Let's look at this concept of number of resolvable spots in a little more detail because it's quite important. The normal definition of resolvability involves looking at a couple of spots. We now know those to have a shape and perhaps some finite radius, which we call the diffraction limit. And seeing how far over you have to move a spot in the image plane to let the two distributions of power be distinguishable. The most common version of that is called Rayleigh resolvability for the Rayleigh, and it's pretty obvious thing to do. If we have an Airy disk that has a null in it, he said let's define resolvability as we move the next spot over such that its peak lands on the null of the first. And if you add the H_squared, so if you add the power of these two spots, you would find that you get a peak, a little bit of dip, and then another peak, so it's just enough for you to be able to tell there's two items there. This would occur for example if you're looking at two stars, a binary star system for example. You just begin to resolve that there were two point sources, not one. What I want to look at here is there's a second, equivalent way to look at resolvability, and it really relates to the equation for the Lagrangian invariant which we just wrote down. Remember it had two terms, u y_bar and y_bar u, and those two terms seemed to flip character when it went from the image plane to the aperture stop. And so, it seems like somehow we can perhaps learn about resolvability or see some concept there in the aperture stop. So, that's what we're going to do here. So first, we're going to calculate if we have an exit pupil, remember that's the image of the aperture stop in image space, we're going to calculate if we were launching a focusing cone of rays from the exit pupil, how far we'd have to tilt this central ray to move over our zero, our diffraction limit, at spot radius. So that's relatively easy. We would say that would be an angle θ, which is simply r_zero, over the distance between the exit pupil and the image plane. We'll call that d. We have an expression in the case of a slit, 0.5λ/NA, that's the size of my diffraction limit spot, over d. And of course, I can now calculate the numerical aperture in the image plane. Remember, numerical aperture is a property of cones of light, not lenses, and that will be the diameter of the exit pupil over two, over the distance to the image plane, the d's cancel out. And I get that the resolvable angle I have to tilt is simply the wavelength of light over the diameter of the exit pupil. Okay, that's nice and simple but what does it mean? Well, now let's do the calculation a little differently. Let's look in the exit pupil itself. What I'm curious there is, how much would I have to tilt the wavefront, and remember the wavefront would be something normal to these rays, such that the the wave that I saw across the exit pupil, remember, this is similar to our spatial frequency concept, that it would have one extra total wave of delay? In other words, that if I tilt this wave, I'd find that the edge of the wave on one side of the exit pupil has moved a total of λ, whatever the local wavelength is if I was in some medium. And what would that angle be? So, if this was a plane wave and I've illustrated a blue plane wave to be perhaps the central wave of going down the axis, and the red being a plane wave representing the central special frequency launched at the tilted angle, how much would I have to tilt in order to see one wave of delay across the edge here? And I find of course that's an easier calculation. That is simply λ_over_d again, and that's the same expression. So, the point is, I can think about resolvability two ways. In an image plane, I can think resolvability is the amount of shift I need to move a spot over such that I can see the difference between two spots. I've moved over by about the spot radius. However, in the aperture plane or probably a pupil, I can think about resolvability here in terms of angle. And resolvable wave in the exit pupil is one that, whether its shape would be here, it's a sphere, but I've added a tilt to that sphere and that tilt is exactly one extra wave of delay in the exit pupil, such that if I took the overlap interval between those fields, I would get zero, much like if I took the overlap interval here, I would similarly get a minimum right here because I've moved over enough to put peak tunnel. So the point is, I can think about this in angle or in space, it's more natural to think about resolvability in space, in the image plane, and angle in the aperture stop plane. So, since number of spots is important, it might be interesting to understand how many resolvable, Rayleigh-resolvable spots I get for a typical lens. This a nice reference I've given you down here at the bottom. The authors took thousands of lens designs out of the patent literature, where all of the radii and indices are specified, and put them into a code like Zemax OpticStudio and calculated the quantities that we've just learned how to calculate, the linear number of spots across one axis of the lens. This is for various kinds of lenses, but let's just look at the simplest case here which is a monochromatic and using only spherical surfaces, calculated the number of spots, field diameter over spot diameter in either the image or the object plane versus the number of pieces of glass, so a singlet, a doublet, a triplet, etc. And they found a fairly straight line, and I think up here at large numbers, they had statistics that were getting a bit thin, they didn't have enough elements. But in general, they found 1,200 spots per element. That's actually pretty good. That means even this singlet has been optimized for what's called the best-form lens. We'll learn about that in course three. But this gives you an idea if you need a lens with a certain number of spots, how many pieces of glass it's going to cost you. Remember that this is across the diameter of the lens, a real lens is typically of circular field stops, and so you'd have to take π times the radius squared where 1,200 is the diameter to come up with the actual usable number of spots. But roughly, you can get almost a megapixel out of even a singlet, and that's fairly impressive. So, this reference is a convenient one to have when you need to get some idea for a particular application how complex a lens you might use. However, it's actually a bit more complicated than that. Those lenses were generally imaging lenses set up for a fairly similar set of conditions. If instead you look at the number of spots for a different kind of lens, a microscope objective, as a function of the requirements in the lens, let's say the magnification, you can find actually precisely the opposite behavior. So, I wanted to include this, so you didn't think you could just use those previous tables and always understand what any lens might do in terms of performance. So here, we have a table that I've pulled from Zeiss, most lens manufacturers would have a similar table, that shows the magnification of the microscope objective starting from one to one, the image on your camera would be exactly the same size as whatever was under the microscope objective itself, up to 100 to 1 which is typically about as big an optical objective as you can buy. There's three different kinds of lenses here, let's just look at the last one. And the point is, this gives us numerical aperture as a function of magnification. So from that, we can calculate spot size. Though it's somewhat variable with magnification, most microscope objectives provide about the same field size at the image plane, and that's because they want to go, they want to all talk to the same camera chip back there and it's not going to change size when you rotate your turret and put different objectives in. So, there's something called the field number. That's a funny name because it's not a number. It's the size of the field, the diameter of the field, the millimeters that'll be a bit bigger than the camera chip that you put back there. So as a pretty good approximation, let's take the field number, the diameter of the field, to be 20 millimeters, then we'll divide by the magnification to find what the field is at the object because that's what magnification does. So, 20 over 2 is 10, 20 over 100 is 200 microns. So here's our field of view at the object, that's what we need to calculate number of spots. We can, for a particular wavelength, let's take sodium D, calculate the radius of the diffraction on the spot size given the NA, so this gives us a diffraction-limited spot radius from about 3.5 microns down to about 0.25, a quarter of a micron. We divide those two and we come up with the linear number of spots across the field. And the point is, it drops pretty dramatically. For a 2x objective, you can get almost 3,000 linear spots. That's probably more than your camera chip might be able to provide because those are typically order megapixel, so maybe a thousand spots. But by the time you get down here to a 63x or 100x, the number of spots has dropped to less than a thousand. So, you're actually limited by diffraction and your ability to see down here because the camera chip is providing more pixels but you're not actually able to provide a diffraction-limited spot size back at the image plane that matches that camera chip. That is despite the fact that the number of elements in this lens is going up quickly as you go up in numerical aperture. There might only be two or three elements down here at the 1 or 2x range and there could be 10 or 12 up here at the 100x range. So this actually has the opposite behavior of the last chart. And the point is, the last chart was lenses under fairly similar imaging conditions and that's why you saw that positive slope. Here, we're putting greater and greater requirements on the lens. 100x magnification is a much harder lens to design than 1X magnification, and so in this case, we see the opposite behavior. So, this gives you some feel for under sort of two distinct conditions what you might expect for number of spots versus number of lens elements.
