Hey folks. So in this video, we're gonna talk about radiation pressure, which is a kind of counterintuitive concept. But we're actually gonna see that we use this all the time in the development of space technologies and other kinds of things. And we also need to know how to solve problems involving radiation pressure. So I'm gonna show you how to do that. It's pretty straightforward. We'll leave if just a few equations. So let's check this out here. So remember like all waves, electromagnetic waves carry energy, we saw that in the last few videos where electromagnetic waves have intensities that we can calculate things like that. But additionally, electromagnetic waves also have momentum, which is a little bit of counter of a counterintuitive idea because we remember that the equation for momentum is that P equals MV. So one of the things you might be wondering is how can light have momentum if it does not have mass? It's one of those things that we just sort of discovered in the early 19 hundreds. After a bunch of experiments, we found that light can also act as if it has mass. It's one of the things that you're just gonna have to trust me about. So if light has momentum, that means it also has to obey conservation of momentum. So basically, what happens is that whenever light hits an object, for example, if you were to shine a flashlight onto a piece of paper, it actually transfers its momentum and it basically pushes an object or pushes objects with a force. All right, now, there's two basic types of situations that you'll see in your problems, one where you have absorbed light and then one where you have reflected light, the problems will almost always tell you which one you're dealing with. All right. So let's look at these two examples or these two different cases here in a case where you have a absorbed lights, imagine that you had a flashlight. So imagine that you have all these electromagnetic waves that are hitting this piece of paper over here. Now, those electromagnetic waves have momentum. So in other words, you have some initial momentum here. And what happens is that the cons the momentum has to be conserved. So when the light gets absorbed into the paper, one of the things it has to do is it has to transfer some of it so that the paper starts moving. So it's gonna have some tiny little velocity. And the only way it does that is if it accelerates the paper and that basically means that it's exerting a force on the paper. All right. So when light hits an object, it actually pushes on them with a force which is kind of counterintuitive. This is actually very similar to a completely inelastic collision. Remember a completely inelastic collision is where you have an object that hits another object, they stick together and they both move as one. The other kind of situation is where you have reflected light. This could be something like a mirror or, or something that basically just reflects all the incoming light backwards. So this is different because here you actually have light beam that's hitting the mirror and it's gonna be going in this direction. And then later on, it's actually gonna go backwards and it's gonna go in this direction. So it's basically just gonna go back the way it came in this case. What happens is the momentum change between the light is actually greater than it was for the absorbed light case. And so basically what happens is that the force that it exerts on that object is even greater. So all I'm gonna do is I'm gonna say that this is F absorbed and this is F reflected. And what we can say is that F reflected is always greater than F absorbed. All right. So this also this piece of paper, this mirror also is gonna gain some velocity, but the force is gonna be even greater. This is actually similar to an elastic collision. Remember, an elastic collision is like where you had like a, like a billiard ball where it hits something and then rebounds backwards. Now, let's go to like talk about the equations because they're actually straightforward, but they're also similar to each other for the uh case of the, of the force and the absorbed light case, it's actually just equal to the intensity times A divided by C. And for the reflected case, it's similar, except there's gonna be a two in front of it, there's two IO A over C. So basically, the relationship between these two formulas here is that F reflected is equal to two times F absorbed. And this is because again, the momentum change of the incoming light is bigger, basically, it just rebounds. Uh And so the momentum shift of the object has to be larger in order to conserve momentum. Now, for the pressure equations, that's why we call it radiation pressure. I remember that the, the, the, the relationship between pressure force an area is that pressure is equal to F over A. So if we just take this equation here and we divide out the area, basically what you'll get here is that this is I over C and for the reflected case you'll, that this is two I over C, all right. So reflected always is gonna have a two in front of it. Otherwise the equations are pretty much the same. Let's just go ahead and take a look at an example here. So we can see how this works. So we've got a laser pointer that we're gonna be shining onto our hands and we're told that the average power output is five milliwatts. So this is our power over here. Now, the beam focuses onto a small area on the palm of our hand. So this is gonna be our area over here. That's our a and we're told that the, our hand completely absorbs the incoming light. So, problems again, are almost always gonna tell you what case you're dealing with. So we're dealing with complete absorption. Now, the first thing we wanna do is we wanna calculate the radiation pressure. So if I'm dealing with an absorption case, then I'm basically just gonna look at these formulas over here. All right. So if, if in part A I'm gonna be solving for the radiation pressure and that means that I'm gonna use the P abs uh absorbed equations. This is gonna be P absorbed is gonna be I over C. All right. Now, remember now I got the uh that C is just a constant over here, but I don't have what the intensity of light is. But remember I can always figure out intensity because intensity is related to power over area. All right. So if you actually rearrange this equation, what you'll see is that P absolute is equal to power divided by area times C. So don't get these two confused by the way, this is gonna be a lowercase P that's P, that's lowercase P for pressure. Uh Not to be confused with uppercase P for power. All right. So I'll try to make those big, so you won't get confused. The power output is five times 10 to the three watts divided by the area which is one times 10 to the minus six. And then C is gonna be three times 10 to the eight. That's the speed of light. If you go ahead and work this out, what you're gonna get is that the pressure is 1.67 times 10 to the minus six or sorry, 10 to minus fifth. And that's gonna be pascal. So this is your final answer for the radiation pressure. That's actually a very, very, very small pressure. If you think about it, let's move on to the second part here, which is, we're gonna calculate now the force that is exerted on your hands. So now we're gonna be dealing with the force absolute over here. Now, there's a couple of different ways you can actually go about this. You could just go ahead and plug it, plug this formula in which is the I A over C that would totally work or what you can do is you could just use the relationship between force pressure and area. So in other words, your force from the absorbed light is gonna be the pressure of the absorbed lights times the area. Remember we always have P equals F over A. So because we just figured out what this is in part A then we can just go ahead and make this a little bit easier on ourselves. So this is gonna be uh one times 61.67 times 10 to the minus fifth pascals times one times 10 to the minus six. That's uh meters squared. And what you're gonna get here is you're gonna get uh 1.67 times 10 to the minus 11 newtons. And that is your final answer. If you use the other equation, we would have, we would have gotten the exact same answer. All right. So if you look at this is actually an incredibly tiny force and that makes sense because the laser, when it shines onto our hand, it's not like pushing our hand away. It's an incredibly credibly small force. That's it. For this one. Folks. Let me know if you have any questions.