Hello, everyone. And welcome back. So when we study mechanical waves, we saw that some questions will not only ask you to draw the functions, but they'll also ask you to write out their wave functions. We're going to see the exact same thing for electromagnetic waves. And so what I'm gonna show you how to do in this video is I'm going to show you how to write the wave function for electromagnetic waves. We're going to see a ton of similarities between how we did this for mechanical waves, but we'll also see some differences that you'll need to know to solve problems. So let's go ahead and just jump right in. So remember that electromagnetic waves are transverse waves, which means that there's something going up and down. Now for mechanical waves, it was the string itself, you whip the string up and down and the string is going up and down for electromagnetic waves. It's a little bit different because it's actually the electromagnetic field itself that's sort of oscillating, it's going up and down like this. Now, the magnetic field was sort of going forwards and backwards like that. And we saw that these things were perpendicular, but they're still transverse, there's still something sort of going up and down or side to side or something like that. All right. So this was the E field and then this was the B field. All right. So because these are also transverse waves that we can also use sine or cosine to describe them, so we can use them sinusoidal functions. All right. And we're going to have to write their wave functions. All right. The only difference between electromagnetic and mechanical waves is that because electromagnetic waves are made up of two oscillating waves, we have an E field and A B field, then we're going to have to write two wave functions to describe them. All right. So we, for, as we're so, whereas for mechanical waves, we only needed one equation for electromagnetic waves, we'll need one for the electric field and then one for the magnetic field. Otherwise, what we're going to see is that the structure of these equations is actually very similar. All right. So let's talk about that. The structure of the mechanical wave equation was we had an amplitude and we had a sine term and then we had a KX minus omega T. What we're going to see for electromagnetic waves is something very similar. We have a sine and a KX minus omega T. And by the way, some of your books might actually use cosine instead of sine. If your book used a cosine for this function, then they're going to use a cosine for the electromagnetic waves. But remember you can use either or the difference really just has to do with where the wave starts. Like the wave started over here, we would use a cosine, but that's all arbitrary, right? So it's totally fine if you see a cosine. All right. Now, this front term here for the mechanical wave was the amplitude and the uh sort of analogy or the, the for mechanical waves was basically how high the string actually was going when you whipped it up and down. For the electromagnetic waves, it's a little bit different because there's nothing really going up and down. What's going on here is that the electric field strength is changing. So what happens here is that this maximum value we're not going to use amplitude, we're actually just going to call this E max. And similarly over here, this is going to be B max. This is sort of like the amplitude analog for an electromagnetic wave. All right. So what goes on here in this front term isn't a, it's just E max and B max. All right. That's really the only difference to these equations. Everything else structurally is the exact same. For example, this K term over here we found was the wave number which was two pi over lambda. It's the exact same thing. It's just that you have a different lambda because light waves are different than mechanical waves. And then for this omega term, which was the angular frequency, it's gonna be the exact same sort of variable here. It's gonna be angular frequency. Just your F is gonna be a little bit different. All right. So the last thing I want to point out here is that the electromagnetic wave, the E and the B field are always what we call in phase. And this just means that they have the same KX minus omega T term, which is actually really useful for us because basically what it means. But once you figure out what the KX minus omega T is for one of the equations, then you figure it out for both of them because it's going to be the same for both of them. All right. So it's going to be the same KX minus omega T, once you've solved it, once you've solved it for both of them. And by the way, this just this phrase in phase just means that they reach their minimum zero and maximum values simultaneously, which we've also seen from the graphs, they hit their zero points at the same point, they hit their maximums at the same point respectively in their cycle and then they just repeat the whole process over and over again. All right. So let's just go ahead and jump into a problem right here. We've got an infrared laser that emits a 10 micro wavelength. This is going to be our lambda in the X direction. So we've got that the E field is parallel to the Y axis and it's got a max value of 1.5. And by the way, when it says that the E field is parallel to the Y axis, what you'll see here is that this, this E field is written with A Y that just means that it's sort of parallel oscillating in the Y axis. All right. So that means we're going to have the same thing over here. The E is oscillating in the Y axis. So the EY of X and T and then on the B fields, we're going to have parallel to the Z axis. And what we actually want to do here is we actually want to write out what their wave functions are. All right. So that's what we're tasked to do here. Write the wave functions of E and B. Now, the first thing you might be wondering is what do I actually pick, do I have to pick a sign or do I have to pick a cosine, which do I use? And so it actually brings me to the last point I want to make in this video, which is that if problems don't indicate a waves start point with either the graph or the text, then you could actually use, use either sine or cosine. Sometimes you've got a problem in which it doesn't really specify which one you're supposed to pick. And it's because it's kind of arbitrary, right? So that's not as important as it was for mechanical waves, you can pick s or cosine for this particular problem. So I'm just gonna go ahead and pick sign. So this is going to be emacs and then I'm gonna have a sign of KX minus Omega T and then I've got B max, this is gonna be signed of KX minus Omega T. All right. And by the way, we know this is going to be a minus sign like it was for mechanical waves because it's traveling in the X direction. And we can assume that that's just positive. All right. So if we take a look at our variables here, we actually have what the maximum value of the electron, the electric field is. So we have what the E maximum is. So that's the, that's what we've had, we've got here. What we really need to do is we need to figure out what the K and the, and the omega are because that's what I need to actually write out the full equation. All right. And you'll notice here that I also don't have what the B max is, but it can also figure that out. All right. So the first thing I'm gonna do is I'm actually going to look at this K term over here and see if I can figure that out. So how do we figure out K? Well, if you remember K stands for the wave number. And there's a special equation for that which is two pi over Lambda. So this is going to be K equals two pi over Lambda. And I actually have what LAMBDA is. Lambda is just 10 micrometers. So this is going to be two pi times or sorry, divided by 10. Uh This is going to be times 10 to the minus six over here. All right. So if you go ahead and work this out, what you're gonna get uh is you're gonna get, uh let's see, this is gonna be, this is gonna be 6.28 times 10 to the minus five. And uh we don't actually need the units for that. All right. So we've got what the K term is. Now, let's go ahead and look at the second term here. We need the omega term. I'm gonna go ahead and work that out over here. So we need omega. Now, remember omega is the angular frequency, which is equal to two pi times the frequency, but it's also equal to something else which is actually much easier to get, which is just a V times K. So in other words, remember that V for electromagnetic waves is actually going to be C. So this is also going to be equal to C times K. Now, just because we figured out what K already is and we know what C is, we can actually just figure it out more directly this way. So this is going to be three times 10 to the eighth and this is going to be times K which is 6.28 times 10 to the minus five. All right. So if you go ahead and work this out, what you're gonna get here for Omega is you're gonna get uh this is gonna be, let's see. Um What I get is one points. Oh I'm sorry, this is the times. Oh This shouldn't be negative. This actually should be positive. Oops. So that should be a negative sign. There should be positive. And what you're gonna get here is you're gonna get uh 1.88 times 10 to the 14 and that is going to be the angular frequency. All right. So that's your angular frequency. The last thing I need to do is now I just need to figure out what B max is. That was my last unknown variable. So how do I figure that out? Well, I can always just relate to B max or any of the magnitude of B to E by using the uh equation E equals CB over here. So I have what B max is equal to. So I have what E max is equal to. So in other words, E max over C is going to equal B max. So that means that 1.5 times 10 to the sixth, this is gonna be five divided by three times 10 to the eighth is gonna give me my B max which is gonna equal 0.005 otherwise known as five times 10 to the minus three Teslas. All right. So what we can do here is now that we have all of our variables, we can write out the final expressions so that EY of X and T is going to be E max 1.5 times 10 to the six times the s. Now you've got the K term which is 6.28 times 10 to the uh this is going to be fifth minus and that's going to be the X. Don't remember. Don't forget the X minus 1.88 times 10 to the 14 T. And then we've got that BZ of X and T is going to equal uh five times 10 to the minus three. This is going to be sign and this is gonna be 6.28 times 10 to the fifth X minus 1.88 times 10 to the 14th T. Remember, we can just plug in the same exact values for this KX minus Omega T term. All right. And this is your final answer, by the way, this is your EY and your BZ wave functions. Anyway. So that's it for this one. Folks. Let me know if you have any questions.