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(I) What force is needed to accelerate a sled (mass = 55 kg) at 1.4 m/s$^2$ on horizontal frictionless ice?
(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?
(I) What is the weight of a 68-kg astronaut ($a$) on Earth, ($b$) on the Moon ($g =1.7 m/s^2$) ($c$) on Mars ($g = 3.7 \,m/s^2$) ($d$) in outer space traveling with constant velocity?
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welcome to the first section in our unit on wave optics. In this section, we're going to be discussing some of the most basic examples and wave optics of interference. Now we're going to start, though, by looking at the Huygens principle. So Huygens principle is Thea underlying principle of all wave optics. What it says is that we have a source that produces a wave front and we can consider each point on this wave front to be producing small wave lits. He called them that look like this. So each point on the wavefront is actually producing secondary wave lits and then to find where the next wave front will be, What we do is we draw a line that is tangent to the tip of all the wave. Let's like this and then to move to the next step, we would draw, wave, let's all over the second line and then again draw the line tangent toe all these wave lits. Okay, so hopefully you get the idea. It's a pretty simple principle. But as we'll see as we look at this interference, um, it is it is very powerful in helping us analyze interference. Now, looking at, uh, interference. Generically, we know that we have a couple of basic rules, the first one being that we have constructive interference and destructive interference. Remember, given a situation where we have two light sources that are producing the same wavelength of light, and they are a distance R one and R two from a point of interest than the difference in path length K. So that's Delta R is the difference in path length for the past R one and R two. That difference in path length must either be a integer multiple of love wavelengths in order to obtain construct of interference or a half into germ multiple of wavelengths in order to obtain destructive interference. And this principle is still true. It's still correct, even for light, as it was for sound, because this is a mathematical truth about waves. Um, but as we'll see, it has some different implications for what we'll do here with optics. The first optical experiment we want to consider is double slit interference. The idea with double slit interference is that you have a source of waves and that source of waves. The waves hit a double slit, so this is a wall with two slits in it. And when it does that, the two slits themselves now act as sources of light producing waves. And when we see how that interferes notice here, what's gonna happen is we're gonna have consistent positions where the waves overlap and what those are as thes our positions. These air going to lead to what are known as light bands, K or interference bands. So these bands or fringes, as they're called in some books are going to represent the different positions at which constructive interference is met, where we have the requirement that Delta R is equal to mm times lambda. So here we have m Z equals zero. Then we have to m equals one. And to m equals to notice that M equals zero is right down the center in the middle between the two slits now thinking about thes the screen here, the filter and then the wall that the light is hitting being separated by a distance. L we can consider what's the mathematics of this specific situation? Drawing out a large perspective here in thinking l to be much larger than d such that we would find the angles between r one and R two and the horizontal will have to be the same. They look like parallel lines as long as l is a far enough distance away. Yeah, that is to say much larger than D. When we do this, what we'll find is that the path difference between R one and R two all happens right here at the beginning because they're parallel After we go past this segment here, what that means is that our one in our two will have the same path. So what is this segment are two minus. Our one is going to be D, which is our high pot news here multiplied by the sign of data as we can see from our graphic here. So we have Delta r equals d sine theta must be equal to M times Lambda in order to have constructive interference. Now we can simplify this further recognizing that theta is generally going to be a very small angle, especially if l is very large. Remember, that's the distance between the filter and the wall. And if we make the small angle approximation, what we find then is that for these angles we will have constructive interference now. It could be difficult to major angles sometimes, but it's not so difficult to measure distances. So some people prefer to instead right out the position instead of the angle. So again, thinking about are the picture we here As we've blown it up, we have a distance l and an angle theta, which means the vertical distance along the wall is going to be l times the tangent of theta. Therefore, we come up with y equals l tangent data. Applying this small angle approximation, we again have tangent of data is approximately Thatta and we confined that the positions of our bright fringes are going to be m times Lambda Times L over d, where all we've done is plug in theano Gle that we found in the previous picture. Now again, the EMS can be 01 or two or any integer. But one thing we should recognize is that these are fringes. So it's a certain range over which we have, Ah, bright fringe. And then there's a space. Before we have another bite fringe. In between the two fringes, there is a dark fringe, and that dark fringe is going to be where the requirement for destructive interference is met. Which again, if we just apply the same logic we did before. If we just say m plus one half instead of em where M is restrained to India Cher's, then we'll find that these are the positions of our dark fringes. So this is very helpful. We have the positions of our bright fringes and the positions of our dark fringes. I should note that if you're having trouble with any of these images, I highly recommend you go and look at your textbook where they'll have very nice, uh, color images drawn by professionals. Um, that will perhaps make a little more sense to you. One note I should make here is that the intensity of the different fringes can be found using this equation. You'll see that there's a lot of variables here that we're kind of taking for granted D and lambda and L and why all of these different things you need to keep track of what they all mean. In this case, we've added a nine on which is the initial intensity of the light coming in before the filter so you can use this actually draw a picture. If I were to have an M equals zero here, I'd have the highest point and then it would come down and then we'd have it coming back up on either side. It would be symmetric. So where a Thai? We have a bright fringe where it's low, we have a dark fringe. So here are some bright fringes and here are some dark fringes. Okay? The other experiment we should take into account is something known as the diffraction grating. The diffraction grating is kind of similar to the double slit experiment. Except we have ah lot of slits in this case. And if we have, say, end number of slits that's going toe, all all of them are going to contribute to what we see on the screen. Now we end up seeing more or less the same pattern of light and dark fringes. Except we have very narrow, very bright fringes compared to the double slit experiment. This makes the diffraction grating especially helpful when we're looking working in a small lab where we can get very obvious results that are easy to measure and easy to see now in this case If we want to define our distance d, we're going to need to define it as millimeters per slits or, in this case, the number of slits per millimeters that would be n divided by the total length l or not. L but we'll call this h the height h of the diffraction grating in millimeters. We'll give you the distance in millimeters one over the distance and millimeters. So using that, we can actually same used the exact same formulas that we have before, except recognizing that we're going to have a maximum intensity equal to end squared, multiplied by our initial intensity high. Not now. One thing to keep in mind here for this fringe is that we are going to have a rather quick fall off with intensity where we're going to have a high intensity in the middle, and then it will fall off as we move away from the M equals one position. The same thing will actually happen with the doubles. The experiment, if you go back and think about cosine squared as why gets bigger here, Cosine squared moves farther and farther away from one. So we will have while the positions air. Correct. The sighs of these bumps will get smaller and smaller like this, so we won't just maintain a constant intensity for all fringes. It will be a small reduction as we move away from the M equals zero position. So these are our first two examples of interference using wave optics. Let's do a couple of examples really diving into each one.
Reflection and Refraction of Light