 Welcome to this last lecture in Physics 1308, Introduction to Electricity and Magnetism. In this lecture, I'm going to cover thin lenses and the human eye. To begin the discussion of thin lenses, we have to look back at the law of refraction, and also known as Snell's law, and the way in which light moves from one material into another. So to refresh here, we could imagine that we have a situation where we have one material, like air, over on one part of our picture, and then there's a boundary between air and another material. So for instance, we can imagine that we have here some glass. Now what distinguishes air from glass for the purpose of optics is the index of refraction of these two materials. So we have N1, which is the index of refraction of air, and that is pretty much very close to 1.000. So we'll just write that as 1.0. And we have N2, which is the index of refraction for glass, and that varies depending on the type of glass, but we can imagine that we have an index of refraction of about 1.6 for the glass in this example here. Now what I'm going to do is I'm going to send in a ray of light and that ray of light is going to originate in the air and it's going to strike the surface of the glass. Some of it will reflect, but I'm interested in the part of the light that is going to refract, that is pass through the interface between the air and the glass and then travel inside of the glass. So for instance, I might imagine sending an array at an angle like this and we know that we then have to consider the relationship between that light ray and the normal to the surface of the glass where it strikes. So for instance, I've drawn a nice red dot here to indicate where the ray strikes the glass and I can just put a little dotted line through here that represents the normal to the surface. So this right here at this point where I've drawn the ray striking the glass makes an angle of 90 degrees with the surface of the glass. Now we can write the angle in the air part of the problem as theta 1. That's the angle between the ray and the normal to the surface where the refraction is going to occur and the ray is going to exit and it's going to exit at a different angle. The light ray will bend. We know from the discussion of Snell's law the light ray will bend because the speed of light inside of glass is different than the speed of light inside of air and this causes the wavelength of the light to shift slightly and the net effect of this is that the trajectory of the light ray changes when it goes from air to glass or if it were to go from glass to air that would also happen. So we have some different angle here and we have the refracted ray. So here is our refracted ray and it's going to make an angle with respect to the normal of theta 2 inside of the glass and Snell's law allows us to relate the two. So Snell's law is simply n1 times the sine of theta 1 must be equal to n2 times the sine of theta 2. Now we can rearrange this equation to write the ratio of the sines in terms of the ratio of the indices of refraction. So let's go ahead and do that. What I'd like to do is I'd like to get the ratio of sine theta 1 to sine theta 2 in terms of the ratio of the index of refraction of the glass over that of the air. So with a simple rearrangement I can go ahead and do this. I can say sine theta 1 divided by sine theta 2 is just equal to n2 over n1 and because of the indices of refraction that I've picked for this example this ratio is just 1.6 because it's 1.6 divided by 1. This then allows us to look at the ratio of the sines. We see that the ratio of the sines of the two angles is going to be 1.6. Now the sine of a small angle is a small number and the sine of a larger angle as you approach 90 degrees for instance is a larger number. So what this tells us is that the sine of theta 1 divided by the sine of theta 2 is a number greater than 1 and that implies that the angle theta 1 is greater than theta 2 because the sine of theta 1 is greater than the sine of theta 2 according to this equation and that implies and if you're not sure about this go and play around with the sine for a little bit this implies that theta 1 is greater than theta 2 which is exactly what I've drawn in this picture. So in this refraction we have that this angle theta 1 is a larger angle than this angle theta 2 because we're going from a material with a low index of refraction into a material with a higher index of refraction and in general this will always be true. If you go from a material with a low index of refraction and you pass into a material with a high index of refraction relative to the original medium in which the light was propagating you will reduce the angle with respect to the normal that the light ray is making. Now let's take the principle that we've just looked at with refraction where you go from a medium with a low index of refraction to a high index of refraction and let's apply this to what's known as a lens. A lens is nothing more than a system of materials designed to change the path that light takes so light will enter the material at one angle and it will exit the material on the other side at a different angle and by adjusting the properties of the material one can cause the light to follow very well-defined predefined paths that allow you to focus light to a point or defocus light away from a point and by doing all of this we can build up optical transportation systems. So let me sketch a lens. I'm going to make the lens slightly rounded on one side just a little bit and then I'm going to have just a really thick bit of material in here and then I'm going to have that same slightly rounded face on the other side. So these slightly rounded faces for the purposes of this course I'm going to keep a nice simple shape. I'm going to make them semicircular. That is, these are part of some very large circle with radius, for instance, for the right-hand side, R2 and then similarly over here you can imagine that this face is part of some other large circle with radius R1. So the radii of curvature of the two sides of this glass system let's just call this glass some index of refraction N2 these are semicircular faces of glass that you can imagine having been cut from a very large sphere of glass with very large radii R1 and R2. Now just like in the mirror case I'm going to define the optical axis to be a line that passes dead center through the lens so let's label this the optical axis. The optical axis is a convenient reference point for images and objects and focal points and focal lengths and all that stuff that we used for mirrors and we're going to repeat that now for not for reflection but for refraction. Now the whole purpose of this is to show you what the transportation of light looks like through this system. So let me send in a ray that comes in parallel to the optical axis so here's my incident ray and it's parallel to the optical axis which is this black dotted line here. Now it's going to refract when it strikes the air glass interface so let's write this N1 and we can call this air we've already handled that situation before that's an index of refraction of about one. The glass is an index of refraction of something like 1.6 it's a bigger index of refraction so we know that once we draw the normal to the interface between the air and the glass whatever the angle is between the incident ray in air and the normal it will be a smaller angle inside the glass. So the normal to the surface of a circular interface is going to lie along a radial line of the circle from which the glass is cut. So this is a very large radius circle so we might imagine that the normal looks something like this it doesn't have, it makes an angle with respect to the optical axis that is not zero but it's very slight as we can see here we know that the refracted ray is going to come out at a very slight angle to the normal inside of the glass so we exaggerate that by having it lie almost on top of that normal see it has a slight divergence there there's my ray and there's my refracted ray so we have an incident ray making an angle theta one with respect to the normal and we have the outgoing ray inside the glass making an angle theta two that's a lot smaller than the angle between the incident ray and the normal on the outside that ray is going to continue to travel inside of the glass and it's going to strike the glass on the other side where it meets the air and we have to just repeat the refractive process again we have to draw a normal to the surface of the air glass interface and our refracted ray will make a larger angle outside of the glass than it did inside of the glass because now we're going from a large index of a fraction material to a lower index of a fraction material and there we go so we'll call this angle theta four and this angle theta three so the relationships that we expect between all these angles is that theta two is less than theta one and theta three is less than theta four so this is the basic principle by which any kind of lens works parallel rays coming in from the left will refract at the first boundary they will bend inside the material as a result of that refraction they will experience a second refraction over on the other face of the material and then they will bend even further and in this particular configuration this results in a parallel ray being bent toward the optical axis on the other side of the material and this is known as a converging or focusing lens it takes light that comes in parallel and it bends it down toward the optical axis on the other side you can imagine that another ray that enters down here at the bottom following a similar set of arguments so here's our normal we get a refraction that results in a smaller angle on this side and then we have a second refraction here and we get some refracted ray on the other side we see that a ray that enters the bottom of the lens on the left side will be bent up toward the optical axis on the right an array that enters at the top on the left side will be bent down toward the optical axis on the right and you can see why this gets its name as being a focusing lens now I didn't do a great job of drawing this if I'd drawn this as a perfectly smooth circle and got my normals exactly right then of course these rays should have converged maybe right on the optical axis but you get the idea this is an example of a converging lens and here you have convex glass faces on both sides and the effect of those convex glass faces is to cause a light rays to bend toward the optical axis if they came in parallel on the left side of the setup that was an example of a converging lens system but we can also imagine a diverging lens system here instead of having a convex glass face we just have to make a slight concave glass face on one side lots of material and then a similarly slight convex face on the other I will draw my optical axis and then I will send in a light ray up here parallel to the optical axis now in this case the circular face bends away from the light ray so the radial line that points back toward the center of the circle points back this way unlike in the case of the converging face where it sort of tilted down here it tilts up we get a slightly smaller angle with respect to the face we get a second refraction over here and this ray will bend away sending a light ray down here again we draw our normal to the surface we have a slight angle here and we have the normal that way and we get an even more pronounced angle out of the other side of the interface this is what is known as a diverging lens and you can see why it causes parallel rays from the left to pass through the glass diverging away from the optical axis on the right and here, whereas the focal point was on the right hand side of the lens before here if we extend our rays back to where we think they came from on the other side we see that they converge actually on the side of the lens where the light originated and this is a very handy point to define real and virtual images for lens systems so let's go back to the concept of images again a real image just like in a mirror a real image for a lens system is located in a place where it physically could in fact be observed with a screen you could put a screen in that location and you could actually see the image of the light from the object on the other side of the lens real images form on the side of the lens opposite the object that is, if I have a light from an object starting out over on the left side I expect for a real image for a lens system that the focused light, the focal point for the refracted light will lie on the other side of the lens so on the right hand side of the lens a virtual image on the other hand it lies on the same side as the object and you can see already that a converging lens forms, has the potential to form a real image on the opposite side of the lens and a diverging lens forms a virtual image on the same side of the lens where the object is actually located if you try to put a screen there you just block the incident light from the object and you wouldn't be able to form the image on the screen unless diverging lenses are extremely important in optical systems and you need both converging and diverging lenses in order to transport light for all kinds of applications like medical imaging and non-invasive, well invasive but non-surgical imaging techniques like endoscopy now before moving on to a demonstration of lensing systems what I would like to do is I would like to discuss thin lenses which are the things we're actually going to use for the rest of the class so I've been showing you essentially thick lenses to motivate the refraction process but we can instead switch to a related class of lenses known as thin lenses and these are lenses whose thickness is much smaller than the distance of the object from the lens that's it so if you imagine having an object like a person standing on the optical axis of a lens and we were to put a thick lens far enough away from them so that the thickness didn't really show we could imagine that a converging lens could be represented simply as two semicircular faces glued together with very little material in between them or similarly we could imagine a diverging lens again here's the person a diverging lens simply being represented as two semicircular faces but concave rather than convex so here we have converging and here we have diverging and it's actually much easier to represent these lenses using a simple symbol and a typical convention for converging thin lenses is to simply draw them like this and for diverging thin lenses is to simply draw them like this so this just represents a thin lens a lens whose thickness is far smaller than the distance of the object the distance is involved for instance of the object to the lens system you can use a line with two arrow heads for converging and a line with two reversed arrow heads for diverging and the idea here is that any light ray that enters will be bent converging lens will bend the light toward the optical axis diverging lenses will bend the light away from the optical axis so you can very quickly sketch these pictures and draw some light rays and you'll be good to go now just as in mirror systems it's very convenient to analyze lens systems in the way that light moves through them using this concept of principle rays so for instance if I imagine that I have an object over here some distance p away from the lens system so this is my object I'm just going to use a simple upright arrow to represent that object in this case you can use principle rays to analyze the transport of light through the system so for example if you have a lens that has a focal point given by f that point might be located over here and because lenses for instance are symmetric it's the same distance on either side so light entering from the left can pass through the focal point on the right lenses are typically some kind of symmetric or semi-symmetric system so the focal point of a lens applies equally on either side of the lens this is distinct from a mirror because light can't penetrate a mirror the focal point lies on either one side or the other of a mirror now the principle rays that we will use to look at the transport of light through a thin lens system one we'll just use parallel rays that's one of the rays we looked at with mirrors and we'll look at it again with thin lenses a parallel ray comes in parallel to the optical axis so if I imagine a ray coming from this point here coming in parallel to the optical axis these light rays will be bent through the focal point of the lens so here is my parallel ray now another kind of principle ray is a central ray this is a ray that passes dead center through the optical axis of the lens so in this case that would be a ray that starts out at this point passes through the dead center of the lens and actually passes through undiverted that's a central ray so any ray that passes through the dead center of the lens right through its optical center will essentially look like it's undiverted and it will continue on its way on the other side of the thin lens system now of course you can also have rays that go through the focus the focal point so we will have focal rays we can look at that as a principle ray that's another ray that you are free to draw that's a ray that passes through the focal point of the lens on the object side and then comes out parallel to the optical axis on the other side so this one is a focal ray I didn't do a great job of trying to draw the focal point as symmetric on either side but basically where these rays converge over here represents the location that is equal to the location that I started with on the object and so I would here draw my image over here and we see that it's inverted so this is a converging lens and we see that when we have the object to the left of the focal point of the converging lens we wind up with an image that is real and inverted now we can continue to analyze lens systems of these principle rays and to do that what I'd like to do is not keep drawing these by hand I'd rather have a computer do all this stuff for me so that my pictures are prettier and more accurate so what I'm going to do is I'm going to switch now from this hand drawing nonsense to using a simulation of optical effects that is provided by the University of Colorado's FET demonstration system so this is the optics lab demonstration system by the FET simulation suite you can see here that we have a very nice system we have an object over here that we can move and I've elected to have the simulation show principle rays so here you see the focal ray passing through the focal point marked with an X of this converging lens it comes out parallel to the optical axis on the other side we see the central ray passing through the optical center of the lens undiverted and then coming through the other side and we see a parallel ray coming in and then being bent through the focal point on the right hand side of the converging lens and they all meet up over here you see when a computer does this it looks a whole lot prettier than what I do I've actually set the refractive index of this lens to be 1.6 and you'll notice it has a certain diameter so that just changes basically the lens so that just changes the diameter of the lens itself and you can affect the radius of curvature of the glass and you can alter the curvature in such a way that you also alter the optical properties of the lens itself so I'm going to leave that at 0.8 meters for the curvature radius that basically means that this glass has a rounded surface that is as if it were cut from a big sphere of glass whose radius is about 0.8 meters now you'll see that I have it set up to show virtual images when those occur we haven't gotten to that quite yet but we will get to that in a moment and what I'm going to do now is I'm going to change our little object here so we have something a little more akin to what we've been using an arrow and I'm going to put that arrow pretty firmly right on the optical axis so I've now laid that arrow right on the optical axis and you see here that my little sketch that I did a moment ago for a converging lens when the object is located to the left of the focal point that's not too bad I actually got that pretty good considering how crummy my drawing was you see that the resulting image is in fact inverted on the other side and it's real it's on the opposite side of the lens from the object which is on the left image is on the right now let's see what happens as I move my object closer and closer to the focal point we see that what's happening is that as I move my arrow closer to the focal point the rays are having to diverge away from one another and never mind this this is just an effect here to show you how the ray would bend if it passed through the lens on that end it's getting bigger the image is getting larger on the right hand side it's still inverted but it's a much larger image and so you can kind of see that if you want to enlarge an image of something if you have an object like writing on a piece of paper over here and your eye is over here on the right and you want the writing on the paper to be bigger you want to move the paper closer and closer and closer to the focal point of the lens system now let's see if I can rein this in a little bit so I'm going to see the refractive index here let's take it up to like 1.8 just so we can see everything still on the screen here alright I'm going to continue to move my object closer and closer to the focal point and right at the focal point we get into an interesting situation at the focal point all the rays coming out the other side are parallel that is they don't focus on the right hand side when you stick an object right at the focal point of a converging lens you can't form an image on the other side we'll see this reflected in a bit when we look at the lens equation the thin lens equation which is very similar to the mirror equation now as I move my object inside the focal point we see that now the rays do converge but they don't converge on the right hand side of the lens rather they point back at a focal point on the left hand side of the lens now we're making a virtual image there's no way I could put a screen over here and get this image to form alright so what's kind of neat about this is that this is essentially the operating principle of a magnifying glass so a magnifying glass uses curved surfaces of glass you can hold it in your hand and the surface of the glass is curved such that the focal point is very far away from the central point in the lens if you lay the magnifying glass over a piece of paper you will see on the other side of the glass a large an enlarged upright image of the writing for instance that you're attempting to magnify a magnifying glass that I've just created here so if my eye is over here on the right hand side I will appear to see the rays from the object converging at a point behind the object but enlarged and upright and that's exactly what we want from a magnifying glass we don't want a magnifying glass to invert text and enlarge it then we can't read it we want it instead to have the text to be upright and that's why we have to put our eyeballs over here and we can see the rays that are refracted they appear to come from an object that's behind where the actual object is located much bigger and upright so this is the operating principle of a magnifying glass you put your object inside the focal length of the magnifying glass lens and you put your eye on the other side and the rays that come out all appear to come from a point back here behind the lens this is a virtual image you put your eye over here and you'll see the rays as if they appear to converge back behind the lens where the paper is located so that's how you get a virtual image so this very nicely demonstrates how you can use a lens and an object to create different kinds of images you can create a real inverted reduced image if you move your object far from the focal point if you move your object closer to the focal point you'll get a real enlarged an inverted image and if you move your object within the focal point you can get a virtual enlarged upright image that winds up being on the same side as the object itself and that's the basics of how a converging lens will work a diverging lens will only ever form a virtual image that lies on the same side of the lens so that's one of the neat things about a diverging lens the converging lens has a much more rich set of structures that much more rich set of outcomes that can occur and that's why I demonstrated here but a diverging lens is also useful but it always forms a virtual image on the same side as the object itself now the workhorse of thin lenses is an equation known as the thin lens equation and it won't look too surprising to you it's simply that one over the object distance plus one over the image distance is equal to one over the focal length of the lens now again sign conventions play a very important role in this business P which is the object distance is always positive now whereas for mirrors where P defined the side where positive image distances were to be defined rather for lenses we have i greater than zero when it's on the opposite side from the object so this is the key distinction between mirrors and lenses for mirrors images are real that is i is a positive number when the image forms on the same side as the object but for lenses the image is real that is i is positive when that image forms on the opposite side of the lens from where the object is located so just to complete this i less than zero you're on the same side as the object and this defines real i positive just like before and i negative defines virtual now for focal lengths f is greater than zero for a converging lens and that's because parallel rays focus on the side of the lens opposite the object so you can kind of see a pattern here when the image forms is positive when it's on the side opposite the object the focal length of a lens is positive when it focuses parallel rays on the side opposite the object f less than zero is a diverging lens and that's because parallel rays focus on the same side as the object so that's the thin lens equation and this allows you to locate images given focal lengths or figure out the focal length of the lens given the image distance and the object distance so this is the kind of thing you'll have to play around with now of course in addition we have the concept of magnification again and just like with mirrors the absolute value of the magnification is equal to the height of the image divided by the height of the object so this is the image height this is the object height and M is a signed quantity and again we have a situation where the sign recalling again that the real images for a converging lens formed on the opposite side of the lens and were inverted so those eyes were positive but to get an inverted image you have to have a negative number so this is just the negative of the ratio of the image distance over the object distance so this gives you both the magnification factor and a sign that means upright or inverted so if M is greater than zero we have an upright image and if M is greater than zero we have an inverted image that's all that sign means otherwise take the absolute value of M and you get the magnification factor magnification factor so it's the same game as with mirrors M equals negative I over P and the absolute value of M gives you the ratio of the heights of the image and the object about that thin lens situation so let's think about a thin converging lens here you have a focal length that's greater than zero so F is a positive number of course we have that P is a positive number and let's look at the lens equation for different situations so the lens equation is 1 over P plus 1 over I 1 over F now let's focus on images so if P is greater than F that is if the object is located very far from the focal point of the converging lens then we can figure out what the image is going to do so we have 1 over I is equal to 1 over F minus 1 over P now let's put in some numbers let's say that the focal length of this lens is a convenient 1 meter and let's put the object distance at a whopping 4 meters away from the lens so that is 4 times the focal length now we can plug into the lens equation we have 1 over I equals 1 over 1 meter minus 1 over 4 meters so we have 1 minus 1 fourth which is 3 fourths so we have 1 over I equals 3 fourths inverse meters or 1 over meters so we can solve I is equal to 4 thirds of a meter so we see that we have a real image that forms as we saw in the computer simulation we have a real image that forms and it forms at a distance of 4 thirds meters which is just slightly longer than 1 meter so it forms a little bit close closer to the lens than the object but still further than the focal length on the other side and we can also see that this is going to be an inverted image because M equals negative I over P and this is going to be negative 4 thirds over 4 which is negative 1 third so we wind up with inverted reduced image compared to the object it's 1 third the height of the object that's the 1 third and it's a negative sign in front of that so it's inverted so we can play around with the equation and very quickly see what we saw in the computer simulation if you put an object at a larger distance than the focal length then you get an inverted image that's real you can continue to play this game so again let's imagine that F is 1 meter and P is now 2 meters so we have 1 over I equals 1 over 1 meter minus 1 over 2 meters and that leaves us with 1 over 2 meters or 1 half meters to the minus 1 so that I is equal to 2 meters so we see here that again we get a real image and it's magnification factor negative I over P is negative 2 meters over 2 meters is negative 1 so when you put an object at twice the focal length when P equals 2F something interesting occurs when you get a real image it is inverted but it is un-magnified it's the same height as the object so that's a special case when P is equal to 2F you wind up with a real inverted image whose height is exactly the same as the object now let's go inside the focal length let's put F equal to 1 meter and P equal to 0.5 meters or 1 half of a meter so now we have 1 over I equals 1 over 1 meter minus 1 over a half is 2 meters to the minus 1 so we wind up with 1 meter to the minus 1 1 meter to the minus 1 minus 2 meters to the minus 1 is negative 1 meters to the minus 1 so I equals negative 1 meters so here we have a virtual image because I is a negative number we can look at the magnification factor this is negative of negative 1 meter divided by 1 half meter which is equal to negative 2 which is equal to positive 2 so here you get an enlarged upright image of the object that's virtual so a virtual enlarged upright image height is greater than that of the object by a factor of 2 and it's upright we have a positive sign on the magnification and it's virtual because it forms on the same side of the lens as the object that is at a negative distance so I hope that this gives you the groundwork for playing around with thin lenses and the last thing that I'd like to do before closing out the formal material in this course is to talk about the human eye