 Ok, great welcome back um we are continuing onward in our discussion of light and optical systems. So, I went over basic properties of simple lenses um we have just talked recently about the eye a human eye as an optical system. And we ended with um explaining that the diopter of the eye I introduced the notion of diopter which is the converging power of a lens or an optical system and I said that the um the power of the human eye is nearly 60 diopters. So, I want to talk a bit about the imaging properties of a lens and you know just in connection with the human eye as I showed last time um remember that the light comes in um it is refracted at many places across the cornea through the um the boundary between the cornea and the outside air the cornea and the aqueous the um aqueous and the lens and the lens and the vitreous. So, an image forms on the retina all the way across this curved area here. So, the retina starts roughly here goes all the way around and stops here with the fovea being the place of greatest acuity where um we receive the highest density of information the part becomes very important. And it is one of the motivations there are others, but one of the main motivations for eye rotations that we perform. So, we can change the the the location of greatest visual acuity um. So, I want to talk a little bit about images and how they hit an image plane or in this case notice is not going to be exactly what I am covering because the engineered systems that we tend to make have have um focal planes and we talk about fixed focal distances here the retina is curved and so there will be a curved um region that will remain in focus on the outside world as a consequence of that. And so I am not going to give sort of all of the geometry of that, but I am at least going to explain what appears in standard engineering systems. So, let us talk about the imaging properties of a standard lens. So, we have the standard kind of simple lens as we have been talking about here and I am going to stop drawing pictures on the board now and just utilize these pre drawn pictures. So, we have a converging lens and imagine there is some object that is distance s 1 away from the lens appearing in front of it um. If we put this object exactly um away at infinity right. So, we move this object really far to the left so that it is infinitely far away we get our parallel lines and then the image or the focal plane appears exactly at the focal length. But we can bring the object in closer so that instead of being at infinity it is at distance s 1 and then it is going to take longer for the rays to converge because the object is closer, but in some cases they will eventually converge and they will form an image that is further away than the focal distance which is distance s 2. So, the expression for this is 1 over s 1 plus 1 over s 2 is equal to 1 over f the focal length and notice in this picture that s 1 and s 2 are greater than f. So, for this particular case we say that on the right side we obtain what is called a real image I am going to make a distinction between real image and virtual image um on the next slide, but this is called a real image which means that if we were to place a sheet of paper or a screen here then an image of the object would actually appear there. So, the light hitting it would form a representation or a kind of rendering if you like on the screen in the focal plane. So, we call that a real image it should be sharp in focus everything should work out very nicely like that and if I pull the object away suppose I hold the screen fixed and then I take this object and start moving it forward or back a bit then this image will go out of focus, but I could then move the screen to get it in focus again and I do that according to this formula. Does that seem fine? If this is a nice kind of sanity check if s 1 equals infinity or take the limiting case where s 1 tends to infinity then what is s 2 in this case f which is good right because that is what is supposed to happen. So, if we have s 1 equals infinity that means the object here has been dragged all the way to the left forever as far as one can go and then some and. So, in that particular case this should converge in the limiting case to s 2 equals f the focal length. So, I just want to point out that relationship that even though we made a big deal out of the focal length the focal point this focal plane all these things we made a big deal out of that it is very important, but there is also a range over which the lens can be useful and can produce real images correct and the same thing is happening in your eye it is just that it is curved. So, same in some way let us say it may happen similar because of the curvature of the retina and in this case there will be a surface a 2 dimensional surface of locations where you will get perfect focusing on to your retina of a real image. So, there will be in fact a real image that appears on your retina right. If you get to another case where let me just draw a simple example here there may be no solution of this equation. So, it may be that for example maybe I have focal length could be one half meter and I say s 1 equals one tenth. So, I get 10 centimeters away from the lens then if I try to solve this equation here I get what 10 plus 1 over s 2 I am trying to figure out where this image would appear in fact let me go back a slide this is what I mean to say I am trying to get this to work out I want to figure out what s 2 is going to be. So, I make the focal length be one half I make s 1 where I place this object be very very close. So, I put it very close to the lens it is so close that the rays are diverging a lot this lens is not maybe powerful enough to converge them back they are just too far spreading. So, if I do that I get this and there is no solution right. So, this tells me that the lens cannot converge the rays. Now, if it cannot converge the rays what if it just does whatever it can. So, it may converge them some amount it may change the divergence a bit and do a little bit of converging, but it is still overall the rays diverge after hitting the lens. We can still do this trick that we did for diverging lenses which is work backwards and see what the rays would converge to on the left side of the lens and we get a picture that will look essentially like this and in this case we are allowing s 2 to be negative. So, we allow s 2 to be negative and we still get a solution for this and this corresponds to a very familiar process called magnification right. I pull out a lens and put it up very close to something and it appears to be larger. However, that is just an appearance to us and in this case we call this larger object that we get. So, again if you are looking into the lens from this direction now, we put an object up very close to it, but from this perspective the object appears to be very large right. So, that is magnification. So, we are using the same kind of lens there is nothing different about it other than just where I have decided to place the object and where I have decided to view the object from. Now, there is one more maybe slightly confusing thing about this which I should point out if you decide to use a magnifying glass to look at an object up close, there is still a real image being produced somewhere this process is still happening in your own optical system of your eye. So, when you see the object appearing very large looking through a magnifying lens as it hits the retina just pay attention that that is still going to be a real image. So, ultimately your eye has to when looking at this picture here and if your eye is over here on the right side looking through this magnifying lens, it is still going to converge the rays anyway. So, that it produces a real image on your retina right, but it gives you the illusion that this object is much larger ok. So, I just wanted to point those things out. Any questions about that?