Polarization & Polarization Filters - Online Tutor, Practice Problems & Exam Prep

Polarization refers to the orientation of the electric field in electromagnetic waves. Light can be polarized by passing through a polarizer, which allows only the component of light aligned with its transmission axis to pass, reducing intensity by half. When polarized light encounters another polarizer at an angle, the intensity is further modified according to Malus's Law: $I_2=I_{1}cos\left(\theta \right)²$. Understanding these principles is crucial for analyzing light behavior in various applications.

Introduction to Polarization

Hello, everyone, and welcome back. So in this video, we're going to talk about a new concept called polarization, which has to do with the orientation of light and also its intensity. So we're going to 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? Well, 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. It doesn't go forwards and backwards or off at some angle. So, basically, what we say is that's the polarization. The polarization of this light here, this electromagnetic wave, is going to be along the z-axis.

Now, obviously, drawing these diagrams and these waves would be super complicated over and over again, so we have 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. That's all polarization is.

Now let's take a look at a different example because there are 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 z-axis, but it kind of oscillates at some angle here. So what 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. We're going to draw this double-headed arrow, and we would just indicate with little dotted line that this angle here is 30 degrees. That's really all the polarization has to deal with.

Now there's also such a thing as unpolarized 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, we're going to see unpolarized lights. Usual examples are going to be sunlight or light that's coming from light bulbs or something like that. Most of the time, it'll actually just tell you if it's unpolarized. But, basically, the way that we represent unpolarized light is by using a double-headed arrow, except in all directions. So usually, what I do is you're just going to see a couple of lines, maybe like 3 or 4 of them, with all of the arrows like this, and this just represents unpolarized light.

Now let's talk about a polarizer very quickly. A polarizer looks like a little circle with a little grating, and it's basically just a filter. A polarizer is a filter that only allows components that are parallel to the transmission axis to pass through. So when you have this unpolarized light that's moving this way and it passes through the polarizer, the only thing allowed to pass through is this component of the light here, the component that is parallel to this transmission axis. What happens to the other components? Well, 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 unpolarized lights get passed through a polarization 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.

Now the other thing that happens is that the intensity also decreases by a factor of one-two, 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 = 1 2 I 0

and we just call this the one-half rule. So whenever you have unpolarized light that becomes polarized, its intensity decreases by a half, and it becomes polarized in the direction of the transmission axis.

Alright? So let's just go ahead and just jump into a really quick example problem here. So for the situation that we just described above here, if the intensity of the unpolarized light is 100, so in other words, if this I₀ is equal to 100, then what is the intensity of the transmitted light? In other words, what's the intensity of this light that makes it out of the other side? So, basically, if we want I, then we're going to have to use our new equation, I equals one-half of I₀. So, in other words, it's just one half of 100, and that just equals 50 watts per meter squared.

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 go through a quick practice problem, and we'll jump into another video.

Unpolarized light with intensity of 6 W/m² 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/m²
• B. 6 W/m²
• C. 3 W/m²
• D. 0 W/m²

Multiple Polarizers & Malus's Law

Hey, everyone. Welcome back. So, in the last few videos, we saw what happens when you have unpolarized light passing through a polarization filter. But many problems, in fact, most of them, are going to have multiple polarizers in a system. So, I'm going to show you how to solve these kinds of problems here because we're going to need a new equation. I'm going to show you how to solve multiple polarizer problems, and we're going to go over an equation known as Malus's law. Let's just jump right into it. Now remember that unpolarized light becomes polarized when it passes through a polarizer.

Basically, what we saw is that if you have the intensity of incoming light as 100, if it’s unpolarized, then when it passes through this first polarizer, everything that's along this transmission axis gets passed through, and then everything that's not part of this axis gets blocked out. So the only thing that survives is basically this vertical component. So you end up with something that kind of looks like this over here and the intensity drops by half. So in other words, I 1 = 50 .

Now we're going to see what happens when you have this polarized light that passes through another polarizer at a different angle. If polarized light passes through another polarizer oriented at a different angle, then two things happen. The first thing that happens is that this new light actually gets polarized again. It gets polarized in the direction of the second or the new polarizer. 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. So then you have new light that passes through at this angle here θ. And the second thing that happens is that because some of these components get eliminated, then the intensity will also decrease again based on the smallest angle that is between the two polarizers.

We're actually going to need a new equation for this, and I'm going to show you what that is. It's that the intensity is equal to I 0 × cos 2 ( θ ) .

This is sometimes referred to as the cosine squared rule, whereas the first rule that we learned was called the one-half rule. The two most important things you need to know is that you always use this equation whenever your initial light is unpolarized like we had over here, and then you use this equation whenever your light is already polarized like you have in this situation.

Let's go ahead and take a look. So we have unpolarized light with an intensity of 100 watts per square meter. It passes through these two different polarizers. One has an angle of 30, and the other one is actually oriented along the horizontal axis. We want to calculate ultimately what's the intensity after it passes both of these different polarizers. The first most important thing to do is actually going to be to draw a diagram.

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 polarizers. So what we need to do is figure out what happens to the intensity at each one of the steps. For each nth polarizer, we're going to use our two equations, the one-half rule or the cosine squared rule. So for each nth polarizer, what we're going to use is I 2 = I 1 × cos 2 ( 60 ) .

This final intensity, when the light exits the final polarizer, is polarized in the horizontal direction, and its intensity is going to equal 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.

Horizontally polarized light is incident on a polarization filter. The initial intensity of the light is 0.55 W/m², but is then reduced to 0.40W/m² 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

Example 1

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 unpolarized light with an average intensity of about 1350 near the Earth's surface. Now the first part here says that if sunlight passes through 2 polarizers that are angled at 90 degrees with respect to each other, we want to find the intensity of the light after passing through the 2nd polarizer. So I'm just going to jump straight into the steps here. First thing we want to do is just draw a diagram and label what our initial light is. So this is gonna be my diagram. I've got my initial light, which is unpolarized. Right? It's gonna kinda look like this, or you can draw it sort of like that, I guess. Like that. Alright? So this is my initial light. This is initially unpolarized, and the intensity is 1350. Alright. So this passes through 1 polarization actually, 1 polarization filter, and these things are angled 90 degrees with respect to each other. Now it doesn't tell us which specific angles, but it turns out it actually doesn't matter because, remember, the only thing that matters between these 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 0 degrees, such that the angle between them is just 90 degrees. That's theta. Right? Okay. So, basically, what happens is when this light passes through the 1st polarization filter, it's gonna be vertically polarized. This is gonna be my I one. And then when it passes through the 2nd polarizer, it's gonna be polarized horizontally, and this is gonna be my I 2. Okay? We want to calculate what this I 2 is, so let's just go ahead and set up our steps for the nth polarizers. There's 2 of them here. This is the first one and this is the second one here. So So let's just go through the first one really quick. So the first one has initially unpolarized light that's passing through, so we're going to use either one half rule or the cosine square rule. Hopefully, you realize that we're going to use the one half rule here. So what happens here is that I1 is just going to be one half of I0. In other words, it's just going to be one half of 1350, and that's going to equal 675. Alright? 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 light, but then gets polarized to a different angle, so we're gonna have to use the cosine squared rule. So this I 2 says this is gonna be I 1 times the cosine squared of the angle. Now remember, this angle here doesn't have to do with the angles of the individual axes, but the angle between them. This one's at 90 and this one's at 0, 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 going to be 675 times the cosine squared of 90. Now remember the rules for cosines. Whenever you have a cosine of 90 degrees, it always cancels out. So cosine squared of 90 is also just equal to 0. So basically what you get when you calculate this is that the intensity of the second light here is going to be 0 watts per square meter. And this actually has nothing to do with the fact that the intensity drops by half or anything like that. This intensity could be whatever number. Basically, what this means here is that whenever you have 2 polarizers angled 90 degrees with respect to each other, this is sometimes called cross polarization, then 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 vertically polarized light that can survive once it passes through the second filter. They kinda just cancel each other out. Alright. 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 going to take a 3rd filter and we're actually going to 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 going to stick another polarization filter in here and see what happens. We're going to see what the intensity of the sunlight is after it passes through all 3 polarizers, okay? So let's just go through the steps again, right? So here we have a transmission axis like this, and I've got initially unpolarized lights like this. This is gonna be my unpolarized light, which is 1350, but now we're actually gonna have 3 polarizers. There's the first one, then there's the second one, and then there is the third one. Alright? 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 kinda like this, such that this angle here is 30 degrees. So this is 90, and this is gonna be 0 degrees. Okay? 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 to the third one, it's gonna be horizontally polarized, and this is gonna be I 3. Alright? So now let's go through each one of our polarizers. This is the first, n equals 1. This is the second one, n equals 2, n equals 3. Alright? So here was here's what happens. So for the first polarizer, we're going to use the one half rule. Again, nothing's changed from before. It's just going to be I one is equal to, 675. So it's just gonna be half of 1350. It's gonna be exactly what it before, what it was from part a. Nothing has changed. So this I one here is 675. Okay. So the second filter, now what happens? The second filter says that I 2 is gonna be I 1 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 first and second polarizer. Alright? Then this angle over here, the angle between the second and third, I'm gonna label this as theta 23, the angle between the second and third. Okay? So what this says here is that I 2 is going to be I 1 times the cosine squared of theta 12. Alright? So this is just going to be 675 times the cosine squared of theta 12. Alright? Now what's that angle? Now remember, just be very careful. You always wanna actually calculate what the angle is between the polarizers. 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, this is just gonna be 30 or, sorry, 60 degrees. Now what you should get, is you should get let's see. You should get 168.75, and that's watts per meter squared. So that's actually what comes out the other side, 168.75. 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 2 polarizers and they completely cancel each other out to 0, when you stick 1 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. Alright? Now let's keep going. We've got one last step here, the 3rd polarizer. This says that the I 3 is gonna be I 2, and, again, we use the cosine squared rule. So this is gonna be I 2 times cosine squared, but now we're just gonna use theta from 2 to 3. We're going to use this angle over here. Alright? So here, what what this says is that I 2 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 0. So in other words, this is gonna be 30 minus 0. 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 a 126.6, and this is watts per meter squared. So this is actually the final answer, by the way. That's what survives after passage of the 3rd polarizer, and it's gonna be whole horizontally polarized. 126.6 watts per meter squared. So, anyway, there's a lot of moving parts in that problem, but it's kinda 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.