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.