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. A certain microscope is provided with objectives that have focal lengths of $16 \mathrm{mm}, 4 \mathrm{mm}$ , and 1.9 $\mathrm{mm}$ and with eye pieces that have angular magnifications of $5 \times$ and $10 \times$ Each objective forms an image 120 $\mathrm{mm}$ beyond its second focalpoint. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable.

a) 641.6b) 42.5

Physics 102 Electricity and Magnetism

Physics 103

Chapter 25

Optical Instruments

Electromagnetic Waves

Reflection and Refraction of Light

Rafael Guillermo G.

September 28, 2021

1

Cornell University

University of Michigan - Ann Arbor

University of Washington

Hope College

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All right. So for this question, he was asking us what is the largest? Oh, yeah. Okay, so we know. Um, I just want to care about a value off magnification. Looking is simply pull absolutely on em here. I'm here since for the by an indication which is equal, absolute value minus V over U, which is equal to absolutely be over you. Okay, he's the image distance you is the object distance. So in this question, you're saying that each object lance's forming an image 1 20 millimeter Beyonce second before. Okay, So the image distance form by the objective lance is much larger when compared to their local. And this only happens when after a distance is nearly equal to the local. That's awfully. Let's. Which means that we basically need to consider the object Distance is equal to the pope away. Okay, so now we have Sachin question with absolute value and equal to be over. So now this star plugging all different focal lens and the image of distant to find out different met the education number. Okay, So well, just one of those you guys know image distance here is equal to a distant par. Let's say L plus the focal eyes. Okay. And in a question, you were saying it is 120 millimeters? Yeah. So it's 100 20 millimeter, because so whatever focal lens you have. Okay, the first let's focus on the bubble ends off 16 year. So we got em along is equal to Are you gonna put absolute value? That is totally fine. OK, we'll be, Which is 1 20 millimeter plus 16 year over 16 millimeter. And this will give us the body was about 8.5. Okay, so for the four homelands with, uh, four million, we will have a magnification number off the over AB wishes 1 20 No mere plus or millimeter over four millimeter. And this will give us 31. What a magnification number, Okay. And the magnification number fall the vocal ins off 1.9 one point. I mean, here will give us We would just want on the millimeter plus 1.9 millimeter over the vote plans with this 1.9. No, I'm here, and this will give us about 64.16 Okay, so now we know that comparing with all its focal lens. The magnification number is the largest one in Foca. Lance is 1.9 millimeter. Okay, Just here. So this is the largest. Okay. Remember the question was saying that eyepiece have been angered. England magnification of five times and 10 times is objective for okay. Which means that you want to master largest. Making the vacation number. Waiting to measure eyepiece was 10 times like a depiction number with the focal lens off 1.9 millimeters, which is will give us an magnification. I'm a 64.16 So the largest magnification number this a ah, Norwood, you say a max means Okay, So the largest negative begins number we go to m three times 10. Okay, this will give us 64.16 times came, which is 641. 26. Okay, so this is the largest magnification number. Okay, Full question be it was asking us what is the smallest over our angular magnification, uh, number. Well, now we know that making the application number for the focal lands off. 16 millimeter is 8.5. And for Oakland's off four millimeter is there. You want the focal lens on the magnification number for a local s over 1.9 millimeter is the largest, which is 64.16 And we know the eyepiece. This a, uh, em fire em for I just have to defer making a vision number, which is five. And tank. So in order to get a smallest one, we need to find a smaller, smallest pair from all these magnification numbers. We find out this in this this morning's want to match. Okay, so if you plugging the value, Okay. Mm. Maimon, Which means this morning's one is equal to and war times M for which is a point by times by and this will give us 42 point by. So this is the smallest over angular magnification number working game. Okay. And these are the solutions for the question. Thank you.

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