13.4 Reflection and Refraction

Ok, today we're gonna look at what happens with EMR waves. We're gonna treat them as waves and we're gonna look at what happens when a wave hits a different surface. So, the first thing we're gonna look at is what happens when it hits a flat mirror. It doesn't have to be in the air, but we typically use mirrors. It could be water or a shiny and smooth surface. The law of reflection basically states that the angle of incidence equals the angle of reflection. So, a couple of key things to remember: we always measure the angle to the normal. The incidence angle is between the ray and the normal, and the reflection angle is between the ray leaving and the normal. The normal is always perpendicular when we're talking about the angle of incidence and angle of reflection. If the surface isn't flat, if it's sort of bumpy, it would still be the same. At the top, we'd have our perpendicular surface and the angle coming in would be going out. But if our surface was at an angle, and let's suppose our light ray comes in at the exact same angle, it would, depending on where it hits, scatter or something different like that. We would get rays coming in and going out at different angles. Therefore, the image that we would see would be blurry and distorted. That's why if you look at, you know, water that's not perfectly smooth, we don't get a clear reflection. It's kind of a blurry image instead. We're gonna look at curved mirrors fairly soon, but right now we're gonna look at drawing objects. A couple of key things to remember with just a flat mirror is that our object is what we're looking at in terms of reflection, and our image is what we're seeing in the mirror. Our image is typically a virtual image because it appears behind the mirror. Normally, we see it as the exact same size and shape, just reflected side to side. The reason for that is just the way the rays go from the original to the mirror. Let's suppose we have this moon-shaped object I see on your t-shirt and you stand in front of a mirror. Your eye would see that image the same distance away. So, whatever distance you are standing in front of the mirror, that's the distance it would appear behind the mirror. It would be the same size, same distance, same shape. But you can see how the ray diagram would work. If we look at the top of the moon, the angle going into the mirror would equal the angle coming out. But we don't see that, we see what appears in the mirror. So, that's why we extend the line down. And the same thing on the bottom. You can see, here her bottom would go up to our eye at the same angle. We have to extend that down and we see the bottom of the image. Pretty simple to draw these diagrams. Like I said, we're gonna do more of them when we do curved mirrors. But for now, the key idea is that the angles will always match, and that's why we see the image as if it was three-dimensional. Let's suppose this moon shape was somewhere out in front of it. When the rays hit the mirror and hit our eye, it's because the rays would sort of criss-cross. That's why we sort of get the left-to-right reverse. A couple of key things: - The flat mirror images reverse left to right.- The image appears to be behind the mirror. That's why we call it a virtual image, and that's why we use dotted lines and a dotted image to show us that it's virtual.- The way we could test that is if we put a piece of paper where the mirror is, we wouldn't see anything. Because we can't see it, we call it a virtual image.- If for some reason the image was projected onto the paper, like an overhead projector or [Music] Something would work, then we'd say that's a real image, and we're going to get to those later on when we do curved mirrors. Our next topic we will look at is refraction. Refraction is what happens when the light rays enter a different medium. So, in air, we say our light travels at 3 x 10^8 and it travels at whatever frequency and speed that it's going on. But if it hits a different, more dense medium, it's going to slow down, which would then cause the wavelength to drop and our angle to get bigger, and so on. So, refraction basically is just the slowing or the change of mediums, which would then make our object look bent or kind of distorted. So, you can see this pencil in the glass, the image kind of bends a little bit, and you can see it's sort of magnified at the bottom. Snell came up with a formula on how to calculate this, and it's basically a ratio comparing the two different mediums. So, we say in this case that, suppose air has an index of refraction of 1, and let's suppose water has an index of refraction of 1.33. 1 would represent normal, 1.33 would be a little bit more dense. And then, our formula is like this, and you only need two parts of it. We could use the angles and the index number of the substance, or we could use the angles and the velocity, or we could use wavelength and velocity, or whatever they need to, to get to the topics. And the only thing you've got to be careful with the formula is, we just call it surface once we lay. In this case, let's suppose we said the air was number one and water was number two. So just plug in our numbers accordingly. You can see that N is the only one that the one and two are backwards. So, the index of refraction, like I said, is just a way of comparing densities of objects. So, something like air is our standard, it's one, and then the more and more dense we get, the higher and higher the number. So, you can see water, which we used in the last question, was 1.33. If we get to something like diamond, we get to 2.42. So, the higher the number, the harder it is for the light to travel through. Let's do an example. It says we've got two indexes, 1.47 and 1.33, and it says it's entering at 43 degrees between glycerin and water. Let's call glycerin number one, and let's call water number two. So, it says it's traveling at it hits at an angle of 43 degrees to the normal. So, what's going to happen when it goes through into the water? If we use our sine beta 1 over sine theta 2, that'll equal n2 over n1. So, we just have to make sure we get the right two angles and right indexes in the right spot. So, we'd have sine of 43 over sine of the angle that we're looking for, and n2 is our water, so that's 1.33 over 1.47. To solve that, all we have to do is cross multiply. Go second sine of our answer, we get about 49 degrees if we round it off. So, you can see in this case, I actually drew it wrong. Our angle of incidence was 43, and our refracted angle is 49. As it hit the more dense subject, more index of 2 object (sorry, should be "object") from glycerin (1.47) into water (1.33), as it hit the less dense object, it's going to speed up, and therefore, our angle gets bigger. It moves away from the normal, and that's a good way to remember these. If you go from more dense to less dense, the wave will speed up and bend away. From the normal, if we went from less dense to more dense, it's gonna slow down and bend towards the normal. So, with the speed of light, if we notice that in our formula we have v1 over v2, the speed of light is going to change when it goes into a different medium. So, our default number that we use is for air, or specifically in a vacuum going through space, so it would be slightly slower in air due to the air particles and things like that. But we still just use 3 times 10^8, and you could actually slow it down by using different substances. So, to get it into more intense subjects, you should be able to get it to almost a standstill. Scientists have actually dropped it to about one kilometer per hour just using refraction. Theoretically, we could go faster than 3 times 10^8 if we could ever find a medium that's less than nothing, less than space, which is pretty tough. So, that's why we use one as our standard, and quite often that's how they'll figure out an index. They'll shine light through it and figure out how the light changes, and then from there they can figure out what the index of refraction is for that object. Let's do another example. It says if we travel, just shine light into an object, and we get to point 2.18, it's asking what the index would be. So, quite often in this case, what we're going to do is compare it to air. So we're going to use n1 equals v1 over v2. Let's call one our air and two would be the unknown substance, sorry, n2 is what we're looking for. And one for air is one, the speed for air is 3 times 10^8 and the speed for this unknown substance is 2.21. All we have to do is cross multiply those and we get an index of 1.36. Okay, so that's a pretty easy calculation to figure out what the speed is going to be in the new stuff and the new object, and then from there you can usually figure out the index. One last thing to look at, remember in physics 20 we said that the wavelength can change in a wave but not the frequency because frequency is usually based on what's creating it. So, the sound of my voice has a certain frequency, it'll change and distort depending on certain situations, but my voice is still making the sound as the same waves. If you're waving a rope up and down, the source isn't changing, it's just how it's transferring is what changes. So, the key thing to remember is that frequency will always be the same. Therefore, if velocity goes up, wavelength will go up. If velocity goes down, wavelength will go down. And we can usually use that to help us figure out certain kinds of things. So, in this case, let's say we've got red light going through water and it goes into glass. I forgot to write down the indexes, but for water, we'll use 1.33 and for Flint glass, I think we use 1.42. And it doesn't really matter what the numbers are, let's just go with this. So, if we use our formula again, n2 over n1 will equal 1 over wavelength 2. So we put in, let's call one our water and two be our glass, we'd have 1.42 over 1.33 equals, water was number one, so that'll be our 700 nanometers. We don't even need to change the units, we can just leave it as is. So, if we cross multiply and solve it, oops, I made a mistake, I think that would be index was supposed to be 2.42. So, let's try that when we cross multiply and that gives us an answer of about 385, approximately. So, if 2.42 was our right number, we'd have 385 nanometers for our [Music]. Wavelengths of 385 would be seen as an indigo blue color, as 350 is violet. So, it would probably be similar to indigo in terms of color. That's my guess. And I think that's it. Yeah, so we'll stop here. You can practice some of these questions in the worksheet.