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32. Electromagnetic Waves

Polarization Filters


Polarization Filters

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Hey, guys, in this video, we're going to talk about this important quality of electromagnetic waves called the polarization, and these objects called polarization filters. Okay, let's get to it now. What the polarization is is it's the angle of the sorry. It's the polarization of an electromagnetic wave. Is the angle of the electric fields relative to self origin relatives at zero degree angle? What do we mean by that angle? I have four figures here where each red double headed arrow represents an oscillating electric sorry and oscillating electric field that's coming right at us. So what we're looking at is there some oscillating electric field and we're looking at it head on. Okay, so it's oscillating up and down, up and down hints the double headed arrow. Okay, now in the first image, although on the right, By the way, each blue vertical line represents that origin. That zero degree angle that we're measuring the polarization from in the left most image. We have clearly a zero degree polarization. Okay? No, sorry. It is polarized because it is at a particular angle. But that angle is zero. It lines up with our origin. Here we have some acute angle. It doesn't really matter what it is. It could be 30 degrees, 45 degrees, 60 degrees, etcetera. It's some degree. Sorry. It some angle that is less than 90 degrees. And the third image. We have a right angle. We have a 90 degree polarization. Now, the polarization is always gonna be between zero and degrees. Because if you continue rotating that polarization sorry that electric field you might get one that looks like this. Don't draw this guy's Just listen for the explanation. You might get one that looks like this. This is still gonna have an acute angle. Okay, I didn't want you to draw because I was just gonna erase it now all the way to my right. Right here. What we have is a whole bunch of electric fields that are all at different polarization angles. Now, truly, to represent this image, we would need a new infinite number of electric fields, all of which have every possible polarization angle. Right. And I could keep drawing more and more of them etcetera. This combination of all of these multiple waves that all have different polarization angles and it comprises all of the polarization angles this we refer to as unp ola rised light Okay, so unp ola rised light is made up of a whole bunch of different electric field orientations that compose all 360 degrees of a circle. Okay, so now we want to talk about what happens when light passes through a polarization filter And what a polarization filter is is it's something that filters out all light except for light off a particular polarization. Okay, now let's consider initially unp polarized light as in the image all the way above me This unp ola rised light right passes through a polarized a polarization filter And this polarization is choosing a particular angle right? It's choosing a particular polarization and all of the light other than the light that is at that particular angle gets filtered out All that light is gone. Okay, so some amount of light from the initially unp polarized light comes through and now has the polarization chosen by that polarization filter. Okay, the question is, how much light do you lose? Well, for polarization and polarization filter, We track it by intensity, okay? And when initially unp polarized light like this unp ola rised passes through a polarization filter. Regardless of the angle of that polarization filter the intensity always drops by half. Okay, now, a little bit of notation This right here was called I subzero or I'm not that variable. I not does not represent UNP polarized light. It's always used for whatever, like passes into the polarization filter, regardless of whether it's unpasteurized light or light, that is already polarized. Okay, because in the next equation, I'm gonna use I not again, but it's going to be for initially polarized light. Okay, I just want you guys to be aware of that notation because your professors, they're gonna use it. Your book is gonna use it. So it's very important that you understand the notation clearly. Okay, so now let's consider what happens whenever initially polarized light passes through a polarization filter or a polarizer. Okay. The intensity drops like so I becomes I not or not becomes I not times cosine squared of five. Okay, now five is what I would call the polarization difference. Okay, So, relative to whatever zero we have, whatever origin we've measured are polarization against. If you have some initial polarization of 20 degrees and the polarizing filter is 45 degrees. The polarizing difference is the difference between those angles 20 and 45. The differences 25 That's what five would be okay so automatically if you have a polarizing filter at degrees to the initial polarization angle, that's always going to result in a complete blockage of light. No light is gonna pass through in that scenario. Okay, let's do a quick example. UNP Polarized light passes through a polarization filter with an angle of degrees relative to the vertical. This newly polarized light passes. There was second polarization filter with an angle of 55 degrees relative to the vertical. If the intensity of the UN polarized light was 100 watts per meter squared, what's the intensity of the light after passing through the second polarizing filter? So just for notation safe, I'm gonna call I not the UNP polarized light. It passes through the first polarization filter, and then I'll call it. I want the results of the first polarization filter. Then it will pass through a second polarization filter, and I'll call it 52 Okay, just for a notation sake, just to keep everything clear So the initially the light is completely unp ola rised. So we know that when it passes through any polarization filter its intensity is going to drop by half. So the first polarization filter I one equals one half I'm not, which is one half of 100 watts per meter squared. That's what it started at right which is simply watts per meter square. Very easy. Unp ola rised light going through a polarization filter always dropped by half. Now the second polarization filter Now we have a polarizing difference. Okay, we have light that is polarized toe 45 degrees relative to the vertical That is the polarization of our light when it leaves the first polarizing filter The second polarizing filter is 55 degrees relative to the vertical. We want to apply the equation I two equals I won cosine squared If I remember that this variable right here which in our equation I called I not is simply the intensity that enters the polarizing filter In this case we call it I won. So the question is, what is our polarizing difference? Well, if our polarizing filter Sorry, our initial polarizing filter was 45 degrees to the vertical and our final polarizing filters, 55 degrees to the vertical. Then our polarizing difference is degrees minus 45 degrees, which is simply 10 degrees. Our initial polarization entering the second polarization filter was 50 watts per meter squared times, coastline squared of 10 degrees. And this whole thing equals 48. watts per meter squared. Okay. Pretty straightforward application of equations. Polarization. Remember, guys, that fi is not the angle of the polarizing filter. It's the polarization difference. It's the angle between that. The initial polarization of the light entering it and the polarizing angle off the filter. Okay, guys, that wraps up our discussion on polarization and polarizing filters. Thanks for watching.

Initially Vertically Polarized Light

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Hey guys, let's do an example on polarization. Initially, vertically polarized light cares an intensity of 0.55 watts per meter squared. If you had an adjustable polarizer, what angle relative to the vertical would you want to align it so that the intensity of the light exiting the polarizer is 0. watts per square meter? So we know that Oh, our light is initially polarized. It's saying it's initially vertically polarized. So if you have initially polarized light entering a polarization filter, we have to use the equation I equals I not cosine squared of five. If the light was initially UNP polarized, thin passing through any polarization filter would just drop the intensity by half. But this is not the case, okay, so you always have to keep track. Is your light initially polarized? Is your light initially UNP polarized? In this case, it's initially polarized and it's passing through and unknown angle that angle of the polarization filter we can adjust all we know it is the input intensity and we want to adjust it to get a particular output intensity. So we know that the initial intensity is watts per meter squared, and we want the output intensity to be 0.2 watts per meter squared. So we want to find what angle gives us this. So what I'm gonna do is I'm gonna divide I'm not over and deal with the square root. So that tells me cosign of Phi equals the square root of I divided by I'm not, which is just the square root of 0. divided by 0.55 And that's about 06 So if I plug this into a calculator to find out what five my polarization differences, that tells me that my polarization difference is degrees. Okay, so how would you align your polarization? Filter? What angle? Relative to the vertical Would you align it at? You would align it 53 degrees from the vertical right. And that's because your initial polarization was zero degrees relative to the vertical, and five is your polarization difference? Let's say this wasn't initially vertically polarized. What if this was initially polarized? 10 degrees to the vertical. What would you Where would you put your polarization filter while 53 plus 10 would be 63 degrees. Okay, that would be one option. Technically, you can also go in the other direction. And 43 degrees to the left of the vertical would also be an option. Okay, so just remember, five is not necessarily the final answer. Five is the polarization difference. The final answer in this case is 53 degrees from the vertical. Okay, we found that by finding five, obviously. All right, guys, Thanks for watching.

In each of the following cases, initially vertically polarized light enters the polarizing apparatus with the same initial intensity. Which polarizing apparatuses will cause the light to exit with the largest intensity, 90° from its initial polarization? 

a) A single polarizing filter, oriented 90° from the vertical 

b) Two polarizing filters, the first 45° from the vertical and the second 90° from the vertical 

c) Two polarizing filters, the first 60° from the vertical and the second 90° from the vertical 

d) Two polarizing filters, the first 30° from the vertical and the second 90° from the vertical.