predictcalculate-you-have-two-lenses-with-focal-lengths-f_1260-mathrmcm-and-f_2204-mathrmcm-a-how-wo

Question

Predict/Calculate You have two lenses, with focal lengths
$f_{1}=+2.60 \mathrm{cm}$ and $f_{2}=+20.4 \mathrm{cm} .$ (a) How would you arrange
these lenses to form a magnified image of the Moon? (b) What is
the maximum angular magnification these lenses could produce? 
(c) How would you arrange the same two lenses to form a magnified image of an insect? (d) If you use the magnifier of part (c) to view an insect, what is the angular magnification when the insect
is held 2.90 $\mathrm{cm}$ from the objective lens?

Nathan Silvano · Accepted Answer

Okay, so in this problem, we have two lenses. The first lens has a focal length off to a point six. Same team enters, and the 2nd 1 has a focal lens off 20.4. Same team enters. And in their first itin of this problem, we have to predict how you we would arrange this lens to have a magnify image of the moon. Okay, so remember that this is the first item I think a Remember that to get her magnify image, we have the folk Oh, objective. Divided by the focal. I piss and always remember that the focal objective needs to be bigger. Then the focal eyepiece. That&#x27;s how we built a telescope. So since we have in this problem that I have to is greater than F one, we can say for sure that the telescope has objective equals the F two, and we have the focal I p&#x27;s You closed the F one and that&#x27;s the answer off the first item. The second item. We have to calculate the magnification. Okay, so we already know how to calculate. It&#x27;s just dividing objective by eyepiece. So we&#x27;re gonna get here 20 point four divided by 2.6. This is going to be equal to seven point 85. And that&#x27;s the magnification off this telescope. The turn itin his detergent itin. In this item, we have to rearrange the lens to Butte a microscope. And in the microscope, we have the inverse of the telescope we have that the objective needs to be smaller, then the eyepiece. So here we have objective equals F one and the piece it close F too. You just need to remember that the microscope works the opposite way off the telescope. Okay. Ah, Now we have to calculate the distance off the image off this off this microscope. So what&#x27;s pushing here? The distance off the image we know using the feelings of creation. The distance of the images. Just one divided by the objective miners one divided by zero, all to the power off minus one. So we&#x27;re gonna get the I equals one divided by 2.6 miners. One divided by two points. Nine oh, to the power off miners. One, This is equals two 25 saying team enters and now we can calculate the Soto magnification. Actually, the last thing we need to calculate the magnification. So let&#x27;s circulate in here. It&#x27;s Toto magnification defined by miners. Do I times and divided by the focal objective and the focal I piss. So this is just miners 25 times 25 divided by 2.6 times 20.4. So this is minus 12. Same team enters. I think you manage 12. This is the magnification. We don&#x27;t have unities. So the magnification is just miners 12 and that&#x27;s the final answer.

William Affel · Answer

All right. So for this question, we have two lenses at our disposal, one lens of a focal length of positive, 40 centimeters and another lens of focal length of negatives. 40 centimeters are questioning. They wanna answer is if we&#x27;re trying to project an image onto a wall that is some distance away from the lens Which lens should be used? And this is a really easy question. Answer. Because all we have to figure out is, uh, which lenses con Ah, vex. Excuse me. Do you want the converging lens? We want the ones that will be able to create a real image. Diverging lenses are able to create real lenses. There are real images. They&#x27;re only able create virtual images. So the positive 40 centimetre one is the con vex lends, and the negative 40 centimetre one is the con cave ones. And we know this because when we set up our lens of quite equation, conclave lenses always have negative focal lengths. So you want the one with the positive focal length the convex lens, because that&#x27;ll be able to create a real image that we can project onto a wall. So we want the one with the positive 40 centimetre with lens deposited for Senator your full length, that is our ones. So now that we know we want this lens, next time you want to know is where do we place the lens and we&#x27;re to create an image that has a magnification factor of two. So it doesn&#x27;t mean we have a magnification of to. It means that we&#x27;re increasing the images of size by a factor of two, whether it&#x27;s real or virtual. Now, we already know that we want to create a riel image, so we know where modifications will be negative. So I would say our modifications gonna be negative to create a real image with a mask, a factor of two, and that&#x27;s going to equal to the negative image distance over the object distance. Now we can drop the negatives from the two and the you have negative image distance and say two is equal to humans. Distance over the object distance, using the fact that our focal lengths 40 centimeters we set up our lens equation over 40 centimeters is gonna be equal to one over dio most one over d i to solve for either D or D I can use this system of equations here. So what we want to solve four is ah di I The distance of the image since that will tell us where we want to put our lens. So I want to say to Dio is gonna be equal d I and we&#x27;re in a substitute out dio for 1/2 d I was a dio it&#x27;s gonna be 1/2 d I So we&#x27;ll say 1/40 is gonna be equal to two over d I That&#x27;s just me substituting in 1/2 d I infer one over Dio plus one over the I don&#x27;t give us three over d I. So now we can say that 40. It&#x27;s gonna be equal to D I over three in the image distance most thusly be 120 centimeters. So this is where replace our lens because it 120 centimeters away from the wall that were to create an image that has a magnification factor of negative too. And that&#x27;s our final answer

Nathan Silvano · Answer

Okay, so and this problem we have a student with to magnify glass. Correct? Yeah. The student has two lenses and the first lens as a focal end off five centimeters. And the second chance as a focal length off 30 dirty ST emitters. So five centimeters and 13 70 metres. Right. Well, there first itin off this problem. We need to explain which off this lances we&#x27;ll provide, um, the greater magnification. Okay, So first of all, we know that the magnification gration for magnification just one plus and divided by f So they lens there has they&#x27;re smaller. Focal lens is going to provide the greater magnification. So a smaller focal lens is F one. So following this theory, F one should be the greater magnification. And what is the second thing we need to calculate? And here, the second item sporting here the second ison. We need actually to calculated the magnification and we know that the near point distance is 25 20 meters. So to the glass one that&#x27;s portend blue. A glass one. We have magnification one off, one plus, but five divided by five. This is just going to be six to the glass too. Think less true. We have magnification to with one less hopes. One plus 25 divided by 13. This is going to be a good to point knife, and we prove that our answer was correct.

Nathan Silvano · Answer

Okay, so and this problem, we have, uh, convex lens. Use it as a magnifying less. We know that this lens has, ah, focal lens off. 15. Let&#x27;s see, 14 think emitters. 14 ST Emitters and their first item off the problem. We went to calculate the magnification m when the image is an infinity. So if the image is that infinity, we can say that the distance of the image is miners infinity. And we know from the chapter that the magnification toto magnification is just and divided by F when the image is a infinity. So in is near point off the off the lens, and the problem tells us that the near point that&#x27;s put in here then your point is just 20 five centimeters and the focal ends is 14. You already know. So this needs to be equal to 1.8. So this is the magnification off this lens. Okay, a 2nd 1 what we went to calculate, we want to calculate how far the object needs to be held from the lens. So we went to discover the zero. Okay, so let&#x27;s calculate this. The zero using the feelings equation just going to be one divided by F minus one divided by the I. All these to the power of minus one. We know that the distance off the images at infinity So this is simply one divided by f the power off minus one, which gives us 14. Same team enters. Oh, sir. So this is the end of the final answer to the first item. We have magnification off 1.8 and the distance of the object off farting sing team enters the second in the second item. We need to calculate the magnification when the image is that near point. Well, let&#x27;s see, I can be the magnification when the image is that the near point is just one plus and divided by F. So this is just going to be one plus 25 divided by four things. So the magnification is just hopes. Sorry. So the magnification just going to be 2.82 point eight of magnification And what as we need to calculate, we need to calculate the object distance. The object distance indecent area, just one divided by F minus minors to minus one divided by the distance of the image reaches at the near point miners near point. So this is going to be one divided by 14. That is the focal int miners, one divided by minors, 25 the distance of the focal aunt. All of these to the power off minus one. So this needs to be equal to eight point 97 Same team enters, and that&#x27;s your final answer. This problem.

Mayukh Banik · Answer

here We have, you know, much much. Get a bad if not equals 36 centimeter. And that&#x27;s why we consider that the object distances infinity on. That&#x27;s why kill not equals f You&#x27;re from here. We can find that the objective is for positive. The eyepiece us the distance we&#x27;ve been there now for President of the eyepiece comes out between 36 minus do. It was four centimeter. So we have. But the idea is to be negative force Saturday because this is an

Daniel Alva · Answer

in this problem, we have a set off four lenses with focal length off one serum ear. Things were a mirrors 57 mirrors in 83 meters. And in the first part of the problem, we want to compute. Where is the largest value off the under magnification off a telescope that we can make with his Ted off lenses? Well, remember that a telescope uses influences. One is the over give, and the other is the eyepiece lenses in order to form the finally much off the object in the value off the younger magnification for a telescope is yes, no. But if off the focal length of the objective over the vocal in off the Ivies, So in orderto making this magnification big its largest value, we reduced the largest value for the objective length India&#x27;s smallest value for the I b there that we can see from the relation the younger magnification increase us the focal length off the object in Greece and also in Greece, us the focal length of the ideas degrees. So we&#x27;re gonna take for your gift The largest body that is 80 centimeters importante IBC the smallest value that is 13 years. But this is my note. 80 meters. I&#x27;m Tory, not serum era. Just miners A. In the second part of the problem, we went to know which lens we used for the objective in which, for the i. D. S so we can see from the previous cocoa Daschle that for the object we used on and off 80 they&#x27;re a mirrors importante IV&#x27;s well in with. Well, they&#x27;re a mirror. Oh, for Colin. In the last part of the problem, we want to compute. Where is the distance between the list the lenses? And we must remember that, you know, let&#x27;s go the distance between the Nance&#x27;s Yes A Just that some on the values of the Foca blend this So we can&#x27;t say that these dancers between the lens you just the objective class d I these for Collins and this is just davey it thus one nearest So this is 81 7 years

Problem

The eye is actually a multiple-lens system, but w…

Problem 86 Hard Difficulty

Answer

a) $f_{b}=f_{2} \quad i f_{e} y=f_{1}$
b) $=7.85$
c) $f o b=f_{1} ; f_{-y}=f_{2}$
d)$=-12$

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James S. Walker

Physics

Liev B.

Farnaz M.

Jared E.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

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a) $f_{b}=f_{2} \quad i f_{e} y=f_{1}$b) $=7.85$c) $f o b=f_{1} ; f_{-y}=f_{2}$d)$=-12$

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) $f_{b}=f_{2} \quad i f_{e} y=f_{1}$
b) $=7.85$
c) $f o b=f_{1} ; f_{-y}=f_{2}$
d)$=-12$

James S. Walker

Physics