Anderson Video - Polarizer

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