Hey, folks. So in this video, we're gonna talk about radiation pressure, which is a kind of counterintuitive concept. 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 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 could calculate, things like that. But, additionally, electromagnetic waves also have momentum, which is a little bit of a counter of a counterintuitive idea, because we remember that the equation for momentum is that p=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 1900 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 objects with a force.

Now there are 2 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. Alright? So let's look at these two examples or these two different cases here. In a case where you have absorbed light, 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. In other words, you have some initial momentum here, and what happens is that the conservation of 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. Alright? 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 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 the force that it exerts on that object is even greater. So what 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}>f_{absorbed}. Alright. 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 what 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 case of the of the force in the absorbed light case, it's actually just equal to the intensity times the area divided by c. And for the reflected case, it's similar, except there's gonna be a 2 in front of it. So there's 2i/c. So basically, the relationship between these two formulas here is that f_{reflected}=2×f_{absorbed}, and this is because, again, the momentum change of the incoming light is bigger. Basically, it just rebounds, 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, remember that the relationship between pressure, force, and area is that pressure is equal to f/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/c. And for the reflected case, you'll get that this is 2I/c. Alright? So reflected always is gonna have a 2 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 5 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 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. Alright? So if in part a, I'm gonna be solving for the radiation pressure, and that means that I'm gonna use the p_{absorbed} equations. This is gonna be p_{absorbed}=I/c. Alright? Now remember, now I've got the, 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. Alright. So if you actually rearrange this equation, what you'll see is that P_{absorbed}=power/area∙c. So don't get these 2 confused, by the way. This is gonna be a lowercase p. That's p. That's lowercase p for pressure, not to be confused with upper case P for power. Alright. So I'll try to make those big so you don't get confused. The power output is 5⋅103 watts divided by the area, which is 1⋅10−6, and then c is going to be 3⋅108, that's the speed of light. If you go ahead and work this out, what you're going to get is that the pressure is 1.67⋅10−5, and that's gonna be pascals. 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 going to be dealing with the force absorbed over here. Now there's a couple of different ways you can actually go about this. You could just go ahead and plug this formula in, which is the ia/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 light times the area. Remember, we always have p=f/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, 1⋅106 1.67⋅10−5 Pascals times 1⋅10−6, that's, meters squared. And what you're gonna get here is you're gonna get, 1.67⋅10−11 Newtons, and that is your final answer. If you use the other equation, we you would've we would've gotten the exact same answer. Alright? So if you look at this, it's 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, incredibly small force. That's it for this one, folks. Let me know if you have any questions.