Now we know that when we begin to analyze a optical system, we start with launching that marginal rate tilting it up until it runs into the aperture stop and that determines the numerical aperture that emerges from our object that gets into the system. We've used the concept on recover right here before but this is such a common mistake I see students making. I want to hit this one more time. Numerical aperture describes the property of cones of light getting off of the object or to the image. It's not a property of lenses. Now despite that, lenses are often labeled with numerical aperture. For example, microscope objectives might have 0.5 NA listed on them. What that means is, if used as designed, with an object that radiates at very large cone of rays, then you will get an NA of 0.5 off of the object and on into your microscope. However, you could have an object, let's say, that emits a smaller cone of light. At that point, you are using that lens at maybe 0.3 NA. The most common mistake I see students using is when trying to understand the resolution or the MTF of an optical system, they simply use the NA written on the side of the lenses, not the NA as the lens is being used. And that's the concept I want to make sure we have understood here. So let's imagine I have an object same as diagram I use before when we were understanding the spatial frequency bandwidth that got off of the object and into the optical system, and blue now we understand this ray is my marginal ray. So I've tilted this ray up until it ran into my aperture stop which happens to be, in this case, the edge of that very lens. And this angle then is the largest angle that gets off my object. And so I can calculate the numerical aperture in the object space by taking the sine of that angle, of course, I multiply it by n because that impacts my waypoint clearly. If I went off through lots of this was a microscope, and this is some small object, then somewhere, an image space, I've formed an image I'd have a very similar diagram on the backwards. But presumably, my image would be much bigger because you like microscopes have large magnification and therefore, because angular magnification is the inverse of transverse magnification, this alpha in image space, the angle that the marginal ray made with the axis, would be much smaller than the equivalent angle in object space. So the numerical aperture in image space would be smaller. That would in turn mean that the point spread function in the image space would be bigger than the point spread function here in the object space. And that seems to make sense. It would be bigger by exactly the magnification. So that's all self-consistent. But notice the lenses aren't really in there, it was how you use them. And here is the specific place people go wrong. This angle is given by, in the paraxial approximation, the diameter of my entrance pupil formally over the distance between my entrance pupil and the object t, not the focal length of the lens. It's only the focal length of the lens if I happen to have put the object at the front focal plane. And that's the mistake people make all the time. So my numerical aperture in object space here, it's the diameter of the entrance pupil over the distance between the object in the entrance pupil. Now it's convenient all times to use f number instead. It's almost exact same concept but these concepts so common. It's nice to have that quantity and essentially it's inverse. f number is simply the distance over the diameter of the lens. The f number of a lens will be f over d. But the f number at which I am using this lens is t over d and that's very simply related to the numerical aperture. Remember that this numerical aperture or if you want to use the f number directly gives me the spatial frequency that gets off the object. And again, I must be using, in this numerical aperture expression, this distance, not some random focal. The focal length for this lens can be a little bit tiny but that isn't important in terms of simply what frequencies get off of this object and through the entrance pupil. So, what sets my cut off frequency? And of course, equivalently, what sets the spot size that I'll be able to see my minimum result of the feature size on this object? Is the numerical aperture of this coner rays not some numerical aperture printed on the side of the lens? So when we say that we can see features down to a size of about lambda f number, the f number here is not the f number of lens. It's the f number of how I'm using the lens. So, this is the two cases that get people confused. If I'm using the lens in an infinite conjugate conditions, I use my lens and get a particular diameter, and then my image or object, doesn't matter which, is at the focal plane, then all of those concepts I previously just mentioned, the distance becomes the focal length. So this is the f number you'll see printed on the side of a lens or the NA you'll see a printed on the side of a lens. It's typically the fastest, the shortest, smallest distance you use lens. It's rare that you would put an object or an image inside the focal lens. It's not impossible but it's unusual. So, the f number here would be the minimum f number of lens you tend to use it or the largest numerical aperture you tend to use the lens. But often, you use these lenses in a finite conjugate condition. And now this angle in this case illustrated for image space would be smaller than this sort of maximum angle you might use. And I worked out for you the expressions of how the f number of the lens would relate to f number of the image space and other object space when you use it on a conjugate condition. I'm not sure that those expressions are that useful. I just calculate the individual condition that I have. But the point is, the f numbers and the numerical apertures, image and object space which are usually different quantities, are not the same as the numerical aperture of the lens or the f number of the lens if it was used in the case where the image or object is at the focal plane. This is the typical number you quote when you talk about the f number of the lens. That mistake, it's made all the time and it's really important because you'll miscalculate your spot size or your MTF if you get that wrong.