Let's now use the graphical ray traces we've done to derive fundamental equations which describe the single lens imaging system. These are the thin lens equations. It turns out, this problem is simple enough. It's described by a small number of variables and a small number of equations. We can basically learn everything there is to know about it. And we can derive all of that from these rays we've just learned how to trace. This will be the basis for building understanding a more complicated systems. So we've already done magnification by using that single first ray that goes from the tip of the object through the center of the lens into the tip of the image. And we see the two similar triangles allow us to define the ratio between the image height and the object height as the magnification. And similar triangles tell us that that's also given by the image and object distance ratio. Remember your sign conventions here. This magnification is a quantity less than zero indicating that the image is inverted from the object. And that's because the object distance here is a negative quantity, as is the image height. So now let's go on to more interesting, or the next set of equations. So, the thing we want to know now is, where is the image? We've already figured out how big it is, now we want to know where it is. We have two different equations, which are convenient to learn this. And we'll start with Newton's form, because it's a little easier to derive from the ray trace. In this case, the distances to the image and the object are defined from the focal point. Well call these z, in contrast with the variables t we used before. So -z is the distance from the object to the focal point. In this case as demonstrated, it's a negative quantity because the object is to the left of the focal point. And z' is the distance from the focal point to the image. So now, we will use rays 2 and 3 here, this line, and this line. And we see that these are similar triangles. What that allows us to do, as I show here coded in blue for these, is discover that the ratio of z to y, the object height, is also given by the ratio of f to this distance here, which is the image height. That's this equation here. Similarly, working on these two triangles, we see that the distance to the image, defined in Newton's way, and the image height is given by the ratio of, in this triangle, the focal length to this distance which is the object height. So, if we take those two ratios and rearrange them just slightly, we see that the magnification pops out. So we get a relationship between the object distance, the focal length, and the magnification. And this is worth remembering, by the way. Sometimes this version of the formula is handy. We get a similar formula for the image distance relative to the focal length and the magnification. And if we multiply those two together, we get a single equation which gives us a relationship between the object distance, the image distance, and the focal length of the system. This is the less common version of the thin lens imaging equation, but it's a handy one to have. And remember, the sunken image again. The z and z' are potentially negative quantities, which is why even for a positive focal length f, we have a minus sign here. Now let's derive the more common thin lens equation, this is due to Gauss. In this case, we're going to use the lens, the object and image distances that we derived before, or defined before. These are the object to the lens and the lens to the image with, again, our sign conventions, which I've noted here by putting a minus sign in front of t. So, it turns out we can derive the equation we want with these new variables from Newton's equation by a couple substitutions. First, we simply relate z, remember the object to focal length, to t, the object to lens through the obvious relationship. And then we use one of those intermediate relationships we had before which relates to z', the focal length to image distance to t' and t, the two distances here. So if we substitute those two things in to Newton's form of the equation, we get an equation which now only has t, t', and f. And a tiny bit of algebra results in what's usually simply called the thin lens equation. This relates the focal length of the lens to the two object and image distances. If you look this up on Google, you will see a very similar form of this equation. But there will not be a minus sign here, so be aware of that. The reason for that is, is most simple physics textbooks or physics websites don't use the sign convention that we're using here. And they don't have to because they only are dealing with one lens in sort of illustrating how it works. We're going to have many lenses in sequence. We're going to have virtual objects and images. Remember, we've always, already seen, for example, if t becomes positive, that simply tells us that instead of a real object, we have a virtual object. We need this sign convention in order to keep track of those sort of things. And therefore we have to pay the price that this thin lens equation is not the one you will see in physics textbooks. It is the one you will see in optics textbooks, because optical engineers need to keep track of all these stuff. Now, for simplicity, I have derived all these equations assuming that we have vacuum in front of and behind the lens. It simply is more convenient and easier to do than for example, having this ray bend. Because this boundary where the lens is is also an index of refraction change. But we're going to need that sometimes. So, let me note that if you go through and include an index of refraction change between materials here, then the equation is slightly modified. We're going to see this form basically everywhere, that you can take the distance t and takes its ratio with the local refractive index. And then you get exactly the same set of equations. So, most of the time, we use this one, but sometimes we need to remember to include the refractive index. Let's pause just for a minute, and define a quantity which, while it's not central to the concept of single lens imaging, is actually quite important for understanding the power that gets through a system. This is formerly called radiometry. If you remember, one of the questions that we wanted to ask of geometrical optics was how bright is the image? This would tell you how long to leave the shutter open, for example, on your camera. In this new quantity, it's really important for understanding that question. The quantity's angular magnification. It's an analogy to the positional magnification. We use that so often and we just call it magnification. And remember, this is where rays land their height, their position, the image height over the object height. And all of these rays start out at some position here, y, and end at some position here, y'. And the magnification is the ratio of both. The angular magnification has something to do with how angles transform from object to image. So to think about that, let's get rid of position for our ray by setting its initial position to zero. Let's launch the ray from the origin. And then shoot it up to any arbitrary position in the lens, any height. I'll call that h, and I could make h anything I want here, including negative. And then let it go on to the image plane. Of course, since the ray started at the object on the axis, it must end up at the image plane also on the axis. And now we'll define an angle u, which is the object angle. And then angle u' which is the image angle. Note that as shown here, this would be a negative quantity, so I've noted that for my sign here. The angular magnification is nothing more than the ratio between these two angles, just as the analogy to our regular magnification. And through just the two triangles that we made with the red ray here, it's very easy to calculate what that is. Now remember, we're doing paraxial, near the axis, optics here. All off these rays, even though I may have shown them as very, very large angles, we're assuming aren't small enough, that we can take all trigonometric functions, and keep only their first order Taylor series term. In the case of sine and tangent, sine theta would equal theta, tan theta would equal theta. So, to be consistent with that analysis, this first order analysis, in which everything works perfectly, because we don't have any of the nonlinearities that come from the other terms in sine theta, the theta cubed and higher order terms. We'll calculate u here just as h over t. And there it is. And then we'll calculate u' from the similar triangle here at h over t'. And thus we see something super important, that the angular magnification of the system, u' / u, is equal to t / t' which is exactly 1 over the magnification. This is far from coincidence. It turns out this can be derived from conservation energy. So it's a very, very fundamental rule. And it's a very important concept to start understanding about optical systems, is that if you take an object here and you magnify it, you make it bigger in the image plane, then the cone of rays that got off of the object will be smaller. They were smaller coming into the image. And of course you see by my red ray continuing here, the rays that continue on from the image will be smaller by exactly the inverse of your magnification. The two quantities are duals, or equivalently, and this starts to get to the conversation of energy part, their product is conserved. There's another detail on that. But that's not a bad way to start. So this is really important intuition. Magnification is greater than one in absolute value, the angular magnification will be less than one. And then also our sign convention works here because this angular magnification would be negative because as shown, u' is negative quantity. So this is a little aside from our thin lens imaging problem. But it's a very important quantity, and very important intuition for optics going forward. Okay, returning to the last of all possible quantities we can derive about a single lens. You might have a design problem where you have to get the object to the image in a fixed distance. You have parts going across an assembly line. The camera has to be mounted a certain distance above the assembly line, so you don't have any choice. You have to fit in that space. And so all of your optics have a certain throw, the sum of the object and image distances as constraint, okay? No problem? Let's just derive or define throw as the sum of the two distances. Now throw is a distance, therefore it's a positive quantity. So we have to keep track of our sign convention here, because t, as shown, is a negative quantity. And now we can just use our Gauss's thin lens equation. Substitute this throw in to eliminate the variable t', and do a tiny little bit of algebra. And we come up with an equation for t given the throw and the focal length. And these are equivalent, so you could do the same thing to come up with an equation for t'. It might be interesting to know, let's say in that assembly line imaging problem, how close can you possibly put the camera and still, with any particular lens, get an image which is in focus. Through that, we'd simply minimize this quantity, and or this quantity. And we'd find the minimum throw is when the total throw is 4 times focal length. That turns out to be the symmetrical system when t and t' are both equal to 2f. That's the one-to-one imaging system because the magnification would equal minus one. And that also happens to be the minimum throw. That's a good thing to know. All right, let's summarize the whole system then. It turns out, the single lens imaging system is pretty darn simple. There are a total of five independent variables that one could describe it with. The object and image distances, the throw, the magnification, and the focal length. There's nothing else to know. You could replace t and t', z and z', your choice. We have three equations which relate those variables, the definition of throw and magnification. And pick your favorite, but one of the two imaging equations, the thin lens equations. So the point is, is if you are given any 2 variables, given out of 5, you can solve for the other 3. So that's why I say this is a relatively simple design problem, that's everything there is to know about it. I just point out our common couple of forms to help with that. Perhaps you're given the focal length and the magnification. That's the example when using Newton's form of the thin lens equation is quite convenient. Because it has distances, focal lengths, and magnification in it. So immediately, if you simply plug in the things that are given, focal length and magnification, you have z and z'. Sometimes you're given throw and magnification, that would be a pretty common thing. You don't really care about distances, you don't care what lens you're using, just you have the external specs. You gotta magnify so much and you get a fit in the space. Again, a little bit of rearrangement of the equation, starting with the definition of throw. And you can come up with an equation for the throw, focal length, and magnification. Notice here, this looks bad because you always worry about things blowing up in the denominator. But in the single lens imaging system, magnification can't be plus one, it can only be negative. So this can never blow up. So, that's everything there is to know about single lens imaging systems. And this is a pretty important set of equations to be very, very comfortable with. Because we're going to build more complex things on these in future lectures.