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An object of height 2.5 cm is placed 45 cm to the left of adiverging lens of focal length 30 cm. A converging lens of focallength 24 cm is placed 12 cm to the right...

Question

An object of height 2.5 cm is placed 45 cm to the left of adiverging lens of focal length 30 cm. A converging lens of focallength 24 cm is placed 12 cm to the right of the first lens. Findthe location of the final image. Specify its distance from thesecond lens and whether it lies to the right or left of the secondlens. Is the final image real or virtual, upright or inverted? Howtall is it?

An object of height 2.5 cm is placed 45 cm to the left of a diverging lens of focal length 30 cm. A converging lens of focal length 24 cm is placed 12 cm to the right of the first lens. Find the location of the final image. Specify its distance from the second lens and whether it lies to the right or left of the second lens. Is the final image real or virtual, upright or inverted? How tall is it?



Answers

A 1.00 - cm-high object is placed 4.00 cm to the left of a converging lens of focal length 8.00 cm. A diverging lens of focal length 216.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. Is the image inverted or upright? Real or virtual?

Here. We're trying to find the AMS Distin's D I. It was one of our F minus one deal. In words of that, it was one of over 20 centimeters, minus 1/25 centimeter inwards off that Equus 100 sent.

High In the given problem, there are the two lenses, the first one is diverging, So it's focal length will be taken as negative -12.0 Centimeter and the second lenses converging so its focal and will be taken as positive plus 12.0 centimetre. Now the gap between them, it's given us Just 9.00 centimetre. Hi dr object, which is scared to the left of the diverging lands. It's given us why And let it be why one And that is given us 2.50 Milli meter by one because there are two lenses so the height of the object for the first lands is assumed to be violent and the stance of object from this diverging lens is that would be excellent for the first lens, this is minus 20 centimetre as the object is to the left of their diverging lands. That's why it has taken as negative. Now, in the first part of the problem for the first glance, forced trance, which is diverging in nature using England's equation, which is a relation among the objective stance in its distance and the focal length. This is image distance from this first trans which is R. S. Mandate given by S. Mandresh -1 by S. one is equal to one by F. one. No plugging in alone values here one by S. Windex. The image distances missing -1 byte minus 20 centimeter is equal to one by upon which is minus 12. So rearranging the terms we get one by S. One dash is equal to -1 by 12 -1 x 20. So this is calculated to be as Mendes is equal to -7.5 centimetre and hear this negative sign shows that this image is to the left of first lens means if we show in a diagram in a rough diagram, the first one is on diverging lands And at a gap of at a distance of nine cm. This second one is to converging lens. This is the object actually stands off 20 centimeter. This distance is nine cm. Then the image is also being formed here to the left. Now this image will serve as the virtual object for the second lens. So the distance of this object for the second lines maybe 7.5 cm 9.00 centimetre. No, this image will serve as object for the second lens. So the object distance for this converging lens, S two And again it will be negative but 7.5 plus 9.00 centimetre. And how did we come out to be -16.5 centimetre And its focal land. Already we know this to be plus 12.0 centimetre hands using the Finland situation. Again, one by S two dash minus one by S two Is equal to one by F two. Now plugging in the known values again, one by S two dash that image distance from the second lines that is missing minus one by minus 16.5 Is equal to one x 12. Hence This one by S two will come out to be one x 12 -1 x 16.5. So finally this S two dash means that our distance of image from the second lens is calculated will be Plus 44 cm may be treated as one of the answer for this first part of the problem. So the distance is 40. So this distance from the first diverging lens so that he stands of image from the first cleanse will be This 44 plus gap between the two lenses, which was 9cm and it comes out to be 53 centimetre. No, in the second part of the problem, this image will be real as it is being formed on the right hand side of the convex lens. So we can say the image the final image is real. No, in the 3rd part of the problem. Using the expression for the linear magnification from both the lands for both the lenses. For first one this is S one dash by S. One and for second as this is a stew dash. Bye as to Therefore the net electoral magnification will be equal to and one in two M two. And that will be given by the height of the ratio of the height of the final wage. Do the post object by two dash bye bye one. Hence plugging in unknown values. This white two dash means height of the final image will be given by Why? One into M. One means S. One dash by S. One Into empty means. As to dash by S. two. So finally plugging in all known values. Why? When this is 2.50 Millimeter multiplied by S. one, Which we have found to be -7.50 centimetre divided by S. One which was minus 20 centimeter. Then for esta 44 cm, Divided by S. two which was -16.5 centimetres. So solving this, we get Why two dash to be equal to minus 2.50 millimeter only means the final image is having the same height as that of the object and their negative sign shows that the images in ward eight. Thank you.

Because this is the diverging thence. Focal length is negative. 15 centimeter image distance is given by five centimeter, so we will use the lance form. The one over ethic was one over object distance plus one over image distance. From here, we would find object. Distance equals one of the F minus one over the I in his distance in verse off that that comes out to be negative. Whether were 15 negative 1/5 in divorce in verse it was that made it can come outside. 1/15 just 3/15. It was negative. 15/4. Sending me did so. Object position Deal equals negative. 15/4 is 3.75 centimeter, which means this is ah, right to an object. Now magnification is given by immense distance over object distance absolute value that is five over 15/4. So that comes out to be 4/3. So magnification is also expressed as image height equals magnification, times object hide. So object Hide is image height divided by magnification. So that comes out to be too divided by the magnification. Becomes 3/4. So that is 1.5 centimeter then. Now we are changing the focus event to last 15. So we are starting part B. Now we're changing the focal lead. Two plus 15. So we had tried to find the image distance right now, and images distance will be one over F minus one of our d, all in verse of that which is 1/15. Thus once over 3.75 in verse and I'm going to use the calculator right now. I give up. Ah, 50 inverse plus 3.75 inverse, That comes out to be surprisingly nice. Three centimeter. So that is the image position now. And this is ah, really image. Really major three centimeter magnification comes out to be immense distance over object distance. So 3/3 0.75 Using that e mais height will come out. Toby magnification dives object height, which is 1.5 Sandy Mido We Jeez. 4.5 divided by 3.75 equals 1.2 centimeter

S. So this is another lens combination problem? I just did 45. So we have the two lenses and no one's converging. One is diverging. And so and they both have the magnitude of their The magnitude of their focus is 12. So we can say F one is 12 centimeters and have to would be negative 12 centimeters and they're nine centimeters apart. Which means first, you think it's Oh, the first is diverging. Mmm mmm, mmm. Mmm. So I'm gonna call this one, actually. Thank you, and I'll go ahead and draw it out. I think it's helpful to do that in these double lens problems. Just tow because you not can't just straightforward use the numbers from the calculations. You have to subtract or add based on where the first image is actually located relative to the second lens. So they're nine centimeters apart. So this distance is nine centimeters and and the object is two and 1/2 millimeters tall and its place 20 centimeters to the left. So let's do why is 0.25 centimeters and I cannot remember these numbers. I wish I could get better at it. Uh, but I can't maybe I will. Maybe I'll get better sleep or something and be able to look at a problem and memorize all the numbers. But it's probably not the best use of mental capacity. Okay, So great. And then we want to get the the how far from the first lines the final image off the final image. Okay, so, um so first we need to find the image created by this first lens. So and then that's gonna be the object for our second lens. Um, and so for part thistles, not really part a butt like the first part we want to find. Um yeah, the image of the first ones. So we want to use our usual thin lens equation whenever asked was one of us prime equals one over after. And you want to solve for us crime so you can bring the one over asked to the other side, and you can take the inverse of both sides. So one over as prime is gonna become one over is just gonna become s prime. And then this you could just raise to the negative first power. And now I'm gonna go ahead and plug this in But for now, I'm gonna use F one, so it's gonna be minus one over 12 minus one over 20. It's one of 12 inverted, and that's gonna be a negative 7.5. So I got negative 7.5. The negative means that it's Ah, I said fake virtual image because the, uh that's what the negative means. And so that means it's to the left of this lens by 7.5 inches. So it's actually this is so the object is, like, kind of way over here. I'm not gonna draw it. But then this is like the image. I'll call it the first image, and this is going to service the object for the second lens. So, um, let's go ahead and find out what images produced by this This image, the image from the first lens. So to do that again, we want to use it, then lens. And then I already derived the general formula for solving for s prime. So I'm just gonna quote it. So I'm gonna write out the numbers you usually don't. D'oh. So for this, our focal length is this focal length. So that's going to be won over 12. So I use the negative sign here. Yes, I did, um, And then we want to subtract one over us. As is gonna be this nine plus 7.5. Because that's the total distance it is from the images from the lens. So that's gonna be 97 16.5. And that's gonna be, um and I'm gonna take that to the minus one. So let's go ahead. Do that one over 12. Number 16.5 rising. That 15.5. Great. And so I got 44 so I got us Prime is equal to 40. Positive. 44 centimeters. So that means it's like to the right of this funds, so I'll go ahead and draw it. So this is the final image. Image number two and let's get its properties. So I mean, the 1st 1 I know, um, should be positive real quick. Um, that's right. Um, actually, I'm gonna pause this. Okay? So we want to get how far it is from the fight from the first lens. So we're given this. Gives us how far it is from the second line. So to get, we need to add 92 of that to get how far it is from the 1st 1 So I'm just gonna say Aye. Us prime, huh? I'm not sure what symbol to use. I'm just gonna not use any symbol. Try to reduce the number of symbols. So we wanna add 44 9 and we get 53 centimeters to the right. So that's your answer for a I'll go ahead and box this and what else is a really virtual I would say it is real because it's to the right of the system, huh? So go ahead, write down that you may confirm down here be riel Bie rial and who is probably gonna run space for? See? What's he asking? The Hyatt. Um, so we wanted to find the height. I'm just gonna use any page we want to use. Find the magnification. So magnification is minus s prime over ass. Where asked Prime is the position of the final image, and then s is the position they Luca s is the position of the object. So the object is positioned 20 centimeters. Okay, on. And Okay. So, Marshall, totally sure what to put for us Prime. But what I think I'm gonna do is find the magnification of the first image and then treat that is the height of the object. And then, um, is the sorry, the object for the second lens and then use that and then find a magnification. I'm not gonna give an overview. It's very confusing to describe this, so I'm just gonna try to maybe number my steps, so it'll hopefully make sense. So number one I want to find the height of the first image. So height of image one. So that's gonna be as prime over s for the first s prime over s so and the negative of that. So that's gonna be the magnification factor is gonna be 7.5 over 20 hopes. So, um, equals 7.5 over 20. And, um, therefore, that's height is equal to y. Prime equals sometimes. Why, and that is equal to the highest 0.25 So, um, what we want to do is multiply a 0.25 times, 7.5, divided by 20. Okay, so that's 7.5 divided by 80. So that's gonna be 0.9375 Okay. Okay. And then I'm gonna take that. So that's our height of a new one of the new object. If we're gonna find the magnification of the second lens so that's gonna be oh minus s prime over s. And then this s is gonna be the position of the first object with respect to the first lens. And then this s prime is gonna be the position with respect of the image. With respect to the second lens, so s is equal to 16.5 and us prime. We got waas, I got, uh, 44 and then we want to do a negative of that. So then we get that it's magnified minus 44 over 16.5 and then that's minus 2.67 Okay, so that's gonna flip this image. And so then therefore, our final y prime, Um, it's gonna be minus 22.67 times point. Owen, I am three 9375 So calculating. That got minus 0.25 so it should be inverted and side. Oh, I'm really sorry. If the notation is confusing, you're gonna kind of have to listen to what I say. I think if you get confused. I really don't have the time to, like the time allotted to come up with a completely self consistent, like notation system. Um, so yes. Sorry for any confusion. Um um okay. And yeah, this is a really image, and we know that it's inverted because of this negative sign, and then this is the final height.


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