predictcalculate-your-favorite-aunt-can-read-a-newspaper-only-if-it-is-within-150-mathrmcm-of-her-ey

Question

Predict/Calculate Your favorite aunt can read a newspaper
only if it is within 15.0 $\mathrm{cm}$ of her eyes. (a) Is your aunt nearsighted
or farsighted? Explain. (b) Should your aunt wear glasses that are
converging or diverging to improve her vision? Explain. (c) How 
many diopters of refractive power must her glasses have if they are
worn 2.00 $\mathrm{cm}$ from the eyes and allow her to read a newspaper at
a distance of 25.0 $\mathrm{cm} ?$

Nathan Silvano · Accepted Answer

Okay, So in this problem, your aunt can only see objects there near than 50 sink emitters from his eyes. So if she can only see image at this distance, that means she only produces the distance of the image is at 15. Same team enters from his eyes for here. Ice. Okay, but she&#x27;s trying to read a newspaper. So this is our object at a distance off 25 senti mentors. And we know she&#x27;s using glasses with given refractive our that. We don&#x27;t know yet. At a distance that&#x27;s called distance D off too. Sent emitters from his eyes. So she&#x27;s writing his glasses at a distance of two centimeters from his eyes. Okay, so the 1st 10 of this problem, the first I think of the problem, we want to discover if you if your rent is nearsighted or farsighted Well, since you&#x27;re in, can only focus, object? There are near his eyes. She needs to be. Let&#x27;s see the first item she needs to be year Site heads. Okay, so this is the answer to the first item. This second item What we want to discover in the second the second item we want to discover what type of glass is what type of lens your aunt needs to wear. Okay. So because divergent linds can produce image off a distance object your aunt needs to be using die virgin lenses. Okay? And the final item of this problem, we want to discover the refractive our off this glasses. We know that reflective our this equal to one divided by focal length. But we also know that the focal lens it&#x27;s we call to one divided by zero plus one divided by the I. So in order to Coca Lee, the reflective power, we just need to use the the zero and do I that we have here. But there is one problem. This distance are related to the glasses. But the problem gives is related to the eyes of the person off your rent. So we need to subtract here by zero, actually, by two, not zero. We need to subtract by two. It&#x27;s just subtract both bye to here by two in here by two. So let&#x27;s calculate this. We&#x27;re going to have one divided bye. Let&#x27;s put in meters because reflective power is in meters 0.2 five minor, zero point zero truth plus one divided bye 0.15 minors little 0.2 but remembered that we always needs to put a minor signing front or here or the rise. The calculation will be incorrect. So the answer to all of this it&#x27;s just going to be minus three point three for Diop. Tres So the refracting our off the glasses in 3.34 doctors and that&#x27;s defined arranged.

Adnan Gill · Answer

so be happy I and then in front of the I, we have the lens and the distance between the I and the lens is two centimeters. The object is 25 70 meters from the I. So what that means that is that d all will be 23 centimeters which is 25 minus 2 70 years. So we have Dio uh is equal to 23 centimeters. Now the eye itself can see objects which are 65 centimeters away, which means that the image should be formed 65 centimeters from the I which would then mean that d I would be equal to 65 take away too. So d I would be 63 centimeters. So now we can start using the Finland&#x27;s Formula One over the or plus one over g I is equal to one over f and the d I would be actually negative 63 centimeters because the object and image are both on the same side off the LEDs. So we have 1/0 630.23 changing centimeters. Two meters plus one over negative 0.63 is equal to one over f. And since one over f is the power so solving this equation gives power is equal to 2.8 die Achter

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.

William Affel · Answer

All right. So for this problem, we&#x27;re now holding our friends glasses a distance of 23 centimeters above a printed image. When we look into the last is unknown focal length, we see that the images direct and a magnification of positive 1.5. So we were looking for the focal length of these pair of glasses. So we&#x27;re gonna use the modification equation first and say that 1.5 equal to the negative image distance over the object distance burn uses to calculate the image distance were to say that negative 1.5 times the object distance. It&#x27;s gonna be equal to the image distance she actually this correctly image distance by plugging in Ah, 23 in for the objective, since the image isn&#x27;t should end up being negative. 34 my five centimeters. And now that we have the image, distance and the object distance, we can use that to calculate the focal length of the glasses. We can say that the lens equation one over F it&#x27;s gonna equal to one over dio most one over the I. That&#x27;s one over half. It&#x27;s gonna be equal to one over our object distance of 23 minus one over R image distance of 34.5. So then we should get a focal length of 69 centimeters. So the answer to our first question is the image really virtual will direct and has a positive magnification. So it&#x27;s a virtual image. What&#x27;s focal length of our, um, a pair of glasses when we&#x27;ve just found that 69 centimeters. And this is a pair of glasses, convex or concave? Well, because we have a positive focal length, it must be a convex pair of glasses, and those are answers.

Adnan Gill · Answer

So we are given the power off the lens, which is one of our f. So you know that one over f is equal to 166 die ocher Then this is her I This is the lens and to our two centimeter apart and the use paper she&#x27;s holding is 42 centimeters from her. I which means that deal is equal to 40 centimeters. So we have to you is equal to 40 centimeters, which is your 400.4 meters. And we need to figure out d I so one over Dio. Thus one over d I isn&#x27;t to one were f So what over d I is equal to one over F minus one ordeal. So one over d I is equal to 1.66 minus 1/0 point for solving that For d. I gets a c I A is equal to negative 1.19 centimeter, which means that the images for 1.19 meters from her I which would be negative 119 70 years. So the distance off image from I would be two centimetres mawr, so that would be 121 70

Nathan Silvano · Answer

Okay, So in this problem, you have your one Albert wearing glasses at a distance off too centimeters from his ice. And we know that your uncle has are near point both throwed 35 centimeters and the Farpoint off. 160. Same team enters. Okay, So first item of this problem, the first item we need to calculate to the refractive power off the glasses so the glasses can correct He&#x27;s my opium. So my okay means that he&#x27;s not able to see far distance. So the refractive powers define us one divided by F. And we know also that one divided by F is one divided by zero plus one divided by the I. Okay, so if we want to be able to see distance or projects, we can say that the zero is infinity and the guy that is, that the distance off the image is just a distance off the far point because we want to see far objects. So calculate in this refractive our needed to correct his vision is just one divided by Let&#x27;s put this in meters 1.6. And here is observation. This distance the distance. Hi. Distance off the image is related to the glasses, not to the eye. So we need to subtract here by zero point zero to it is the distance off the glasses from his eyes and always remember that the distance of the image is negative. So the answer to this is going to be minus zero point 63 Die. You hoped terse, And that&#x27;s the answer to the first item of the problem. This second item off the problem, we need to calculate that the object distance ah, which the image is strength is 35 centimeters. So basically, we went to calculate the distance off the object when we are and when we are the near point. So let&#x27;s see the distance off the objects using the fuel integration just going to be one divided by F minus one divided by D I. All this to the hour off minus one. We already calculated the one divided by F his minor 0.63. So this is just going to be minor. Zero point 60 tree miners one divided by zero point 35 miners, 0.2 Again we for minor sign in front. All this to the power off minus one. So the answer to this is just 41 0.7. Same team enters, so this is the answer. The second I turn off the problem and the third item of the problem, we actually have to considerate that distance. Toe Uncle&#x27;s Albert Ice. It&#x27;s too so intimate. Er&#x27;s longer than the distance to the lens. So actually, we need to put in here. Let&#x27;s see the true distance. It&#x27;s going to be the zero. We were already calculated well, US two centimeters. So it&#x27;s just what do you want? 0.7 plus two. That&#x27;s equal to 43 0.7. Same team enters, and that&#x27;s the final answer. Just a problem. Thanks for watching.

Problem

Predict/Calculate The relaxed eyes of a patient h…

Problem 38 Hard Difficulty

Answer

Related Courses

Physics

Related Topics

Discussion

Top Physics 103 Educators

Marshall S.

Farnaz M.

Aspen F.

Meghan M.

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

Marshall S.

Farnaz M.

Aspen F.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Physics

Marshall S.

Farnaz M.

Aspen F.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Marshall S.

Farnaz M.

Aspen F.

Meghan M.

James S. Walker

Physics