In a previous lecture, we summarized all of the properties of a single lens imaging system, finding there were only five possible variables and you could write down all possible equations used to describe and understand that system. We now want to use the concept of the conjugate matrix to do roughly the same thing for the two lens system. Obviously, there's a few more variables, not very many more. And this will form, basically, a reference for you that allows you to work with two lens systems or any system that has two different separate elements and find the focal lengths and other properties of that system. I will work in the scale variables here, that is, I'm assuming that all distances are in a vacuum. If there is, for example, a refractive index between the two surfaces, you could place D by D over that index, as we showed before. So I'm simply going to go through a two lens system here, where we have a first power and a second power separated by a distance, D. We'll characterize the distance to the object by a minus t_one, as we've been doing before, in analogy with the single lens system and the distance to the image by t_two_prime, the distance between the two surfaces, BD, and we can now go through and simply apply all the conditions we just learned to derive useful properties to these systems. The first thing, of course, is to calculate that conjugate matrix, M, by transferring from the object through the first lines between the two lenses, refracting again through the second lens and transferring on to the image. And when you write out those matrices, this is what you get for that conjugate matrix, and this now should express all the properties of these two lens system. So it's a start. Let's find the effective power of this system. Remember that N_21 here is the power from the form with the conjugate matrix. That's this term right here. And from that, we find that the effective power. One over the effective focal length is the sum of the powers minus a correction term. And I mentioned earlier that we use this term earlier, and we didn't have quite the tools then to derive it, and now we do. And again, if you were in a system that had a refractive index between surface one and surface two, you'd have D over N here. It's always a good idea to go back and check your limits, and the easy one, in this case, is to set D equals to zero. That would be the case of two lenses in intimate contact, and we recover the formula we derived very early in this course, that the powers of two lenses should add. Again, from the form of the conjugate equation, we should be able to derive the magnification of this two lens imaging system. And that comes out of that term right there, a little bit of a simplification to use the previous result and the total effective power. And we come up with a fairly simple expression for the total magnification. Once again, let's check that by setting D equals to zero. What we can compare that most conveniently with the Newtonian form of the thin lens equation, and that gives us a form of the magnification. And indeed, that agrees if D equals zero. Finally, let's find the image distance. This would be the equivalent of the thin lens equation. But now, for this two element or thick lens, we can find that from the imaging condition that is setting this term equal to zero, that's this term right here. Again, I've done a little bit of math to factor out a term in the form of a magnification that we just found. But again, that's a fairly nice and simple expression. Once again, let's set D equals to zero and see if that recovers the thin lens equation, and indeed, it does. With, of course now, the sum of the two lens powers, V_one plus V_two, in place of just the simple focal length we have before. We can find the front focal plane by setting N to two, this term equal to zero, and so, that finds the distance here. Remember, as written, this would be a negative quantity because we read it as t_one, not t_zero_prime. And if either the distance were set equal to zero, bringing the two lenses together, or if we simply set the lens power of phi_two equal to zero, then we recover that the front focal distance is minus the focal length. Notice that that's true even if there is a material here. So, if this was a plano-convex lens with a convex front surface and a planal back surface, the front focal distance is still minus_F, and we've seen that once before in one of the examples Similarly, we can find the back focal distance or back focal plane by setting N_11 equal to zero. Once we get a analogous expression, it's very similar to the previous one. If, now, either the separation between the lenses or the front power was equal to zero, which we could do by making the front surface a planal surface, then the back focal distance turns into the focal length. And that should make sense because we should have a symmetric result, in comparison to the last one. Now, we have the tools to find the principal points because since we have found the front focal distance and the back focal distance, that is the distance from the focal plane to the vertex and the vertex to the image or back focal point, we can simply add and subtract one focal length because that should identify where the principal points are. So if we simply take the front focal plane distance and add one focal length, go through and think about the sign convention here to make sure you believe that's the right way to do it. We can find that the distance from the vertex to the front principal plane, which I labeled delta here, with the signed convention that delta is positive. If the principal points to the right, it's given by a very, very simple expression. And as it should, if either the separation between the lenses or the power of the second lens equals zero, then delta collapses to zero, seeing that the principal plane lies on the surface. So it would be the case for a thin lens if D equals zero, or a convex or concave planal lens if phi_two equals zero. And if I may, same thing for the back focal length of distance, we can take this back focal distance measured from the vertex forward, subtract one focal length, and that ought to be equivalent to doing this projection to find the intersection of the incoming axial ray and the exit ray, so that would find P_prime, the rear focal distance. That's the perfectly analogous expression, that if there's a minus sign, because delta_prime is measured backwards from the vertex. Because, again, as the same logic as before, if the lens becomes thin, then we should find delta_prime equals zero, that is the principal plane and the single thin surface overlap. Or if we had a planal front surface, then the principal plane also overlaps with the back surface. And you should trace that ray forward and graphically and make sure you believe that. Finally, I'll simply mention that there are a couple of other useful matrices, not nearly as useful as the conjugate matrix or the system matrix, but every once in a while, these are nice to have. I'll give them to you so you have them in your notes. You can simply find all of these by just doing the same, sort of, analysis we've done up to now. For example, if you move from focal plane to focal plane, you find that the matrix simplifies to this very simple form. And there's actually some interesting understanding in this. Notice we just transform angle to position and position to angle, because we have only the off-diagonal terms, and the ratio is nothing more than the power of the system. And that expresses that what lenses do, if operated between their focal planes, is it's a Fourier transform formerly, position becomes angle and angle becomes position, and that's kind of a useful thing to have. Every once in a while, you might want to move between the principal planes. And that's the matrix for that. I didn't include it here, refractive index difference between the incident and exit materials. And finally, every once in a while, you might want to move between the vertices, the intersection of the surfaces with the optic axis. And that's the form for that, like in tau is the reduced distance, which is the actual distance, VV_prime, or D, in our previous notation, divided by the index. So the point is that for all of these, that we now have a tool to advance through multi-element systems, this ABCD matrix. And by constructing the system, or particularly the conjugate matrix, we can ask questions about the system. What's its focal length? What its magnification? Or if we're doing design, we can determine, we can constrain, we want the magnification to be as a particular value. And we might get a rather complex algebraic expression from that, but that expression would constrain the variables from, if you have three degrees of freedom and you put a constraint on the magnification, now you have two. If you put a constraint on the focal length, now you have one. It's a way to very systematically design these complex systems where you use these constraints to choose, or at least restrict to a subspace, the variables you have available to you to meet the design criteria. And that's what we'll do in the capstone for this class.