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.