predictcalculate-microscope-objective-a-is-labeled-15-times-and-objective-b-is-labeled-25-times-a-wh

Question

Predict/Calculate Microscope objective $A$ is labeled $15 	imes$ and
objective $B$ is labeled $25 	imes$ (a) Which objective has the longer focal
length? Explain. If the image formed by the objective is designed
to be 16.3 $\mathrm{cm}$ from the lens, calculate the focal length of the
(b) $15 	imes$ and (c) $25 	imes$ objectives.

Nathan Silvano · Accepted Answer

So in this problem we have two microscopes, uh, objectives. And the first objective has a magnification. Let&#x27;s go and a off 15 times. And the second objective has a magnification. Let&#x27;s go and be of 25 times. Both of the objectives are beauts to work at a distance that&#x27;s called D I off 16.3 ST Emitters. Okay, so there first I think of the problem, we need to explain which of the objectives has the greater focal length. Well, to answer this, we just need to remember that the magnification of the objective is inversely proportional. So the focal lens off the objective. So if we have a shorter for Holland&#x27;s going to have are greater magnification and vice versa. So the answer to this one has to be m a actually, that sports objective, eh? Objective A. Because objective, eh is the shorter for the shorter magnification. So there needs to be the greater focal length. Well, now the second item, we actually have to calculate that the focal objective length left. So let&#x27;s see to the 1st 1 that&#x27;s call this one deep. Hey, it&#x27;s just going to me f A course to minors 16.3 divided by minus 15. So this is equals to 1.1. Think Team enters and the focal lens off the B 2nd 1 It&#x27;s Toby Miners 16.3, divided by minus 25. So this is zero point 60 65. Same team enters as we can see. The objective, eh? Has the greater for Colin, and that&#x27;s the final answer.

Dading Chen · Answer

All right. So the question was asking us what is over our magnification. So the overall magnet, the magnification is equal to, uh, ish magnet magnification motive about each other. And we know the magnification for the objective Lens is 62 the magnification for the eyepiece is 10. So over magnification is equal to, and it was a 62 times 10 which is 6 20 So this is the overall magnification. And the question be was asking us to determine, uh, each for Hollande&#x27;s. Okay, so the four hole ends for the eyepiece is equal to 25 centimeters over the magnification, okay. And this will give us increase. That F one is the focal. An eyepiece. You give us F one equals or 25 centimeters over 10 which is 2.5 centimeters. Okay. And the focal lens were they on? Objective lens is a bit different here. Okay, That said, the magnification in these cases m zero for the objective plans and this would give us amateur was equal to B, which the image distance over the focal lens off objective bench was just activities case. So I have to will be equal to be over m zero. Okay. And member, it was saying that objective lands and the eyes are 16.5. Sending meter apart is the senate. You&#x27;re sending me there apart, which means that he which is the image distance, is equal to leave apart distance minus the focal lens of the eye. Just f one here and this will give us the image. Distance is go to 16.5 centimeters minus 2.5 seven years. So be, which is the image. Distance will be equal to 14 point all centimeters. So the four hole lands for the objective lands will be going to B, which is 14.0 centimeters overly magnification for objective lens which is 62. And this would be bust 0.22 six centimeters. And this is the focal lens for that objective. Bless for the last question. He was asking us what is the objective distance? Okay, so we know that one over f is equal to one or V bus. What was you affects The focal lens B is the image distance you is the objective distance objected. So you can be goto we have over reminds at so the focal. As we were using his case, he&#x27;s, uh, have to which is still the focal lengths off the over jacket plans. Okay, it&#x27;s given has 14. I&#x27;m I&#x27;m sorry. Schema as 0.226 centimeter. Okay. And we know the image distances. 14 going all same years. This will give us our object. Distance is equal to he does act, which is 14.0, centimeters times, joe. 0.2 to 6 centimeters over B minus half, which is 14.0, centimeters minus zero. Going to 267 years. This will give us the object. Distance is equal to, uh, double check. It was Ah, about 0.23 centimeters. Okay. And these are solution for this question. Thank you.

Nathan Silvano · Answer

So the first item of this problem, we want to calculate it. The focal length off eyepiece lens of a microscope, knowing that this lens has a magnification off 10 times the object. And what as we got, we know that as a standard, we can always use a near point as 25 same team enters, it is a C. So if we want to calculate it for Cole, and I just need to remember off the magnification equation, which is and divided by F when the image is at infinity, which is the case, So this is just going to be the focal. And if we want to calculate it, equal to end divided by mm. So it is just 25 divided by 10 so they focal ends is 2.5. Hopes five. Same team it there&#x27;s and that&#x27;s the answer to the first I 10 of the problem. The second I turned off the problem, I couldn&#x27;t be the eye and be We want to calculate the focal and off the objective lens. The objective lens this time well, calculate the focal length of the objective. We can always remember the formula, which is minus D all right, divided by the magnification off the objective, the D. I is just the distance off the image to the lens, and the problems tells us that this distance is just 163 Millie mentors, and the magnification is 25 times. So let&#x27;s calculate. This is just going to be minus 16 tree divided by 25 miners. 25 actually, because is this standard that the magnification for the objectivist negative? So this is just going to me 6.5 millimeters, And that&#x27;s the answer to the second item of the problem, and that&#x27;s all thanks for.

Dading Chen · Answer

All right. So the question was asking us what is the image distance for this question? So we know one over half is evil. One over. Reports of all over you use the object distance. Okay. And since you were saying that the I&#x27;ll just to review is placed two point only made her beyond a focal point. Which means that the object isn&#x27;t is equal to the focal lands class the beyond distance, which is 10 point on May meter. Last two millimeter, you could trawl name here. Okay. And the focal lens off the objective lamps is 10.0, millimeter. And I&#x27;m sorry. Here should be Evel on the wide right. Heavy. So the image distance we can we go to you task fo divide by U minus fo, which is 12 millimeter times time pulling. Oh, you familiar? Over 12 millimeter. Minus 10.0, millimeter. And this will give us 60 millimeter for the image distance. So the object the objective lens will form is first image as 60 millimeter. Okay, So whole question be it was asking us what is the overall magnification. Well, we know the over minus magnification is equal to 250 million mirror Tosti image distance over, uh, the focal lengths off. The objective times the focal lens off the eyepiece. So if we like no value, we&#x27;ll get em. Is equal to 250 being manner Times 60 millimeters divided by the focal length off the objective lands, which is tempering. Oh, I am here times the focal length off the eyepiece, which is 15.0, millimeter. And this will give us the over magnification is equal to 100. Okay, so these are the solution for the question. Thank you.

Mayukh Banik · Answer

this test between the lancers is given by you knocked us f e. It was 64 5 dust. We weren&#x27;t a it was 18 19.3 brandy. Now so does bang. Education is even by negative off l over if not lives. And over f e ik was negative is not minus if not over. It&#x27;s not our objective science and over or me It was negative off 16.5 minus 7.5, divided by five guys 25 centimeters over 2.8 centimeters. So we have magnification equals negative to 86 Now, in order to find the object distress and yet be not it was one of, if not by a lot of Are you not? It was off. That equals one over five millimeters minus one of 1, 16.5 centimeters, 1 65 m In less of that, it was fight. Born 16

Nathan Silvano · Answer

all right. In this problem, we have ah, medium power. Lentz. A medium power objective lands at their heads a focal objective. Let&#x27;s write this down. We have a focal, our objective off three point 75 Millie mentors. And we have a magnification off miners. 50. Okay, the problem wants to discover, but she did. Problem wants to discover the working distance of these objects, events and the focal length of the eyepiece that will provide overall magnification off miners. 1 25 Okay, so let&#x27;s think about the first part of the problem. We want to discover the working distance. The working distance is just Let&#x27;s put in here. The height and a working distance is just the distance of the object related to the focal objective lens. So we know that we have the focal objective and we have the magnification. If we use the feelings equation, we&#x27;re going to get one divided by half. Miners equals one divided by zero, plus one divided by the eye. The problem here is that we do not have the distance of the image. But we have the magnification, and we know from chapter 26 that magnification is just miners. The I divided by the zero. So from here, we can take that D I is equal to minors. Mm. Multiply by the zero substituting in dis a creation we have one divided by f equals one divided by f equals one divided by the zero miners. One divided by mm The zero. Therefore, we can put one divided by the zero in evidence That&#x27;s continuing here. So we&#x27;re gonna have one. Divided by f equals one divided by zero the multiplies one miners, one divided by m. And from here we can take that d zero. The distance of the object is equal to the focal objective. Multiply Wang miners one divided by m and we can calculate this just going to be three point 75 the multiplies one miners one divided by minus 50 and this is equals two three point a tree millimeters. That&#x27;s the answer to the first. I tend of the problem. Now for the second part of the problem, let&#x27;s think about the total magnification. Get Toto. Magnification can be right. As the magnification of the objective multiply by the magnification off the eyepiece. So we know that the magnification of the eyepiece is just the near point, divided by the focal lens off the I piss and they stunned her. Near point off a telescope microscope is always 25 cent team enters so we can say that their focal eyepiece it&#x27;s just going to be M zero times and divided by Toto magnification on the problems is that the total magnification is just miners 125. So we have minus 50 times 15 25 actually divided by miners. 125. This is going to be equal to 10. Same team enters and that&#x27;s the focal ends off the eyepiece, and that&#x27;s all Thanks for watching.

Problem

A compound microscope has an objective lens with …

Problem 59 Easy Difficulty

Answer

a) objective A
b) $1.1 \mathrm{cm}$
0.65 $\mathrm{cm}$

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James S. Walker

Physics

Christina K.

Farnaz M.

Zachary M.

Aspen F.

Wave Optics - Intro

Multiple Aperture Interference - Overview

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a) objective Ab) $1.1 \mathrm{cm}$0.65 $\mathrm{cm}$

Physics

Christina K.

Farnaz M.

Zachary M.

Aspen F.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Christina K.

Farnaz M.

Zachary M.

Aspen F.

a) objective A
b) $1.1 \mathrm{cm}$
0.65 $\mathrm{cm}$

James S. Walker

Physics