a-pair-of-eyeglasses-is-designed-to-allow-a-person-with-a-far-point-distance-of-250-mathrmm-to-read-

Question

A pair of eyeglasses is designed to allow a person with a far-point
distance of 2.50 $\mathrm{m}$ to read a road sign at a distance of 25.0 $\mathrm{m} .$ Find
the focal length required of these glasses if they are to be worn
(a) 2.00 $\mathrm{cm}$ or ( b ) 1.00 $\mathrm{cm}$ from the eyes.

Nathan Silvano · Accepted Answer

Okay, So in this problem, we have a person wearing glass at a distance diferent his ice. We know the disc glasses have a focal land such that we can see or project at a distance off 25 meters from his eyes. And these glasses produces an image of his far point. We know the Farpoint is just 2.5 meters because the problem gives us his information. Okay, so what we have to doing here, we need to calculate the first item, the focal length when we are using the glasses at a distance off too 70 meters from for for his ice. Okay, so we know that the distance off the object in the distance off the image needs to be calculated related to the glass is not to the ice. So since were, since he is using his glasses on a distance of two centimeters, we need to subtract zero point you two in here and 0.0 to in here. So, using the Finland&#x27;s equation, which is one divided by zero plus one divided by the eye Oh, with the power of minus one, you can calculate now. So let&#x27;s do this. The focal length. It&#x27;s just going to be one divided by 25 miners, 0.0 tooth plus one divided by 2.5 minus 0.2 Remember Now we have to put a minor sign in front of the distance of the image on the rise. The calculation will be incorrect. All of this to the armor of minus one. Okay, so let&#x27;s see. This result gives us minus two point 75 meters. That&#x27;s the answer. So the first item, the second item and we need to calculate again the focal length. But this time we are using the glasses are a distance off once in team manager, away from his eyes. So I was going to be this same formula. But this time, if Alito change here 25 miners 0.1 loss, one divided by 2.5 minor zero points. You don&#x27;t want from minor signing front. All this to the power off my nose Swan. So this needs to be equal to minus two point 77 meters, and that&#x27;s final answer through this problem

Keshav Singh · Answer

the corrective lens must form an upright virtual image at the near point of the I. That is Q easy code of minus 60 centimeters, in this case, for objects located 25 centimeters in front of thigh P is equal to 25 centimeters. So from the Finland&#x27;s equation, one of a P that&#x27;s one of a que is equal to one over F, we can find the focal length off the corrective lens. So if from above, with a bit of mathematics is they put two p Times Cube overpay plus Q. And we know these values. So that&#x27;s 25 centimeter. So he suppressed units, yet times minus 60 centimeters over 25 centimeters, minus 60. Center in. Yeah, once we get the focal length off the lens to be positive 42 nine centimeters, It&#x27;s another way of the focal length we can work on the power. Our fill ins P is equal to one over the focal, Um, meet us so that&#x27;s one. And the focal length is meters zero 0.4 to 9. Me, too, instigate power to be positive to 0.3 doctors

Netra Sharma · Answer

we have a near point off a person, which is 60 centimeter. And if the person wants to be able to see an object from a distance of 25 centimetre, he&#x27;s gonna have to use a corrective. Lance. So first of all, 25 centimeter is gonna be the object of sense, right? 25 centimeter. And if you he uses the corrective lis, the ways of virtual image has to be form, actually, at the 60 centimeters distance at the near point. And that means Q one has to be negative. 60. I will just remind you, this negative sign means that they mazes in front of the lens and his first rule and upright. Okay, so from here, we can actually find the focal length of the lengths that he needs to use. Okay, So all we need to do is use the lens equation for a part, ay, one of her F because one of her p plus one of our cue So this will give us this is going to give us f because Q duck p over que plus p right. So that means negative 60 times 25 over negative 60 last 25. So this is going to be 42.86 Senator. Okay, so what&#x27;s this? This is the focal length. Focal length is 42.86 inhibitor. I&#x27;m just gonna write it down in terms of meter. Zero point for two eight meter. Now for part B. We just used the definition of the power. Okay, so power is actually reciprocal of the focal length. This the capital P. By the way, power is reciprocal of the focal length, and that is going to give us 1/0 0.4 to 8 meter, which is gonna give us 2.33 dieters.

Adnan Gill · Answer

so for party, the distance off the image from the lens would be negative. 67 minus 2.2 67. Because that&#x27;s her off your point and minus 2.2. That&#x27;s the distance between the lands and her eyes and negative because it is a virtual image, so negative 64.8 centimeters. Similarly, the object would be the distance off object from the lens would be 25 minus to fight, too, because the distances given from her eye. So we have to subtract the distance between the I and the lens, so that&#x27;s equal to 28 22 point hit centimeters. Now we can use the England&#x27;s formula that says one over M physical 21 over GOP plus one. Already I school one over F is equal to 1/22 10.8 minus one hour, 64.8, so solving that for F gives F is equal to 35.2 centimeters for partly Similarly, we find the I that would be cool to negative off 67 minus 3.3, cause now the lens is warned. It is 3.3 centimeters from the eye, so that would be negative. 63 point seven centimeters d o will also change accordingly. So 25 minus 3.3, which is equal to 21.7 centimeters. And again we can use the mainland&#x27;s Formula One over air physical toe one over D four plus one over T I, which is equal to one over 21.7, minus 1/63 0.7, so that gives F is equal to 34.9 70 meters.

Kristela Garcia · Answer

So you&#x27;re eyeball acts like a lens. And in this case, we wanted to determine what the focal length off this lens is for an object that&#x27;s infinitely far away at 500 centimeters and at 25 centimeters, if each of these air going to produce an image at a distance of 2.17 years. So to do this we can use the mirror or lens equation, which is one over d o plus one over D. I equals one of her f again, if you have for part A something that&#x27;s infinitely far away, actually, we can go straight to. So it just focal. So notice here that I&#x27;m in dressing the fraction on the right hand side, which means we do have to inverse the entire left hand side. So this becomes one of her d o plus one over t I. That way we actually come out of step for ourselves. Now we can go ahead and plug in my numbers. So this is one over one over a very, very large number infinity. So as this fraction gets infinitely as this number gets infinitely large, the entire fraction becomes infinitely small. Basically zero And if our images 2.1 centimeters away, then our focal will be 2.1 centimeters. Now we can actually apply the same equation, just plugging in our difference distance of the object. So for part B, I have 1/1 over 500 plus 1/2 0.1 gets us a new focal of 2.9 centimeters. And for part C one over 25 plus 1/2 0.1 gets us eight new focal of 1.94 And so your eyeballs actually changing its focal lengths to adjust for these different distances.

Netra Sharma · Answer

So in this problem, the near point. Often I is 75 centimeter. So for part A If we want to find the corrective lens to be able to see at normal near point of 25 centimeter, then when the object distances that normal near point was his 25 centimeter or 0.25 meter. Then the Emmys distance has to be at negative. 75 centimeter negative. 75 centimeter. Okay, which is negatives 0.75 meter. No. The image of the reason why you see this negative here is because the inmates form is going to be first of all and upright. Right? So from here, we can actually find the power of the list. That corrective Lis is going to be won over F because one of her p plus one of a cue this is going to give me que plus p over cute RP and this is gonna be negative. 0.75 plus 0.25 over negative 0.75 times 0.25 and this would be positive 2.67 doctors. So, in part B, the person who made the list mess it up. Okay. And the Liz had different focal. And instead, and he was able to see the person who had the new point at 75 Centimetre was able to see on Lee at the object like the near his near point was at 26 centimeter instead of being 25 centimeter. So we want to find out what the error in power was. All right? So when the object, his near point was at 26 David, then that means this is 0.26 meter than the actual power of the lens Waas, one of the F, which is going to be que plus p over Cube Duh p. I&#x27;m gonna call this P prime and set. Okay, So that one and now p prime is 0.26 meter. So from here, we can actually see that this is going to be positive. 2.51 Dieters. Okay. Now, to find the error Delta P, which is the error in the power, it&#x27;s going to be the absolute value off p minus p prime, which is going to be 2.67 minus 2.51 And this is going to give us your 0.16 doctors. So the power was sirrah 0.16 die after slower then the person who designed the lists actually wanted it to be.

Problem

Predict/Calculate Your favorite aunt can read a n…

Problem 37 Medium Difficulty

A pair of eyeglasses is designed to allow a person with a far-point
distance of 2.50 $\mathrm{m}$ to read a road sign at a distance of 25.0 $\mathrm{m} .$ Find
the focal length required of these glasses if they are to be worn
(a) 2.00 $\mathrm{cm}$ or ( b ) 1.00 $\mathrm{cm}$ from the eyes.

Answer

a) $-2.75 \mathrm{m}$
b) $-2.765 \mathrm{m}$

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a) $-2.75 \mathrm{m}$b) $-2.765 \mathrm{m}$

Physics

Liev B.

Farnaz M.

Jared E.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Liev B.

Farnaz M.

Jared E.

Meghan M.

a) $-2.75 \mathrm{m}$
b) $-2.765 \mathrm{m}$

James S. Walker

Physics