equation, so these distance is related to glasses and not to the ice. So the distance off the object that we have in here off 25 sentiment er&#x27;s we need actually to subtract off the distance off the glasses and the same to the near distance, because in your distance is just the distance of the image, plus the distance of the glass to the ice. So the distance of the image we can calculate from here. So let&#x27;s move on. We know. Then that&#x27;s the distance of the image is going to be minus 67 miners, too, which gives us miners 65. Same team enters so we can calculate this. Now we&#x27;re going to get that The reflective power is equal two, one divided by minors. You know 10.65 waas one divided by zero point 23 which is the distance of the object. When we subtract the two centimeters, that gives us a total off two points, 81 day after&#x27;s. And that&#x27;s the precise, exactly value off the reflective power of these glasses. And as you can see, it is actually greater, then 2.53 doctors, and that&#x27;s your final answer.

Nathan Silvano · Answer

Okay, So in this problem of persons running less with refractive our off two point 75 i officers Ah, the glasses are being used at a distance. The off too. Saying team enters from her eyes and we produce image. When we see all project at a distance off 25 centimeters from here, I Okay, so first of all, what&#x27;s we need to calculate in this problem? First of all, we need to explain referring to the example 27 6 Um, we need to explain if the refractive power, if the near points off these glasses is going to be greater or less than 57 centimeters. Okay, So first of all, because we are if referring to this example and the refractive power off this lens is greater, then the reflective power off the example 27 27. That&#x27;s last six. The near point off this person is going to be greater Damn 57 17 meters. That&#x27;s the first answer. Now we have to calculate the near point of this person properly. So let&#x27;s begin thinking that we to calculate the near point, we&#x27;re gonna need the feelings of creation for the distance of the image. So this is just going to be one divided by half miners, one divided by zero all to the power off minus one. But first of all, let&#x27;s remember that the distance off the object is always related to the glasses and not to the ice. So we need to subtract the distance off the glasses. But we just truce ing team enters, and now we have the proper distance off the object. So let&#x27;s calculate this. You&#x27;re gonna get, uh, two point 75 miners, one divided by zero point 23 out to the power off minus one. This is gives us minus 62.6 ST Emitters. So the near point is the distance of the object, plus the distance of the glass. So the near point is just 62.6 plus to sing team enters. That&#x27;s gives us 64 0.6 and team enters. So we prove that the first answer is correct because the near point is greater than 57 centimeters and we calculate precisely what is this value? So thanks for watching

Nathan Silvano · Answer

Okay, so in this problem, it would. We have to look to the example 27 4 because in the example, we have a person with ah Farpoint. Let&#x27;s call F one off. Three 23 Same team enters This person is wearing glasses at a distance off to ST Emitters in front of his eyes, and the glasses has a focal left. Let&#x27;s go F one off. Let&#x27;s see miners 321 minus 321 Same team enters. Okay, Now we have to consider a second person, this time with a Farpoint off 353 saying team enters the same distance off the glasses. The glasses are used the same years it&#x27;s in distance, and we have to explain if the focal left off the glasses of this second person is going to be greater equal or less. And if the first person Okay, so first of all, because we have a longer far points to the second person, the image does not need to be move closer, but rather far away. So since we need to move the image farther away, we need a focal length that is bigger. So have two needs to be greater then f one. That&#x27;s the answer to the first item. Now we have to calculate the actual value off this focal lens. So let&#x27;s think how we calculate the focal lens is with the if England&#x27;s equation. So we&#x27;re gonna have to because one divided by d I. Plus one day one divided by the zero one If I did buy the zero and since we need to move this, we need to be able to looked distance objects. We want to see the objects that infinity So we can see that d zero is plus infinity and the I is just the Farpoint plus the distance off the glass in front of his eyes. Negative. So this is just going to be a good to the power off minus one. So let&#x27;s calculate this. We&#x27;re gonna have, uh, a focal length that is equal one divided by E i, minus one, divided by infinity, all to the power of miners. One. We know that division by infinity is zero. So this is simply their focal. Lent is just minus 35 one ST Emitters, which it&#x27;s bigger than treat two. Let&#x27;s see here 321. Same team enters. So we proved that our first answer is correct. And that&#x27;s your final answer. Thanks for watching.

Nathan Silvano · Answer

to solve this problem. To solve this problem, we&#x27;re gonna have to look to the example 27 4 But in this problem we have a person that were glasses with a focal lens that&#x27;s putting here or focal lens off minus 301 cent team enters and the glass are years at a distance off two 70 meters in front of his eyes. Uh, this person and focus objects that infinity. Okay, so let&#x27;s see what the problem once. So we have to explain if the Disperse on Farpoint is going to be less equal or greater than 323. Okay, Wow, If we look to the problem their example, we can say precisely that the Farpoint let&#x27;s put itin A. Therefore point is going to be less then 323 sent emitters because they smaller focal length brings the image closer to the eye. So the focal lens off the example is miners 321 So, since our focal length is smaller than into the example, we gotta need a far put, we&#x27;re gonna have a far point. That is also last entry 123 and that&#x27;s different part of the grass chin. Let&#x27;s move on to the second part where we&#x27;re gonna need to calculate the Farpoint. We know that the Farpoint is the distance where the images produced, plus distance off the glasses to the eye, which in this case is just too. So let&#x27;s calculate the distance off the image related to the glasses. This is going to be one divided by F miners, one divided by zero, all to the power off miners. One. So since we want our focus objects at infinity, we have a D zero off infinity. So I&#x27;m gonna have one divided by half miners, one divided infinity. This is precisely zero. So we can say that the distance of the image is equal to the focal length, which is equal to minors 301. Same team enters. Therefore we can see that they&#x27;re Farpoint off. This person is just 301 but less too. 300 er entry, same teenagers. That&#x27;s the final answer. Next reward

Nathan Silvano · Answer

Okay, So to answer this problem, we need to know that the eyes of a patient in focus on distance objects when they are relaxed in this condition the refractive power off the eyes minimal So before calculated anything we know that when the eyes are relax it the objects that we can see are in the infinity. So the problem tells us that the distance from the retina where the images produce is just two point for Santy mentors, let&#x27;s rewrite. This is just true point for same team enters. Okay, So if we want to know if the refractive power the problem gives us is for a person nearsighted or farsighted, we need to calculate supposing that this person is far sighted, she can see really far. Okay, so let&#x27;s think about we know that the Finland&#x27;s a creation disperse, um is just going to be one divided by f equals one divided by the zero blows, one divided by D I. Well, the zero is infinity, so we don&#x27;t need to consider so one divided bye f. It&#x27;s just going to be equal to one divided by 0.0 24 So this is close to 41 0.7 Diop doctors. So if ah far sighted person, when the eyes are relax Onley produce a reflective power off 41.7. So this person cannot be a far sighted person. This person needs to be near cited person. So the answer to the first itin of the problem is near sighted. Okay, So what? As we have to cope elite we need to calculate differ, point off this patient Well, we know that the Farpoint needs to be d zero. So using different lens equation forth, e zero we have one divided by F minus one divided by All right, All these to the hour off, minus 11 divided by F We already know that is the refractive power off this patients before you with five minus one, divided by 0.2 for all this power of mine is run. So the answer. It&#x27;s just going to be 15. Same team enters and that&#x27;s defined Our answer

Nathan Silvano · Answer

Okay. So in this problem we have a librarian that has a near point off 0.5 and meters and a five point off five meters. Okay? And we know that he cannot see distance closer to 0.5 our distance farther off five meters. He can only see between this range. And we need to design a by focal glasses with two different refractive powers in order to correct both his near sightedness and farsightedness. So let&#x27;s begin the problem. The first item off the problem. We need to tell what kind of lens we&#x27;re going to use to correct his near sightedness. Well, near sightedness is when the person can only see close objects. You can only he cannot see distance objects. So to direct this kind of problem, we need Die Virgin lens to correct his near Sinus because he reproduce image of a distant object. Ah, he&#x27;s uncorrected Farpoint. And to correct his farsightedness, we&#x27;re going to need a comb virgin lens because the correct the converging lens I used to correct the vision off people who can on Lee see distance objects and cannot see clothes closer objects. So now let&#x27;s calculate that the refraction, the reflective power, the reflective power off each one of those lenses. So let&#x27;s see, he ordered to correct the Farpoint. The actual Farpoint of this person is five meters Now. We need a Farpoint so this person can be able to see objects at infinity. And we know that the refractive power is just one divided by F and the actual Farpoint of this person when he produced the image is at five meters. But remember that distance off the image always need to be related to the glasses and not to the ice. Five meters is related to the ice, so we need to subtract the distance of the glasses off 0.2 Now we have that reflective power is just one divided by the zero plus one divided by the I. This is going to be one divided by infinity, plus one divided by miners. Because this is this is of the image owner converts converging type of lens five miles, you know point 02 So finally we have the reflective power off the off. The first lance is going to be miners zero point two Diop tres. Now that&#x27;s calculated the reflective power or the converging lens to correct the near point. We also have one divided by F one, divided by zero plus one divided by the I. The zero is ready. Images now farm. So is the near point. The standard near point is 25 centimeters, or 0.25 meters. Again, we need to subtract the distance off the glasses. Zero points 02 and the distance off the image is just 0.5 minor, 0.2 So putting the values for Miller we have 0.25 mile, 0.2 minus one divided by minus 0.5 miners. You know, points, you know to and this gives us are effective. Our sports in here. Refractive power fraud ical virgin lands off positive 2.3 Diop tres And that&#x27;s the final right.

Problem 107 Hard Difficulty

Answer

a) greater than 2.53 diopters
b) 2.81 diopters

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James S. Walker

Physics

Elyse G.

Marshall S.

Zachary M.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

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a) greater than 2.53 dioptersb) 2.81 diopters

Physics

Elyse G.

Marshall S.

Zachary M.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Elyse G.

Marshall S.

Zachary M.

Jared E.

a) greater than 2.53 diopters
b) 2.81 diopters

James S. Walker

Physics