(II) Two lenses, one converging with focal length 20.0 cm and ...

Question

(II) Two lenses, one converging with focal length 20.0 cm and one diverging with focal length -10.0 cm, are placed 25.0 cm apart. An object is placed 60.0 cm in front of the converging lens. Determine (a) the position and (b) the magnification of the final image formed. (c) Sketch a ray diagram for this system.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A designer is creating a new telescope that has two lenses. The first is a converging lens with a focal length of 18.0 centimeters. And the second is a diverging lens with a focal length of negative 8.0 centimeters. And these lenses are placed 28.0 centimeters apart from each other. If an object is 45.0 centimeters away from the converging lens, where is the final image formed? And what is the magnification also create a ray illustration for this particular lens layout? Awesome. So it appears the final answer that we're trying to solve for for this particular prompt is we're asked to figure out what the magnification of this particular lens combination is. And we're also asked to create a ray illustration for this particular lens layout. Awesome. So with that in mind, let's look at our multiple choice answers to see what our final answer might be noting that all of our magnification values are a negative value. So A is 0.20 B is 0.50 C is 0.60 and D is 0.90. OK. So first off, let us write down all of our known variables. So we know the focal length of the converging lens, which we'll call this F one and F one is equal to 18.0 centimeters. We also know the focal length of the diverging lens which will call this value F two and F two is equal to negative 8.0 centimeters. We also know the distance between the two lenses which we call this value D and D is equal to 28.0 centimeters. And we also know we're told that the object distance which is the dis so it's the object distance from the converging lens which we'll call this do one and do one is equal to 45.0 centimeters. Fantastic. So now our first step is we need to calculate the image formed by the converging lens. So as we should recall a note, we need to use the lens equation for a converging lens. So as we should recall our equation to help us solve for the, the image formed by the converging lens. So once again, we're using the lens equation considering a converging lens. So we need to take one divided by do one plus one, divided by D I one, it is equal to one divided by F one where once again do denotes our object distance and D I denotes our image distance. So we don't get confused. Let's make I a little bit more curly. Awesome. Awesome. So once again, this is really important. Do is our object distance and D I is our image distance. OK. So now we need to rearrange this equation to isolate and solve for D I one. So when we use a little bit of algebra and we officially get D I one all by itself, we will find that D I one is equal to do one multiplied by F one, all divided by D 01 minus F one. Fantastic. So moving right along here, now we can plug in all of our known variables to. So for the numerical value of D I one, so D I one is equal to when we plug in our known variables, 45.0 centimeters multiplied by 18.0 centimeters. Fantastic. All divided by 45.0 centimeters minus 18.0 centimeters. Which when we plug that into our calculator, we will find that it's equal to 30.0 centimeters. Fantastic. So as we should note and recall the image formed by the first lens is 30.0 centimeters behind which I'll use a capital B, I'll put it in quotes. Capital B denotes behind the converging lens. So now our second step is we need to determine the object distance for the diverging lens. So the image formed by the converging lens will act as the object for the diverging lens. And since this image is 30.0 centimeters behind the converging lens, and the distance between the lenses is 28.0 centimeters. As we should recall and note the object distance for the diverging lens will be. So do two is equal to 28.0 centimeters minus 30.0 centimeters which is equal to negative 2.0 centimeters. Awesome. So note that the negative sign in this case indicates that the object is virtual, which will use a V for virtual or the diverging lens. Awesome. So moving right along here. So now our third step is we need to calculate the image formed by the diverging lens. So once again, we need to recall and use the lens equation. However, though we're dealing with the diverging lens, so we need to note that we're using the equation that states one divided by do two plus one, divided by D I two is equal to one divided by F two. Awesome. So now we need to rearrange this equation to isolate and solve for D I two. So we use a little bit of algebra, we will find that D I two is equal to do two, multiplied by F two, all divided by do two minus F two. So when we plug in our known variables. We will find that it's equal to negative 2.0 centimeters multiplied by negative 8.0 centimeters divided by negative 2.0 centimeters minus negative 8.0 centimeters. So that will mean when we plug this into our calculator, our final answer will be equal to 2.67. And of course, its units of measurement will be centimeters and this is when we round to two decimal places. So the final image form is 2.67 centimeters beyond the diverging yet lens. So beyond so behind B Awesome. So now we can officially start to solve for our final answer that we were ultimately asked to solve for. So we're trying to calculate the total magnification. Now, so as we should recall and note the total magnification is the product of the magnification of both lenses. So as we should recall, the magnification of each lens is given by the following equation which states that M one is equal to D I one divided by do one. And, and since we have to do for both of the lenses, M two is equal to D I two divided by do two. Awesome. So now we need to substitute in our known variable. So M one is equal to 30.0 centimeters divided by 45.0 centimeters and M two is equal to 2.67. So 2.67 centimeters divided by negative 2.0 centimeters. Awesome. So that will mean respectfully that M one is equal to 0.67. Awesome. And M two, when we plug that into the calculator or into our calculator, we will find that that's equal to negative 1.34. And this is when we both round both of our magnification values to two decimal places. So that will mean that ultimately, our total magnification will be that MT total magnification is equal to M one multiply by M two, which will mean when we take this and we plug in our known variables, that would mean that MT is equal to M one which is 0.67. And we need to multiply it by M two which is negative 1.34. That will mean that our total magnification value when we round to two decimal places to match all of our multiple choice answers. That will mean that our final answer has to be approximately equal to negative 0.90. And that's it. That is our final answer for the total magnification hooray. We did it. So now to do our diagram here, since we were asked to draw a diagram of the lens layout, I've gone ahead and put together a fancy diagram for us to refer to. So as we could see, we have our converging lens first which is shaped like a football and then going from the left to the right of to the right of our converging lens, we have our diverging lens which is shaped like an hourglass and we have our focal length F one which is equal to 18 centimeters. And then we have F two which is equal to eight centimeters. And then we have our object which is denoted by a black dot which is labeled O and then following, going from the left to the far right, we can see we have our image one and then our image two. So it goes image one first with a black dot and then image two denoted by a second black dot And from our diverging lens to our image two, it's 2.67 centimeters. And from our converging lens to our image one is F equals 30 centimeters. And then from our converging lens to our diverging lens, it's F equals 28 centimeters. And of course, from our object to our converging lens is 45 centimeters. Awesome. And then we can use our black bolded lines with our arrows to denote the ray path or the light path. So from the object, you can go up strike the converging lens and then it goes straight towards the diverging lens where they both have an apex where they meet where our images are. And that's it we've officially solved for this problem. Hooray. So looking at our multiple choice answers, the correct answer has to be the letter D which once again is negative 0.90. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Lens Formula

Magnification

Ray Diagrams

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