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(a) Determine the distance from lens 1 to the final image for the system shown in Figure $26-71 .$ (b) What is the magnificationof this image?

(a) see work(b) $-0.12$

Physics 103

Chapter 26

Geometrical Optics

Wave Optics

Rutgers, The State University of New Jersey

University of Washington

Hope College

University of Sheffield

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All right, So for this question, we have a dual lens system where the first lines of the football length of negative seven centimeters Second lands is a football length of 14 centimeters. The distance between lenses is 35 centimeters and the distance of the object in front of the first lenses 24 centimeters. And what we want to know is what is the distance from the first lens to the final image? In order, find out where I have to use our lens equation a couple times. We'll have to find the distance from the first lens to our first image and then use that as our object for the second lens and then find the distance of the second image from second lens and then do some math in order to find the distance from the first ones to that final image. So we want to do is use the for lens equation first for the first lens, so we're gonna say that one over negative seven is going to be equal to one over 24 are object distance plus one over our image distance, and if we do their calculations correctly, we should get an image distance of negative 5.4 centimeters so I can write that down and that should be placed right about here. That's, um de I now image of our the distance of our first image from Lens one. And that's gonna be also d o prime. That is where our object is for our second lens. So this is going to be equal to Well, actually, I take that back. This is the location of our objects for our second ones. But dio prime or the new Miracle Distance is gonna be equal to D l minus d i. And because d I is negative, the negatives will cancel out and we should get an answer off 40.4 centimeters. This is Dio prime or the distance of our new objects in front of the second Latin's this distance there. So we're to use that to find the distance of our second image in from the second one's. But he's in lens equation again. We're gonna say 1/14 the focal length of our second lens is gonna be equal to 1/40 0.4 plus one over d I prime we were using Dio prime. And if we do our calculations correctly, D I prime should be equal to 21.47 years. Now if you want the distance from the first lens to D I prime we got remember that 21.4 centimeters is the distance from the second lens to the image That place is it right over here is his d i prime. So you say that the total distance D is gonna be equal to the distance between the lenses this distance plus de I prime the distance from the first ones to the image of second ones to the image. Excuse me, and that should be equal to the distance from the first ones to the image, and that will give you a numerical value of 56 0.4 centimeters. But the next thing I want to know is what is the magnification between the original object in our final image? D i prime. In order to find that out, we're simply going to have to use the magnification equation twice. We can say that the total magnification em some tea would be equal to medication, cause we're the first ones multiplied by the magnification caused by the second lens. So what is the medication caused by the first lines? Well, it's going to equal to negative. D I over dio plug in the values that we know de eyes negative 5.4 centimeters. The negatives will cancel out. Will get 5.4 divided by idea, which is 24. You should get a magnification value off zero point 2 to 6. So it's M one. Let's calculate them, too, and she was gonna be equal to negative. D I prime over dio prime. We'll plug in our values d I prime we said Waas 21 Point Force will have negative 21.4 over Dio Prime we calculate to be 40.4 and that gives us on em to value of negative 0.53 If you multiply these two numbers together and one times two, we should get a total magnification of negative 0.12 And that's our final answer for the night

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