 In the previous video I used straight tracing to find the images of a conversion lens and the one of a diverging lens. What I used were the three principal rays, the ones that are parallel that go through the focal point, the ones that go through the focal point that become parallel, and the one that goes through the center undiviated. Here's another way of finding the image which is by using the thin lens equation form of us and we should get the same results as the ones I measured here for all the distances. So let's start with the converging lens. Converging lens, I can use 1 over P plus 1 over Q is 1 over F. Now what is P? P is the object distance. I measured this one here as being 0.8 inch and I put it as positive as it's a real object. So I can replace my P by 8 inch. My image distance is actually the one I'm going to be looking for and I measured it already as 2.8. I put it as positive because it was a real image on the right side. We should get a positive answer as well. The focal distance for the converging lens is plus 2 inch, plus meaning it's actually a converging lens. Plus 2.0. So now all we have to do is solve this for Q. So I have to take this one over to the left side. 1 over Q is 1 over 2.0 minus 1 over 0. Therefore, how click the image distance should be 1.0 minus 1 over 8.0. Let's see what that gives. It's divided by 2.7. How is it 2.7 inch? If I look at this, this is the answer and it matches the one I measured, plus 2.8. For the units in the lens equation, interestingly it actually doesn't matter if you use SA units or not. As long as you have been using inches, as long as you use the same units for all of them, for the formulas here, it doesn't matter. For the power, it doesn't matter. There you need to put your focal distance in meters to get your power in diopters. But for those up here, you can put whatever unit you want as long as you have the same units anywhere. So our result now is that Q is plus 2.7 inches. So let's see what we should get for the height of the image. The magnification of the lens is height of the image over height of the object, which is equal to minus my image distance over my object distance. So my magnification for this lens here should be minus Q, which was my 2.7, over my measured object distance, so 8.0. Which should be equal to, let's take a calculator, 2.7, and divide it by 8, 0.337, and this 0.34 should be equal to image distance over object distance. So height of the image, sorry, distance, height over the height of the object, which I measured here to be plus 2 inches, true. Therefore, the height of the image for my converging lens should be 2 times 0.34, which is equal to 0.675, so 0.68 inches for the height of the image. When I was measuring it, I got minus 0.8. Wait, did I forget minus? Yes, I did. There was a minus here, so there's a minus here, so there's a minus here, so there's a minus here, and here. What does that mean, the minus? The minus actually indicates that the image is flipped, so the object got flipped becoming the image, so it's going to have a minus. Now let's do the same thing for the diverging lens. So for the diverging lens, the formula is the same, 1 over P plus 1 over Q is 1 over F, the P being my object distance, which I had measured to be 8.0. My image distance is the one that I want to calculate, and my focal distance here was given as minus 2.0, 9 minus, the minus actually indicating that I'm using the left focal point and not the right focal point, meaning this is actually diverging lens. When I have minus and negative focal distance, I have a diverging lens, and for the positive focal distance, I have a converging lens. So same steps to solve it, so I get 1 over Q is minus 1 over 2.0 minus 1 over 8.0, so Q should be 1 over 0.0 plus 1 over 8.0, and the whole thing minus, just taking the minus out, we see what that gives me, 2 plus 1 over 8 gives me minus 1.6 inches. What I have measured was minus 1.7, so again, within uncertainty, I have the same value. What does the minus mean? The minus means I have a virtual image. So minus means for virtual image while here, when I have a real image, I have a positive object distance. For the virtual image, I get a negative object distance. Now let's look at the image height. So I don't think I need this anymore. So I'm using my magnification equation, so m is minus Q over P, which is equal to minus 1.6, so plus 1.6 over P, which was 8.0, which gives me plus 0.2. And now my magnification, which is plus 0.2, is equal to image height over object height. So I have 0.2 is equal to the image height, which I'm looking for, over the object height, which I've measured to be 2 inches, so 2.0. So 2.0 and 0.2 should give me an image height, which calculates as plus 0.4 inches. And what I have measured was 0.5, so I think I'm pretty close, which will give me uncertainty on the vast significant figure I have the same as. Here I've got a plus. We followed the diverging lens for that situation that we had, so the converging lens for the situation we had. We had a negative image height, which meant the image was flipped. Here I've got a positive height, meaning the images are bright. This is what I'm seeing also on my drawing.