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.