a-if-you-treat-a-10-times-eyepiece-of-a-microscope-as-a-magnifying-glass-that-produces-an-image-at-i

Question

(a) If you treat a $10 	imes$ eyepiece of a microscope as a magnifying
glass that produces an image at infinity, what is its focal length? (b)
The microscope is designed for an image distance of 163 $\mathrm{mm}$ from the
objective, and the objective is marked $25 	imes .$ What is its focal length?

Nathan Silvano · Accepted 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.

Farhanul Hasan · Answer

all right. We&#x27;re given an objective lens with focal length of 12 millimeters objects placed eight centimetres from it, so object distance is 12 plus a look. Simon 0.8 point eight millimeters from its object distances. Truffles 80.8 just 12.8 millimeters on so image distance and part is just whenever focal length, minus one over object distance. So that&#x27;s when I was 12 minus one over 12.8, giving us image distance of 1 92 millimeters or 19.2 centimeters. Barbie. We want magnification and magnification of the objective lenses were after because the real image is formed by the objective lens. Okay, the eyepiece. We see our virtual image. So in this case, it&#x27;s just making a vest primal dress. And so that&#x27;s negative 1 92 millimeters over the 12.8. And so that gives us a 15 note. That magnitude is all we&#x27;re interested in. If we&#x27;re looking at just the orientation of the image, which we are in this case, I&#x27;m so in part. See, uh uh, um, the overall magnification is just linear magnification. Our objective. Len&#x27;s magnification times, angular magnification. And so this is just an absolute value. So that&#x27;s 15 M one, as we know is 25 centimeters over Focal Inc. Ah, F and F is 2.5 centimetres or 25 millimeters. So, um, so we get 15 times 10 and so the overall magnification is 150 times, and that&#x27;s it.

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.

Bruce Edelman · Answer

Okay, so we&#x27;re doing Chapter 34 problems. 59 here. So it says the focal length of an eyepiece microscope is 1.8 centimeters, and the focal length of the objectives 0.87 years separated a distance of 19.77 years. As all shown in this figure over here. So treating all the Windsors as Stine lenses part ay asks what is the distance from the objective to the object be viewed. So first, we need to realize that for this microscope system that final images at a distance infinity because that&#x27;s how we define is that so if we&#x27;re working backward from the park the second one&#x27;s case, if we plug in our equation, we see that the object distance is equal of the folk. Oi! So the object distance is 1.8 centimeters to the left of eyepiece. So this is object is also the image formed from the objective so we can set up next things. Let&#x27;s write this down. So I&#x27;m gonna call this image one who is not. The final images is the image between them. So we&#x27;re gonna call this right here. This this is Mitch Warden 1.8 centimeters away to the left of the second winds. And that makes it 17.9 centimeters to the right of the other leads. So this being object distance for the second case on this object distance for the second case on this, the image distance for the first case. So now you know where the image is formed from the first lens. We working backwards again. So we&#x27;re trying to get to where the object is so we can use our Finland&#x27;s equation. Rearrange for the object. So we&#x27;re trying to sell for this, which is s prime f over s prime this Beth. So we can put that in. So 17 point times 0.8, Teen Court one Ms is equal to 8.377 Oops. Sorry. That&#x27;s not centimeters that this actually we&#x27;re doing in centimeters. So we did the rest incident here, so keep that in. This is 0.837 seven years or 8.37 No meters. Cool. So that&#x27;s the distance. It is around 8.37 millimeters to the left of the first. So part B asks what is the, uh what is angular magnification of side. Wrong question. What is the magnitude of the linear magnification produced by the objective? So to do that, we obviously know we can remedy the image distance over the object distance here, So that is negative. 17 point now over point A 37 So we plugged that in, and that becomes a magnification negative. 21.4. Awesome negative because it is inverted. So part see asks what is the overall back angular magnification of the microscope? So from this we can we can use the equation 30 or 23 relating the total too linear of the objective in one. Call it in 12 times into where into is given as 25 centimeters over left. So school unclog Madden 25 centimeters over 1.8 centimeters. So this is 13.9. So that means our overall magnification is negative. 21.4 times 13 point. Not so if we plug that in, we get a negative 297 for a magnification. So yeah, this microscope magnifies roughly 300 times. It&#x27;s very, very good

Farhanul Hasan · Answer

all right the distance between the objective and the eyepieces. Given those 19.7 centimeters on DH, the focal length of the objective is 18 millimeters or 1.8 centimeters. And so to get the distance of which image formed from the lens, you have to subtract those too, and you get 17 points nine centimeters for the image distance we have focal length of 0.8 centimetres. Object distance is just one over point eight centimetres, minus one over 17.9 centimetres, giving us object distance of 8.37 millimeters or 0.837 centimeters. Great. And so that&#x27;s part a on part B. You just want the magnitude of the objective lenses magnification and so that&#x27;s just negative. Asked Prime over s on dso That&#x27;s negative. 17.9. Yes, this will be the absolute value. Negative 17.9 of the point 837 And so the magnitude of this magnification is 21.4 Finally in part. See, we have this formula that overall magnification is the many. A magnification of the magnet of the ah looks many a magnification of the ah objective times the angular magnification And so that&#x27;s 21.4 times 25 centimeters over F f being the focal length of the eyepiece. Remember that? So this is 21.4 times 25 centimeters, over 1.8 centimeters, and so that gives us on. So that gives us a negative to 97 time, and that&#x27;s it.

Problem

The medium-power objective lens in a laboratory m…

Problem 62 Medium Difficulty

Answer

a) 2.5 $\mathrm{cm}$
b) 0.652 $\mathrm{cm}$

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a) 2.5 $\mathrm{cm}$b) 0.652 $\mathrm{cm}$

Physics

Marshall S.

Farnaz M.

Zachary M.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Marshall S.

Farnaz M.

Zachary M.

Jared E.

a) 2.5 $\mathrm{cm}$
b) 0.652 $\mathrm{cm}$

James S. Walker

Physics