All right, that was all warm up for one of the most important concepts certainly of this class and perhaps the whole specialization. And this goes back to our postdoc's sad tale, and other hints we've had before, but now we're going to prove it. So let's look at a Lambertian source and let's ask how much power would be captured by a lens a certain distance away, and that lens would have a certain aperture on it. That's probably the way to answer that question I posed at the beginning of this module. Let's assume that filament as Lambertian, we have some element of a lens or some sort of aperture, an entrance pupil of our optical system, a distance away, how much power do we gather up? Well that's not so hard. We have some increment of total power here, d_Phi, that would be given by whatever the radiances at some angle, where we're observing at because we're going to have to look over a range of angles. We're going to have some unit, again, remember that radiance is power per unit area per solid angle. So we'll multiply by unit area. We're already used to the fact that the unit area we seem to see of that source is projected, so there's a cosine theta. And then, we're going to accept some solid angle d_Omega. So now we just have to start doing the math. Again, you can start seeing this is why we need these tools. Well, the solid angle is simply some patch of area that's going to be probably on our lens over the distance from the source to that patch of area, r_squared. And we're already used to the fact now that r_squared is going to be projected. So let's calculate the area dS through a little bit of basic trig. This is just spherical coordinate math. And so the area of the the patch here, this dS, is given by r_squared sin_theta d_theta d_phi where d_theta and d_phi are two spherical coordinate system angles. The r_squared's cancel out and I finally get that a little chunk of power comes off my source that's got to be particular radiance as a function of angle and that chunk of power depends on the area that I'm looking at, some dA or a finite source here, and the two set of angles in a spherical coordinate system I'm going to integrate over d_theta and d_phi like cosine times sine, which I've replaced here with a trig identity. All right, with that little bit of trig done, now I can just turn to the simple calculus. I can integrate up to find the total power that I'm going to receive. I'm going to integrate all the way around in d_phi, that's my symmetrical axis spinning around to the normal, to the surface, and in theta up to some finite angle alpha. That alpha is the same alpha on the first slide in this model. It's the numerical aperture or sine of alpha would be the numerical aperture. It's the angular extent of my entrance pupil as seen from the source. So that's gonna I'm going to calculate now a whole cone and see how much power is in it. And so I plug in my little increment d_phi and I do my integral and that's easy. I assumed now that my source is Lambertian. I left the radiance as a function of angle up to now just in case you had a non-Lambertian source, but we'll go ahead now and assume it's Lambertian so L of theta simply equals L, that's what a Lambertian source is. Let's take my area A and, again, from the first slide in this module, assume that it's an entrance pupil with a height h and so I'll plug in A is the area of a circle of radius h. And what I find then is that, and I'll take sine_squared alpha. Well, that's convenient, sine of alpha is at least an index equals one material, the numerical aperture. And suddenly, I see the Lagrange invariant alpha out there. The height of my chief ray and the numerical aperture defined by my marginal ray and that's an important thing. So what we find is the following. The amount of power you get into an optical system to within a factor of a constant out here is given by a super simple expression which is the radiance of the Lambertian source L times the Lagrange invariant, and it's H_squared based on Lagrange invariant is a one-dimensional thing, and now we're talking about a two-dimensional area so it makes sense that it's H_squared. So now you know another reason that you really like to know the Lagrange invariant of your system. If you multiply that squared times the radiance of the Lambertian source you're looking at, that's the number of watts you get into the system, and that's a pretty important thing. And we've already learned that H is conserved as you go through an optical system. So you might think naively, but you won't anymore, that in our first problem with my light bulb at the beginning of this module, that maybe I could gather up a certain amount of power with some special magical system, the black box with the question mark in it between the light bulb and my optical system, and then later on, I could improve that, I could make the system somehow have more radiance, I could get more power, and I can't because this quantity, H_squared is conserved, L is a quantity that describes the light source, so the total amount of power can only go down, I could have a Fresnel reflection or something like that. H_squared and H can't go up. So once you pick a limiting aperture somewhere in your system, your aperture stop, and that's going to determine what your field, H_squared, you're done. That's the amount of power you get through the system. You could for example have somehow designed a highly efficient condensing lens system at the beginning to capture light off of your Lambertian source with a bigger H_squared, but somewhere in your system you have to get through a whole, see my postdoc's sad tale, that limits your Lagrange invariant. The power will fall right there to be whatever is the limiting Lagrange invariant in your system, and then you will never be able to make it go back. So this is the fundamental concept of radiometry and how light sources and optical systems interact, is that Lagrange invariant times your Lambertian light source radiance gives you the power you get to the system, and that's all there is to know. You can't do better.