Polarization & Polarization Filters - Video Tutorials & Practice Problems

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1

concept

Introduction to Polarization

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5m

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Hello, everyone and welcome back. So in this video, we're gonna talk about a new concept called polarization, which has to do with the orientation of light and also its intensity. So we're gonna talk about a couple of conceptual things you'll need to solve problems and we'll also go through a pretty straightforward equation. So let's just jump right in. So what's polarization all about back when we first studied the electromagnetic waves, we said the electric field oscillates in one axis and the magnetic field oscillates in a different axis. And that's basically what polarization has to deal with the polarization of an electromagnetic wave is always just going to be the direction or the axis that the electric field is oscillating along. So for example, for this wave over here on the left, the electric field oscillates purely on the z axis. So basically just goes purely up and down, it doesn't go forwards and backwards or off its same angle. So basically what we say is that's the polarization, the polarization of this light here of this electromagnetic wave is gonna be along the Z axis. Now, obviously drawing these diagrams and these waves would be super complicated over and over again. So we have a very compact ways of representing this by using what's called a polarization diagram. Basically, what this looks like is it just looks like a double headed arrow that points along the appropriate direction. All right, that's all polarization is. Now let's take a look at a different example because there's many different possibilities for the polarization angle or direction. Let's take a look at this wave over here, this wave points along the same exact direction. The only difference is it kind of looks like the first diagram if you were to sort of tilt it a little bit. So if you were to sort of tilt it by 30 degrees, now what happens is the electric field oscillates not purely along the, along the z axis, but it kind of oscillates at some angle here. So we'd say is that this angle, this polarization is 30 degrees with respect to the z axis. All right. So basically, it would just look like this, you're going to draw this double headed arrow and we would just indicate with a little dotted line that this angle here is 30 degrees. That's really all the polarization has to deal with. All right. Now, there's also such a thing as un polarized lights, which basically just means that the electric fields don't point in a specific direction but actually just many random directions. Now, in a lot of problems where you're gonna see un polarized lights. Usual examples are gonna be sunlight or light that's coming from light bulbs or something like that. Most of the time, it will actually just tell you if it's un polarized. But basically the way that we represent un polarized light is by using a double headed arrow, uh except in all directions. So usually what I do is you're just gonna see a couple of lines, maybe like three or four of them uh with all of the arrows like this and this just represents un polarized light. All right. So that's the basic difference between polarization or polarized light and un polarized light just has to do with the angle that the electric field makes. Um or which direction it's oscillating along. All right. Now, a lot of times what's gonna happen in is you're gonna have un polarized lights that then becomes polarized. And the way it does that is by passing through what's called a polarizer or a polarization filter. So let's talk about that really quickly. A polarizer kind of just looks like this. It's like a little circle with a little grating that looks like this. And it's basically just a filter polarizer is a filter that only allows components that are parallel to the transmission axis to pass through. So in other words, what happen is this little polarizer has a specific direction has a specific axis that are only light of that component is allowed to pass through. So what happens is when you have this un polarized light, it's moving this way and it passes through the polarizer. The only thing that's allowed to pass through is this component of the light here. The component that is parallel to this transmission axis. All right. And what happens to the other components will they basically just get absorbed or they get blocked? So these components now are no longer allowed to pass through. So what happens here is that when un polarized light gets passed through a polarization or filter or a polarizer, then it becomes polarized. And basically what it looks like here is the only component that's allowed to survive is the vertical component here. OK. So it becomes polarized. Now, the other thing that happens is that the intensity also decreases by a factor of one half. All right, because you're now removing a lot of the light components. So basically, there's actually a very simple equation for this. It's that I is equal to one half of I knots and we just call it the one half rule. So whenever you have un polarized light that becomes polarized, its intensity decreases by a half and it becomes polarized in the direction of the transmission axis, that's really all there is to polarization. All right. So let's just go ahead and just do jump into a really, really quick example problem here. So for the situation that we just described above here, if the intensity of the un polarized light is 100. So in other words, if this I knot is equal to 100 then what is the intensity of the transmitted light? In other words, what's the tr the intensity of this light that makes it out of the other side? So basically, if we want, I, then we're gonna have to use our new equation. I equals one half of I knots. In other words, it's just one half of 100 that just equals 50 watts per meter squared. All right. So that's all there is to it guys. If you have 100 watts per meter squared of intensity, then at the end, you have only 50 watts per square meter that makes it through the other side. That's it for this one. Folks. Let me just uh go through a quick practice problem that we'll jump into another video.

2

Problem

Problem

Unpolarized light with intensity of 6 W/m^{2} is incident on a polarizer. If the polarizer's transmission axis is at an angle of 45° above the horizontal, draw a diagram of this system and find the intensity of transmitted light.

A

12 W/m^{2}

B

6 W/m^{2}

C

3 W/m^{2}

D

0 W/m^{2}

3

concept

Multiple Polarizers & Malus's Law

Video duration:

9m

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Everyone. Welcome back. So in the last few videos, we saw what happens when you have un polarized light passing through a polarization filter. But many problems, in fact, most of them are gonna have multiple polarizer in a system. So I'm gonna show you how to solve these kinds of problems here because we're gonna need a new equation. I'm gonna show you how to solve multiple polarizer problems and we're gonna go over an equation known as Malice's Law. Let's just jump right into it. All right. Now, remember that un polarized light becomes polarized when it passes through polarizer. All right. So basically what we saw is that if you have the intensity of incoming lights as 100 if it's un polarized, then when it passes through this first polarizer, everything that's along this transmission axis gets passes through and then everything that's not part of this axis gets blocked out. So the only thing that survives is basically this vertical components. So you end up with something that kind of looks like this over here and the intensity drops by a half, some of the words I one was equal to 50. Now we're gonna see what happens when you have this polarized light that passes through another polarizer at a different angle. All right. So if polarized light passes through another polarizer oriented at a different angle, then two things happen. And by the way, sometimes the second polarizer is called an analyzer. All right. The first thing that happens is that this new light actually gets polarized again, it gets polarized in the direction of the seconds or the new polarizer. So in other words, what happens is this new light when it comes out of the second polarizer is actually going to be oriented along this transmission axis over here. Now, you might be wondering how is that possible Patrick because shouldn't this polarizer basically just block out all of this stuff and let nothing through. That's a great question. Basically what happens is it's kind of like how we decomposed forces into components. You could say that this light here has a component that is along this direction, the transmission axis. So what happens is all of this stuff gets sort of blocked out and then all of this stuff is what gets passed through. That's basically what happens here. All right. So then you have new lights uh that passes through at this angle here theta. And the second thing that happens is that because some of these components get get eliminated and the intensity will also decrease again, based on the smallest angle that is between the two polarizer All right. So we're actually gonna need a new equation for this. And I'm gonna show you what that is. It, the intensity is equal to I knots times the cosine squared of theta. All right. So this is sometimes referred to as the cosine squared rule. Whereas the first rule that we learned was called the one half rule. And the two basic, the two most important things you need to know is that you always use this equation whenever your initial light is un polarized like we had over here. And then you use this equation whenever your light is already polarized like you have in this situation. All right. So basically in this diagram, for this situation, you use this equation and then for this situation, you're gonna use this equation. All right, that's really the main idea behind multiple para polarizer type problems. So actually going to show you a step by step process on how to solve them. But actually, we're just going to jump right into our problem. So we can check this out. All right. So let's go ahead and take a look. So we have polarized light with an intensity of 100 watts per square meter. It passes through these two different polarizer. One has an angle of 30 the other one has is actually oriented along the horizontal axis. We wanna calculate ultimately what's the intensity after it passes both of these different polarizer. So the first most important thing to do is actually gonna be to draw a diagram. All right. So you can draw the diagram kind of like how we've been doing before uh so far, which is just draw it sort of like at this angle here, it's kind of like a little bit three dimensional. So you can see some perspective and basically what you're gonna do is the initial light is gonna be un polarized. So in other words, it's gonna kind of look like this. You're gonna have light in all directions with the little arrows and we can label this, this is gonna be our I zero and we know this is gonna be equal to 100. Now what we have now what happens is it passes through two different polarizer that are at different angles. So this one is gonna be the first polarizer and this one will be the second. So the first one we're told is angled at 30 degrees relative to the vertical. So in other words, this is the vertical over here and this is gonna be 30 degrees. Now, the second one is gonna be oriented along the horizontal axis. Now, horizontal, you might be thinking it's gonna look like this and you could totally draw it that way. But from the perspective, it's kind of a little bit weird because this would actually kind of look like this. All right. So this one's gonna look like this, this one is gonna look like this, I'm actually just gonna make this super clear by saying that this is zero degrees or you can also just say this is horizontal. All right. All right. So now that we're done with step one, we can move on to the next step here. Ultimately, what we're trying to do remember is we're trying to figure out the intensity of light over here after it passes both of the polarizer. All right. So we need to do is figure out what happens to the intensity at each one of the steps. So the second step is that for each nth polarizer, we're gonna use our two equations, the one half rule or the cosine squared rule, I might be looking at these and, and see that they look a little bit different from how we've written them up above. And it's for a very important reason, the most difficult part of these types of problems is the fact that the I that you use in one equation will actually become the I knots that you use in the second one. So if you're not careful, it's very easy to sort of confuse the different types of variables that you're working with. So what I always like to do is I like to label each of my intensities. The initial one is always going to be I zero. Whereas the after the light passes the first polarizer, that's I one and then after it passes the second polarizer, that's gonna be I two. That just means that it's really, it's impossible for me to get my variables confused as I'm working with the different equations here. So that's what I want you to do. So for each nth polarizer, we're gonna use that same exact process. So in other words, this is gonna be I one and this is gonna be my I two. So we need to do is figure out what happens after the first polarizer. So in other words, this is gonna be the first polarizer here. Now, we just have to figure out which one of these equations that we use. And it really just comes down to whether the light is un polarized or if it's polarized. So what do we do? What do we do? So in other words, to figure out I one, this is gonna be what happens uh when the initial light that's un polarized passes through the filter. And remember the rule is we always use the one half rule whenever the light is un polarized. So that just means that for this first polarizer, we're gonna use the one half rule over here. This just says that I one is gonna equal one half of I knots. So this is gonna be one half of 100. And we've already seen that that's just 50 watts per meter squared. All right. Now, that's not our final answer because remember we want what happens when it's all the way uh uh when it, when it passes the second um polarizer. So now we just keep going, just, just do it multiple times for each nth polarizer. So now we just go to the second polarizer over here. And now we're going to label this as I two. So which equation do we use? Well, now what happens is once this light passes through the first polarizer, the intensity drops to 50 it's gonna be along this axis over here. So this light is already polarized once it passes through the second polarizer. So that means we have to use the cosine squared rule. So for the second step or for the second polarizer, we're gonna use cosine squared. So what the equation is is that it's I two is gonna equal I one times the cosine squared of this theta angle over here. All right, see how by labeling these intensities, it's very difficult for me to get confused. It's very easy to keep track of which intent I'm dealing with. All right. So now we just need to sort of plug in into our equation. So we have that I two is equal to, this is gonna be I one which we know is 50 we have the cosine squared. And then what's the angle that we plug into our formula here? What happens here is remember that this bliss light is polarized at 30 degrees. So if you sort of draw this angle over here do we use 30? Do we use zero? Well, what happens here is that remember this angle or this initial polar is 30 degrees. So in other words, this thing over here is 30 but that's not the angle that is between the polarizer. That's only just the angle with respect to the vertical. So we actually do not use 30 degrees here. We also don't use zero degrees because that's the orientation of the second polarizer we have to do is we have to figure out the angle that is between these two axes over here. It's not 30 it's not zero. It's actually the one that's uh that's sort of in between 90 uh and 30. Sorry, it's 90 minus 30 that's just in equal theta equals 60. So that's actually really, really important here. It's always gonna be the angle that's between the polarizer. So that's what we plug in to our formula. So just be very careful here. I've had written out for you just so you really, really understand this, be very careful when you're determining theta. It's always the angle that is between the polarizer and it's not the angles of either one of them. All right. So you always have to figure out what that theta is. So if you go ahead and work this out, what you're gonna get is 50 times and if you go ahead and work out cosine squared of 60 you're actually just gonna get 1/4. And so what you're gonna get here is a final intensity is 12.5 watts per meters squared. So this is actually your final answer. So what happens here is when this light exits this, um this final polarizer, it's gonna be polarized in this direction. The horizontal and its intensity is gonna equal uh 12.5 watts per meter squared. That's it for this one. Folks let me know if you have any questions and I'll see you in the next one.

4

Problem

Problem

Horizontally polarized light is incident on a polarization filter. The initial intensity of the light is 0.55 W/m^{2}, but is then reduced to 0.40W/m^{2} after passing through the filter. Calculate the angle of the transmission axis of the polarization filter with respect to the horizontal.

A

43.34 degrees (0.756 radians)

B

37.69 degrees (0.658 radians)

C

31.48 degrees (0.549 radians)

D

Undefined

5

example

Example 1

Video duration:

9m

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Everyone. So let's take a look at this example problem. There's actually a lot going on in this problem. So I'm just gonna jump right in. We're told that sunlight is initially un polarized light with an average intensity of about 1350 near the earth's surface. Now, the first part here says that if sunlight passes through two polarizer that are angled at 90 degrees with respect to each other, we want to find the intensity of the light after passing through the second polarizer. So I'm just gonna jump straight into the steps here. First thing we want to do is just draw a diagram and label what our um initial light is. So this is gonna be my diagram. I've got my initial light, which is un polarized, right? It's gonna kind of look like this. Uh or you can draw it sort of like that, I guess uh like that. All right. So this is my initial lights. This is initially un polarized and the intensity is 1350. All right. So this passes through one polarization actually one polarization filter and these things are angled 90 degrees with respect to each other. Now, it doesn't doesn't tell us which specific angles, but it turns out it actually doesn't matter because remember the only thing that matters between the two polarization filters is the angle that is between them. So what I can do here just really simply is I can say that the first one is sort of vertically polarized and the second one is going to be horizontally polarized, right? So this is just gonna be on a polarization of 90 degrees. And then this one is just going to be zero degrees such that the angle between them is just 90 degrees. That's the right. OK. So um basically what happens is when this light passes through the first polarization filter, it's gonna be vertically polarized. This is gonna be my I one and then when it passes through the second polarizer, it's gonna be polarized horizontally. And this is gonna be my I two. OK. We wanna calculate what this I two is. So let's just go ahead and set up our um steps for the NTH polarizer. There's two of them here. There's, this is the first one and this is the second one here. So let's just go through the first one really quick. So the first one has initially un polarized light that's passing through. So we're gonna use either one half rule or the cosine square rule. Hopefully realize that it, we're gonna use the one half rule here. So what happens here is that I one is just gonna be one half of I zero. In other words, it's just gonna be one half of 1350 and that's gonna equal 675. All right. So this I one here is 675 when it passes through. Now, let's take a look at the second polarizer. Now, the second polarizer has initially polarized lights, but then it gets polarized to a different angle. So we're gonna have to use the cosine squared rule. So this I two says this is gonna be I one at times the cosine squared of the angle. Now remember this angle here, it doesn't have to do with the angles of the individual axes, but the angle between them this at 90 this one's at zero. So the angle that's between them is 90 degrees. So that's what we plug inside of our formula here. So this just says that this is gonna be 675 times the cosine squared of 90. Now remember the rules for cosines whenever you have a cosine of 90 degrees that always cancels out. So cosine squared of 90 is also just equal to zero. So basically when you get, when you calculate this is that the intensity of the second light here is gonna be zero watts per square meter. And this actually has nothing to do with the fact that the intensity drops by a half or anything like that. This intensity could be whatever number. Basically what this means here is that whenever you have two polarizer ankled 90 degrees with respect to each other, this is sometimes called cross polarization. And that means that it actually cancels out all of the lights that's passing through it. One way you can think about this is that the first filter polarizes it vertically. And then with the second filter, there is no components of this vertical uh vertically polarized light that can survive once it passes through the second filter. So you kind of just cancel each other out. All right. So that's really important here. That's the answer to the first part. Now let's jump into the second part here because basically what the second part says is that we're gonna take a third filter and we're actually gonna sandwich it in between the two filters, right? It says that a third filter with a transmission axis of 30 degrees to the horizontal is inserted between the first two. So we're gonna stick another polarization filter in here and see what happens. We're gonna see what the, what the intensity of the sunlight is after it passes through all three polarizer. OK. So let's just go through the steps again, right? So here we have a transmission axis like this and I've got initially un polarized lights like this. This is gonna be my un polarized light, which is 1350 but now we're actually gonna have three polarizer. There's the first one, then there's the second one and then there is the third one. All right. And just as before what's gonna happen here is we're gonna have this first one is gonna be at 90 degrees. The third one is gonna be at the horizontal like this. And the second one is actually gonna be 30 degrees with respect to the horizontal. So basically, this is your horizontal and it's gonna be oriented kind of like this such that this angle here is 30 degrees. So this is 90 this is gonna be zero degrees. OK. So what happens here is that after it passes through the first filter, it's gonna be vertically polarized. That's gonna be I one. Then when it passes through the second one, it's gonna be polarized at this new angle here, which is 30 degrees, that's I two. And then when it goes through the third one, it's gonna be horizontally polarized and this is gonna be I three. All right. So now let's go through each one of our polarizes. This is the first and equals one. This is the second one and equals two and equals three. All right. So here was, here's what happens. So for the first polarizer, we're gonna use the one half rule again, nothing's changed from before. It's just gonna be I one is equal to uh 675. So it's just gonna be half of 1350. It's gonna be exactly what it was before what it was from part A nothing has changed. This I one here is 675. OK. So it's the second filter. Now what happens? The second filter says that I two is gonna be I one times the cosine squared of theta. Now, what's really tricky about these types of problems is when you have multiple angles is that also you can get confused with your angles. So it's a really good idea to label them the angle that is between these two first filters. I'm gonna call theta 12 because it's the angle between the 1st and 2nd polarizer. All right, then this angle over here, the angle between the 2nd and 3rd, I'm gonna label this as theta 23, the angle between the 2nd and 3rd. OK. So what this says here is that I two is gonna be I one times the cosine squared of theta 12. All right. So this is just gonna be 675 times the cosine squared of theta 12. All right. Now, what's that angle? Now, remember just be very careful. You always want to actually calculate what the angle is between the polarizer really, really important here. So don't just stick in 90 don't just stick in 30. You have to figure out what's the angle between them. So what happens is if you sort of like project this out, this is the 90 degrees. This is actually the angle that you need, not the 30 degrees. So theta 12 is actually just 90 minus 30 which is 60 degrees. That's what you plug into this formula, right? So just be very careful. 675 times the cosine squared of uh this is just gonna be 30 or sorry, 60 degrees. Now, what you should get uh is you should get, let's see, um you should get uh 168.75 and that's watts per meter squared. So that's actually what comes out the other side, 1 68.75. So here, what you can see here is that once you've inserted this third polarizer, because it's not at 90 degrees, there's actually some amount of light that survives. So, whereas initially before you had these two polarizer and they completely cancel each other out to zero. When you stick one in the middle, you actually now have some of the light that passes through, which is kind of weird. It's kind of counterintuitive, but that's actually what happens. All right. Now, let's keep going. We've got one last step here, the third polarizer. This says that the I three is gonna be I two. And again, we use the cosine squared rule. So this is gonna be I two times cosine squared. But now we're just gonna use the, from 2 to 3, we're gonna use this angle over here. All right. So here what uh what this says is that I two is going to be 168.75 times the cosine squared of this angle theta over here, this angle is gonna be the angle between this 30 degrees and zero. So in other words, this is gonna be 30 minus zero. This just equals 30 degrees. That's what we pop into this equation here. So this is gonna be the cosine squared of 30. And what you end up getting here is 100 26.6 and this is watts per meter square. So this is actually the final answer, by the way, that's what survives uh after it passes through the third polarizer, and it's gonna be ho horizontally polarized 126.6 watts per meter squared. So anyway, there's a lot of moving parts in that problem, but it's kind of just tedious setting up the diagram and everything. But hopefully it made sense. Let me know if you have any questions and let's move on to the next video.

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