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Predict/Calculate REFERING TO EXAMPLE $27-4$ In Example $27-$$4,$ a person has a far-point distance of 323 $\mathrm{cm} .$ If this personwears glasses 2.00 $\mathrm{cm}$ in front of the eyes with a focal length of$-321 \mathrm{cm},$ distant objects can be brought into focus. Suppose asecond person's far point is 353 $\mathrm{cm} .$ (a) Is the magnitude of thefocal length of the eyeglasses that allow this person to focus ondistant objects greater than or less than 321 $\mathrm{cm} ?$ Assume theglasses are 2.00 $\mathrm{cm}$ in front of the eyes. (b) Find the required focallength for the second person's eyeglasses.

a) greater than 351 $\mathrm{cm}$b) $-351 \mathrm{cm}$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

Simon Fraser University

Hope College

University of Sheffield

University of Winnipeg

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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'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's go F one off. Let'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'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's the answer to the first item. Now we have to calculate the actual value off this focal lens. So let's think how we calculate the focal lens is with the if England's equation. So we'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's calculate this. We'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's bigger than treat two. Let's see here 321. Same team enters. So we proved that our first answer is correct. And that's your final answer. Thanks for watching.

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