when-an-object-is-located-38-mathrmcm-to-the-left-of-a-lens-the-image-is-formed-22-mathrmcm-to-the-r

Question

When an object is located 38 $\mathrm{cm}$ to the left of a lens, the image is formed 22 $\mathrm{cm}$ to the right of the lens. What is the focal length
of the lens?

William Affel · Accepted Answer

all right. So for this problem we have an object that is 38 centimeters in front of a unknown lens lens of unknown focal length. And the image that forms on the right side of the lens is 22 centimeters away from the let&#x27;s. And so we&#x27;re wondering what is the focal length of the lens and the lens telling us, though, end up telling us what kind of lens it is. So let&#x27;s set up our lens equation one over f There&#x27;s gonna be equal to one over Dio plus one over D I. We can plug in our knowns. One over f is gonna be equal to one over 38 plus 1/22. We upload that into our calculator. One over f is gonna equal to approximately 0.0 7177 I&#x27;ve took the inverse of that f should equal about 14 centimeters. I noticed that we didn&#x27;t have to convert two meters. Use this equation on. We still found our final answer in centimeters

Adnan Gill · Answer

So we have a lens and we have an object in front of the lens and we have also got an image. And the question Saiz that behind off the image is half the height off the object. And also the image is important. So that means that the magnification is negative. 1/2 So way know that magnification is also equal to the negative. D I over to your and negative 1/2 is equal to d I over dio. What that means then, is that deal is equal to twice off d I. So if we say that this distance here is ah Dio, then this distance here is 1/2 off deal and if the question also says that the object and the image they are 90 centimeters apart, so do you hope that&#x27;s 1/2? New York is equal to 19 or 1.5. Deal is equal to 90 so Dio is equal to 90 divided by 1.5, which is equal to 60.0 centimeters. So that would be the answer to part a. D. O is equal to 60 centimeters. So now they know that this distance here is 16.0 centimeters, which means there&#x27;s distance. Here is early points, you know, centimeters. Now the next thing is we need to find the full collect. So for party, we will start off by saying one over F is equal to one over the I plus one over Dio or one over F is equal to 1/30 plus 1/60 and solving back for F G S F is equal to 20 centimeters.

Dading Chen · Answer

Okay, So this is the equation forties question here, which is one or F equals warm body parts were the I after still focal lengths is the object distance and the is the image of distant which undetermined. If we move while over the I to the left, I would have while the I used to go to our minds were already so are the eyes you, um the minus at over after So the I is equal to have the over demands half so well. We know the focal lens half is equal to 38.1 centimeter. And that object, this is the issue 60 bonos engineer. Therefore, we have the image distance D I is equal to 3 38.0 centimeter times 60.1 centimeter and over 16.1 centimeter, minus 38.1 centimeter. And this would give us d I, which is the image of distance. It&#x27;s about 104 centimeter

Mayukh Banik · Answer

the object size is irrelevant information. Over here, it is given that the object district&#x27;s equals 20 centimeter because the object is placed 20 centimeters in front of the ConvergEx dance. Now really images formed and the images 10 centimeter away from the dance. So evens distances given to be 10 centimeter with appointive sign because that is a really image. Now we will be using the thence. For me, though one over f equals one of our be all. Thus one of what d I. It was one of our 20 just one of a 10 that becomes 3/20 so focal that is found to be 20/3 centimeters on order 6.67 said emitted.

William Affel · Answer

All right. So for this problem, you have an object that&#x27;s sitting 31 centimeters in front of a conclave lens as a focal length of 39 centimeters. We want to know exactly what the image distance is and what the image magnification is. And in order to find these two things, we&#x27;re gonna have to use the image, uh, the lens equation. Excuse me and magnification equation. So let&#x27;s set up our lens equation. First, we&#x27;re going to say that one over f is gonna be equal to one over the image distance plus one over the object distance and a compliant or known values. We know the focal length to be negative 39 centimeters. We know the object distance to be 31 centimeters. We don&#x27;t have to use meters thes equations because they&#x27;re just ratio equations. So you centimeters and we&#x27;ll say that one over the image distance. It&#x27;s gonna be equal to one over negative 39 minus 1/31. And if we take this difference here or the summation rather and then find its inverse, we can say that d I is gonna be equal to negative 17 centimeters. Now we find the magnification will use the notification equation, saying that M it&#x27;s gonna be equal to negative distance of the image over distance of the object you can plug in are known values. So instead, modifications would be equal to negative negative 17 over the distance of the object. We&#x27;re just gonna be 31 and they will give us a magnification of 0.55 and these are two answers.

Adnan Gill · Answer

So we start with the Finland&#x27;s formula to find over New York just one over D. I. Isn&#x27;t that the one where F. So we&#x27;re told that the object is 18 centimeters from the lens and we do not know where the image is formed. However, we do know that the focal length is 12 and since it&#x27;s a diverging lens, therefore the focal, it will be negative. So we can now say one over D. I is equal to negative 1/12 minus 1/18 and that gives you the G. I is equal to negative 7.2 centimeters. So from here we can find the magnification, which is negatively. I over do York so negative off negative 7.2, divided by 18. And that gives 0.4. Now the question says that we want to introduce the size off the image by a factor of two, which means that our desires magnification. Let&#x27;s call that M Prime is equal to 0.4 divided by two, which is a 0.2. So the magnification that we&#x27;re looking for is a 0.2, and therefore magnification is which is negative. D I over deal. That means we can say that D I will be equal to negative is your point. Do deal? So now we&#x27;re going to use this set of values to go Go ahead and calculate the new value for Dio. So now we have the England&#x27;s formula Gangwon over Deal Plus one over the I is equal to one over F and one over deal yours what we&#x27;re trying to find. And we just said that the new D I would people to negative 0.2 times Dio is equal to negative 1/12. It&#x27;s still the same thing lens, so we can write this as one or GOP. We can factor one over deal and then that would be We will be left with one minus 1/0 10.2, and that&#x27;s equal to negative 1/12. So that gives us negative four. Ordeal is equal to negative 1/12 and that gives us Dio is equal to 48.0 17 years

Problem

An object with a height of 2.54 $\mathrm{cm}$ is …

Problem 70 Easy Difficulty

When an object is located 38 $\mathrm{cm}$ to the left of a lens, the image is formed 22 $\mathrm{cm}$ to the right of the lens. What is the focal length
of the lens?

Answer

As the focal length is positive the lens must be convex. Thus, the focal length of the lens is
$[14 \mathrm{cm}] .$

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James S. Walker

Physics

Andy C.

Farnaz M.

Zachary M.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

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As the focal length is positive the lens must be convex. Thus, the focal length of the lens is$[14 \mathrm{cm}] .$

Physics

Andy C.

Farnaz M.

Zachary M.

Meghan M.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Andy C.

Farnaz M.

Zachary M.

Meghan M.

As the focal length is positive the lens must be convex. Thus, the focal length of the lens is
$[14 \mathrm{cm}] .$

James S. Walker

Physics