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So let us now come to the solutions to the exercises in this section about electromagnetic waves, so



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I hope you took the time and solved these tasks.



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So if you haven't done so, please take a sheet of paper and read the tasks in the previous lecture.



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Now I'm going to present you the solutions.



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So first of all, we had a task where we should plot the real part of these two electromagnetic waves,



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especially of the electric fields, for four different times.



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And from that, you should then determine the polarisation and the corresponding magnetic field.



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So let us begin.



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We have these two functions, A and B, and I want to start with A..



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So what I'm going to do here is, first of all, I want to explain to you what this coordinate system



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on the left.



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So we have here our base vector K and so in general, this is a dot product of the key vector and the



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position vector are so since here it is K times X, we know that our character is along the X direction,



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so this means our wave propagates along X..



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This is why I've chosen this axis here.



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Furthermore, here we have to position X equal to zero.



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And then here I want to show you where X is the space vector divided by four, ten by two or three over



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four.



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And then here the full, uh, the full lambda full wavelength.



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OK, so now let's calculate, um, our real part of the electric field.



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So in a the real part of E.



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Is just one of square root of two Epsilon zero, and then we have here a zero, of course, and then



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for the Y component, we have one times cosine of this argument and then one times I times sign of this



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argument.



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So the real part is just as cosine function.



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So we have a cosine of K, X minus Hormiga T and for the for Z component we get three times cosine.



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The argument plus I times I times sine which is minus sine.



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And this is the report minus sign Kate X, minus Omega T.



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Now let's go ahead and calculate this wave for the time T equal to zero.



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So here we have four four X equals zero and T equals zero.



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We get cosine of zero, which is one.



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So are our Y component is one and our Z component is zero because the sign of zero zero.



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Then if we go one quarter of a period, the cosine turns into a zero and the sign turns into one.



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But we have a minus sign so we get minus one.



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So this means we have here a minus one.



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And then for half a period, the cosine turns into minus one designed to zero.



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And then for another quarter period we go here.



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And then finally we are back at the beginning.



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So this means let me try to draw the wave on this tablet here.



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It's a bit difficult, I think on your sheet of paper, it's much more easy.



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So let's go ahead.



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Try to connect these these dots.



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So we get something like like this one.



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And that's OK.



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Yeah, it doesn't look look too good, to be honest, so here we have, um.



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Yeah, we we can try it like it.



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Like this.



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And then we got here.



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Um, yeah, like this and.



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OK, so I think you you can see how this works and you can really see how the electric field wraps around



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in the circle.



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So we already know of a solution for for the second part.



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So this is a circular polarization.



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For the other times, let's think for a second.



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So if our time propagates or is continued by one period over four, then we increase this argument here.



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So this means our callsign and our sign are shifted into the negative direction.



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So this means that we start not at this point here for zero, but we start at this point here for zero.



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So this means our whole wave is then shifted, so to say.



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So you can you can really see how the wave propagates along this direction, because this is our character.



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And this means you have to shift this wave along this direction when you propagate over time.



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So I will not draw the other images here.



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I think you can really imagine what it looks like.



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And also we have these animations in the previous sections.



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So let's continue with no, it would be with the second wave.



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So here once again, we have to use the real part, of course, because this is the physically relevant



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quantity.



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So, by the way, yeah, this is here.



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This is a vector.



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So what we get in this case is one of five is zero and then a vector.



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So this time our Y factor is obviously along the Z direction.



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So this is why I've drawn this coordinate system like this.



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This time we have here X and here Y and Kate is along.



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This is the direction, which is the direction of the propagation.



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All right.



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So now let's consider the real and the imaginary parts.



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So here this time, it's a bit more difficult because we have in this vector one minus two AI, which



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is a real and imaginary part.



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So this means we will also get two parts here.



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The first one will be one times the cosine.



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So this is just one times the cosine of casette minus Omega T, then we get another term, which is



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minus two times I times I times sign of the argument.



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So this means we get um.



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Yeah, we get two times.



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The sign.



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Yeah, exactly, because minus two times I squared plus two times, that was a sign of Casey minus Omega



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T. And then for the Y component, we get two times cosine.



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And of course, here we got four times sine Casey and we get T zero and that one.



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OK, so now we can write this down a bit of a different way, we write down that this is equal to one



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over five e0.



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Zero two, and then we can use a relation which allows us to tell what is cosine of this argument plus



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two times sign of this argument, and this is something that you do not have to know, but you can look



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up on the Internet, for example.



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So I use Wolfram Alpha and look at what this is.



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In this particular case, this is just the square root of five times the sign of this argument, Casey,



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minus Omega T, and then we get another constant, which in this case is if you look it up, it is the



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arcus tangles of one half.



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But for us, it doesn't really matter what it is.



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Let's just say it is some constant because our cost of one half is just some number.



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It doesn't really matter what it is.



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And here, of course, we get the same thing.



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So Casey minus Omega T plus this constant.



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OK, now we can start drawing and for for a moment to make it a bit easier.



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I just want to show you here that you get a general idea what this wave looks like, look looks like.



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So I will not consider this arcus tangency.



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I will get to zero in the end.



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It is just a phase shift.



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So what what we get here in this case is we get for the time equals zero and the Z component zero,



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we get an argument of zero.



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So the sign in both cases is zero.



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So we get really zero here.



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And then one quarter of the wavelength further we get a one because the sign has done his maximum.



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So it turns into one.



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So this means we get for the X and for the Y component, the maximum.



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So let's say we get this one here.



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Then for half the wavelength we get again zero because design is zero.



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And then four.



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Yeah, three of four of the wavelengths we get minus one and again, zero.



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So let me try to connect this.



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I hope it doesn't look too ugly.



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OK.



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Looks a bit ugly, but OK, and yeah, and so this time you can really see how the wave is linearly



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polarized and how it the electric field just oscillates in this plane that is here and yeah.



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Spent by this latest line between the X and Y axis.



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And of course, there's the axis.



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And now what we only have to consider here is this Arcus Tongans are four and half, which means we



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get an additional phase shift.



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So this means actually our whole wave is a bit shifted.



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And then when we consider the different times, you can once again see how this whole sine function



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is shifted.



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Yeah.



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So this means here for this position, we do not get zero, but we get a negative value.



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So this means our whole wave has propagated along the Z direction.



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So we have already discussed this.



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The wave and A is circularly polarized and the wave and B is linearly polarized.



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So we know our answer for you for the second task.



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So we know that determine the polarization we get in a circular and and B, we get linear.



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All right.



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Now test three in test three, we have to calculate the corresponding magnetic field.



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So how is that done?



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You may ask because we only have the electric field.



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How can we do it?



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And what I want you to do is I just want to, uh, want you to think about the orientation of the magnetic



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field.



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And what we have learned is that the magnetic field is perpendicular to the character and it's perpendicular



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to the electric field and also B and perpendicular to each other.



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So this means we must be able to express to be field.



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So this is proportional to the vector product of K and E, given the fact that both of them are perpendicular,



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which must always be the case.



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And it's it is the case here.



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So all we have to do here really is to calculate the um.



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Yeah.



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The vector product.



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So in a what we get is OK vector, we have discussed this already, it's along the X direction so we



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can just write one zero zero times and then we must just take this vector here in the beginning of the



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end of, uh, characterizing the direction of this EESA, we have one over the square root of two and



156

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then we have zero one and I and now if we calculate this, we get for the X component zero for the Y



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component, we had zero minus I and for the Z component we get one.



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So this means our B of our A. is I'm sorry I forgot one of a square root of two.



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It is one of a square root of two and then the vector zero minus I one times B zero to account for the



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unit and then the exponential function.



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So that's the same argument, of course, as previously for the electric field.



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And what we get in this case is, again, a circularly polarized wave, which is always perpendicular



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to the electric field that we have drawn on the previous slide.



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So if you want, you can go ahead and add this magnetic field with a different color in your sketch.



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Now, for B, we must do the same thing.



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So again, the cross product of the vector and of this vector characterizing the polarization or the



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the orientation of the electric field.



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So this time the vector is oriented along the Z direction.



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So we get here zero zero one and what we calculated, we get one of five and then for the X component



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we get zero minus two minus minus four.



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I for the Y component we get one times one minus two ie minus zero.



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And for the Z component we get zero point zero.



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All right.



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So this time the B field is given by one over five, minus two plus four.



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I one minus two I zero times be zero times exponential function I K, Z, minus God.



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OK, so these are the two solutions.



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And also you can see here that like for the electric field we have here, this vector, when we take



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the real part and the imaginary part, we see that both of them are parallel to each other.



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So this means we get a linear polarization also for the magnetic field.



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So once again, you can go ahead and at the magnetic field into your sketch using a different color.



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And also something that I have forgot to to mention here in task number B or task number two, we are



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just looked at our sketches to determine the polarization.



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But of course, you can also look at this vector here where you see that when you take the real part



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of this vector and the imaginary part of this vector, that they are perpendicular.



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So this is why it's circular.



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And here you take the rear part of the vector and the imaginary part of the vector and you see that



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they are parallel, which is why they are linearly polarized.



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And here it's the same thing.



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All right, so this brings us to our last task, which is very easy once you have solved the third task.



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So here again, we need the relation that the electric field is always perpendicular to K. and is always



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perpendicular to be.



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00:16:41,910 --> 00:16:46,740

And we must explain why the following electric field cannot correspond to an electromagnetic wave.



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And I can already tell you the answer.



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It is because in this case, the electric field is not perpendicular to K, and we can check this by



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just calculating the orientation where we can just take one eye minus one and some factors which we



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do not have to write down here.



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And then let's calculate the DOT product with the K vector, which in this case is along.



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Why?



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00:17:13,590 --> 00:17:21,240

And this is of course one time zero zero plus eight times one is EI plus minus one time zero zero.



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So we get EI, which is obviously different from zero.



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00:17:24,850 --> 00:17:33,990

So this means the electric field is in this case not perpendicular to K and therefore it does not correspond



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to an electromagnetic wave.