32. Electromagnetic Waves

Polarization Filters

Physics32. Electromagnetic WavesPolarization Filters

All right, so let's move on to number four. Four is a polarizer question and it says, an ideal polarizer with its transmission axis rotated to 30 degrees relative to the vertical is placed in a beam of unpolarized light of intensity ten watts per square meter, after passing through the polarizer, what is the intensity of the beam? So, unpolarized light is like sunlight, and now we're going to send that sunlight through a polarizer. So remember we were drawing our polarizers like circles, with an arrow to indicate the transmission axis, and it says that it is 30 degrees relative to the vertical. They want to know how much is coming through the other side. So, if this is the intensity of the sunlight coming in, we want to know I on the other side of this polarizer. So how should I approach this problem? Somebody give me a thought. Eudora I see you hiding back there, what should I do for this problem? >> (student speaking) You can use malus's law. >> Okay, we can maybe use malus's law. All right, what is malus's law? Let's talk about malus's law for a second. Malus law says the intensity coming out is I naught cosine squared of theta. Okay, and when we talked about this in class we drew the following picture. We said, oh here's our sunlight, there is some intensity of the sunlight coming in, I'm gonna put my first polarizer up and that means there is I naught coming through that polarizer and I'm gonna draw the second polarizer at an angle-theta, and if I want to know how much is coming through, then Malus's law holds. I equals I naught cosine squared theta. Okay, this is the one that we did in class, but that looks a little different than this, doesn't it Eudoria? Because here we only have one polarizer whereas in malus's law, we drew it with two polarizers. So what's the purpose of that first polarizer in this picture? What's the purpose of this polarizer? >> (student speaking) To direct, or let through, one direction of like -- Exactly, to only let through vertical polarization, okay. That's not what this arrow is indicating so let's draw with a dashed line, this is the continuation of the beam, right? It keeps going through and if I think about the polarization that came through after this first polarizer, it's vertical, that's what that arrow represents. Okay, and that's what we called I naught, and now when you go through the second polarizer -- how much goes through the second polarizer, it is I naught times cosine squared of theta. But, how much of I sun gets through the first polarizer to I naught? Do you remember what we did in class? There's some factor that I need to include here. Is it all of I sun that came through, or is it some fraction of I sun that came through? Well, I sun is unpolarized. Okay, it has a polarization like this, going up and down, but it also has a polarization like this, going into and out of the board. This guy only picks out one of those. It picks out the vertical, it eliminates the horizontal, so what factor should I put here? >> (student speaking) one half. 1/2, it's gonna pick out half of the light, right? Since this is vertical, that's half of the light, the other one horizontal gets extinguished at that first polarizer. Alright, so that looks pretty close to this except our polarizer is not vertical anymore, right? It's off at this angle 30 degrees, and so the question is does that matter at all? If I hold up a polarizer and I look at the Sun, which you should never do of course, because it's a little bright, even if you cut it down with a polarizer it's still very bright, but let's say you do that, let's say you look up at the Sun with your polarizer -- with your polarized sunglasses, and now you turn your head sideways. Does the Sun get any dimmer? No, it doesn't, because it is unpolarized. Okay, it doesn't have any particular polarization, it has a combination of vertical and horizontal and you can combine those to make everything in between. So this is in fact exactly the same as this. Okay, because if I rotated my Sun by 30 degrees, it's gonna look exactly the same as this problem over here. So what's the intensity? It's Isun over 2. And they told us what Isun was, it was 10 watts per square meter, so we are just going to divide that by 2 and we get 5 watts per square meter.