Repeat Exercise 34.41 using the same lenses except for the fol...

Question

Repeat Exercise 34.41 using the same lenses except for the following changes: The second lens is a diverging lens having a focal length of magnitude 60.0 cm.

Accepted Answer

Welcome back, everyone. We are making observations about an optical apparatus here. Now, we are told that it is composed of a convergent lens, L one with a focal length of millimeters and a diverging lens L two with a focal length of negative millimeters. Now, what we are going to do is we are going to place an object with a height of four millimeters 25 millimeters to the left of our L one. And we are tasked with finding two separate things here. We're tasked with finding what is going to be the final image or the location of the final image produced as well as the height of the final image produced by the optical apparatus. Now, before we get started, I do wish to acknowledge our answer choices in the center of the screen here, these are going to be the values that we strive for. So without further ado let's go ahead and begin. Well, in order to find our values here, we are going to use this equation that relates the object's distance, the image distance and the focal length. So we have one over the distance or position of the object plus one over the position of the image is equal to one over our focal length. Now, what I'm gonna do first is I'm going to figure out what is going to be the image position and the image height after just going through our convergent lens L one. So let me go ahead and change colors here for that. So rearranging our equation up top right here, what I get is that the position of our image after going through the first lens is going to be the position of the object times the focal length of lens one divided by the position of the object minus the focal length of lens. One. Now what does this give us here? Well, this gives us 25 times divided by 25 minus 20. And what is this going to give us? Well, this is going to give us 100 millimeters. Now, in order to find the height of our image after lens one, we are going to need to know the magnification constant here. So how do we find that for lens one? Well, all we do is take the negative of the image height or the image position divided by the object position. So this is negative 100 divided by 25 giving us negative four. Now that we have found that M for lens one, we can go ahead and find the position or sorry, the height of our image after lens one So this is going to be M one times Y one which is equal to negative four times four, which is equal to negative 16 millimeters. Now here is going to be our key here. Our image of lens one will be the object of lens two. So what does this mean? Well, let me change colors here. Again. What this means is that the position of our object for lens two is going to be the distance between our two lenses minus the position of the image of object one. So what does this give us? Well, we're told that the two lenses are 100 and 50 millimeters apart, then we're gonna subtract 100. And this gives us that the position of our object for lens two is 50 millimeters. Similarly, we are going to have that the height of our image of object one or for lens one is going to be the height of our object for lens two, which is just equal to negative 16 millimeters right now that we have those values. Let's go back to our original equation. What I'm gonna do this time is I'm gonna rearrange it one more time. Now, this time solving for the image position of our object for lens two similar to the first half of our problem. This is going to be S two F two divided by S two minus F two. Now, what does this give us? This gives us 50 times negative divided by 50 minus negative 20. What this gives us is negative 14.3 or 14.3 to the left of our diverging lens. Great. So then in order to find the height of our final image or the height of our image after lens two, we once again need to find that magnification value. So this is gonna be the negative of the position of our object for lens two position of our image apologies for lens two divided by the position of the object for lens two. So what this is give us this gives us negative negative 14.3 divided by giving us 0.286. So now finding our final image height, what we have is M two times Y two which is equal to 0.286 times negative 16, which gives us negative 4. millimeters or 4.58 millimeters. And the image is inverted. So we have found both the final position of our image as well as the final height of our image. And both of these answers correspond to our final answer. Choice of B. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.

Repeat Exercise 34.41 using the same lenses except for the following changes: The second lens is a diverging lens having a focal length of magnitude 60.0 cm.

Key Concepts

Lens Types

Focal Length

Lens Formula

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