In this course, we're going to learn a series of increasingly sophisticated design tools. We start with the simple ones because you can use them to rapidly sketch out possible designs for simple systems. This is the first such tool, and it's actually my favorite. Graphical ray tracing can be done with nothing more than a pen and the edge of your driver's license and the back of an envelope. And with it, you can actually lay out and understand a lot of optical systems. As you get many, many elements it gets a little tedious and we'll need something a little more sophisticated. You can also derive the number of the simple lens equations that we're going to use. And I'll show you that in this lecture. So this is a fantastically powerful tool, and the beautiful thing about it is it's extraordinarily simple, just a couple of rules. How we trace rays, and those rules describe the propagation of rays, which in turn, are solving Maxwell's partial differential integral equations. So we're actually solving PDEs with nothing more than drawing lines, and I just find that amazing, so I really love this. So let's lay out our problem. Let's say we'd like to do a ray trace of a single lens imaging system, a nice simple type problem to start off with. We'll represent the lens as a flat surface and the two little arrowheads here represent the lenticular, the pointy bits of a lens. If this was a negative lens, I use an arrow that's pointed the other way, the little triangle here is inverted. We'll label the front and back focal distances with, I like to use circles, but you can use whatever symbol you like. Be careful, make these the same distance, that's important. Precision is actually important here. Remember, you're solving a differential equation. And finally, we'll put an object on the diagram. The object we'll draw is just an arrow, and the tip of the arrow will be the place we'll launch rays from. This is a radiating, spherical wave, it's our field point. And we'll use an arrow so we can keep track of what's up, because at some point, it might point down, and that'll be important. So first rule, a ray that comes off of the object and goes through the center of the lens, the place the lens intersects the optical axis. Comes out the other side of the lens without bending, it's undeviated. The reason for this is, if you think of the shape of a lens, on the axis the surfaces are actually perpendicular to the axis. And so the lens is actually right near the axis, basically a plate of glass. And if you've ever looked through a plate of glass, you know the rays don't bend. Snell's law tells you you refract in, but then n sine theta is conserved, and you refract right back out at the same angle. So rule number one, and you always should do this ray first because it's the easy one. You should ray right towards the center of the lens, and it comes right on out the other side without changing direction. That's cool, and notice by the way, we've labeled the ray height. I'm sorry, the object height, and the object distance just to start getting in the habit of the coordinate system. And again, the object distance is -t here because, from the perspective of the lens, the object is at a negative distance in this example. Let's shoot a second ray. In this case, the ray is launched parallel to the optic axis, it intersects the lens perpendicularly. And a ray launched parallel to the optic axis would be like you're focusing some distance, some object which is very, very far away, like the sun. All the rays would be coming in parallel, and they would of course all converge on the back focal point here. That would be where the bright spot of light is, that you would fry things on the sidewalk if you were focusing the sun's light. So a ray that comes in parallel, comes out, and goes through this back focal point. And of course, we can keep that ray upright ongoing. And something interesting happens over here, is the two rays intersect. But let's not jump to conclusions yet, let's put our third rule in place. The third ray that we shoot, the third rule, is the second rule, only backwards. We have a ray that goes through the front focal point. And therefore comes out parallel to the axis, it's just like rule number 2 except we're running the light backwards. So any light that comes from the front focal point, careful, front focal point, comes out of the lens parallel to the optic axis. And if you are careful, and I encourage you to be careful, then all three rays intersect at some point. And that, of course, is the place we call the image. The spherical wave was launched from this field of point, and it converged again back at this image point. I've gone ahead and labeled our next set of quantities, the image distance, t prime, and the image height, y prime. It's negative because the image is upside-down, that's kind of interesting. Maybe you didn't know that. If this is the cornea of your eye, this is the screen that you're looking at right now. The screen is actually inverted on your eyeball. And you learned this the moment you opened your eyeball as a baby and, well now, that that looks good to you. But if we flip that, and there'd be optical systems which would allow us to have the image upright. The world would look upside down to you for a while, until your amazing brain would get used to it. But it's important to keep track of whether the image is upright or inverted. And because of that, let's define our first optical quantity, the magnification of a system. Which is defined as the image height, y prime, over the object height, y. And because of our sine convention, the magnification here happens to be < 0. That sign tells us that the image is inverted. And the magnitude of the magnification tells us how much bigger or smaller the image is than the object. We can also do some math with this system, and that's fun, so let's do that. If you look at the triangle formed by ray 1, and the optic axis. You notice that there's two similar triangles on the object and image sides of the lens here. And because the magnification is y prime over y, it's also t prime over t, the basis of these two triangles. So we have just learned how to calculate the magnification of a system. It's the image distance over the object distance. And because the object distance is negative, this quantity is also in this example less than 0. So this is an important quantity and we've just learned that we can derive this equation, the equation for the magnification based on image and object distances by just doing this little hand retrace. All right, let's get two more rules, these are less frequently used but they're good to have. The rules 1, 2 and 3 trace very specific rays. You can't choose what these rays are. You gotta obey the conventions, two middle of the lines parallel to the optic axis or going through the front focal point. What if you want to trace an arbitrary ray? And the 2nd, 4th and 5th rules allow us to deal with that. So rule number 4, let's say we'd like to trace some ray I've drawn off of the axis here, but it could be anywhere. But it goes off into some arbitrary direction, what do we do about that? Rule number 4 says that if two rays are parallel on the object side of the system, then they come to a focus. They intersect at the back focal plane. So this would be like, we have rays coming in from some object, very, very far away, thus, parallel. And of course, then they will focus back here, and that means they intersect back here. So now all we have to do to figure out where ray four goes, is pick a parallel ray up in front here. That is, rays 1, 2, or 3, that we know what will happen. So I've picked a ray parallel to ray 4 that goes through the front focal plane because that's one I know what to do with. It comes out parallel to the optic axis back here. And now I can draw ray 4 by going from this point through the intersection of my auxiliary ray with the back focal plane and I can now draw ray 4. By the way, we just learned something else. We already knew that this was the image plane from what we did previously. And since it's the image plane, if I come off the object at the optical axis, I must get back to the optical axis for the image. So if we know where the object and image point are, that's another way to draw a ray. But the key here is, I want to draw an arbitrary ray 4. I just find a parallel ray up in the object space, that is one of the rules I already know, and the two intersect back here. Ray 5, rule 5, is the same thing, only backwards again. Let's shoot an arbitrary ray off the object, let's say this guy here. And what ray 5 says is, is that if I have two rays that go through the front focal plane and intersect, then they come out in the image space parallel. So once again, I just have to find a ray that intersects with my ray, number 5, in the front focal plane. And is one of the rays I already know how to draw. Well, let's use the ray we just found before. I know that if I come off of the object at the axis, I will intersect the image at the axis. So if I draw a ray that comes off of the object at the axis, intersects the ray I'm interested in, number 5, at the front focal plane. Well, I know how to draw that one, it goes here. Then, ray 5, I can now draw this guy, by simply transferring the angle, make a parallel ray out here. And if I'm careful, once again, that should go through the top of, or the field point of my image here. So you don't use those so often, but as you get multiple image systems, or multiple lens systems. Sometimes you end up with rays that aren't 1, 2, or 3. And rules 4 and 5 here allow you to do arbitrary ray tracing. So now let's apply this to negative lenses. Exactly the same rules apply, but you simply switch the front and back focal planes in all the rules. And to make it obvious that the front and back focal planes aren't the same as for a positive lens, notice I've used a tittle inward-pointing triangle now. I've drawn the focal points as xs, but the basic idea is this. First of all, ray number 1 goes right through the center of the lens, no problem there. That's the same as before. Ray number 2 is a ray that starts out parallel to the optic axis. Before, it would have tilted down and gone through the back focal plane. But now it tilts up and goes through, you see the dotted line here. It's as if it came from a front focal plane. It appears to emerge from that front focal point. So instead of bending down, it bends up. And we know where it bends up from. The line goes through this front focal point. Similarly, a ray that is shot towards the back focal point bends and comes out parallel, that's ray 3. And once again, you could do the generalizations of 4 and 5. So you apply all the same rules, you just switch the front and the back focal points. And often you have to do these extensions to figure out where the rays came from, or are going to. We do that all the time in ray tracing. Notice, just for the fun of it, our sign convention is helping us out again. Magnification, again, is defined as simply the image height over the object height. Well, where is the image in this case? Well, the image is the place that all of the rays that came off a point from the object intersect. And notice they no longer intersect over here to the right. But if we extend each of them back, we find that they intersect to the left of the lens. This is called a virtual image, and is the thing we'll discuss next. This image is not one that you could put a piece of paper up, and get the light to show up here. Because the piece of paper would have to go here on the left side of the lens, and would block all light coming from the object. But if you put your eyeball over here, and looked back through this lens, you would appear to see the object here. That's where it would look like the arrays are emerging from, so it is a virtual image. Our sine convention now is that the object height at y is positive. The image height y prime is positive. The object distance, t, is negative. The image distance from the coordinate system of the lens is also negative now. It's on the left side. So if we use the same equations we derived before, the image height over the object height is a number greater than 0, and that's correct. The image even though it's virtual is upright. And conveniently this really illustrates why the sign convention is your friend. t prime and t are both negative, and once again, that correctly calculates the magnification and its sign. So this really illustrates why this sign convention is a useful thing. So next we'll turn to virtual images and objects. And understand this virtual word in more detail.