Radiation Pressure - Online Tutor, Practice Problems & Exam Prep

Radiation pressure arises from the momentum of electromagnetic waves, which can exert force on objects upon absorption or reflection. When light is absorbed, the force is given by $f=\frac{I}{c}$, while for reflected light, it is $f=\frac{2I}{c}$. The pressure equations follow similarly, with absorbed pressure as $p=\frac{I}{c}$ and reflected pressure as $p=\frac{2I}{c}$. Understanding these concepts is crucial for applications in space technology and optics.

Radiation Pressure

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.

A radio transmits a wave with intensity 27.0 W/m² towards a flat surface (perfectly reflecting) with area 2m². Calculate the force and radiation pressure on the surface.

• A. Pressure = $1.8\times10^{-7}Pa$ ; Force = $3.6\times10^{-7}N$
• B. Pressure = $9.0\times10^{-8}Pa$ ; Force = $1.8\times10^{-7}N$
• C. Pressure = $1.8\times10^{-7}Pa$ ; Force = $9.0\times10^{-8}N$
• D. Pressure = $9.0\times10^{-8}Pa$ ; Force = $3.6\times10^{-7}N$

Example 1

Alright, folks. So hopefully you got a chance to tackle this problem on your own. If you got a little stuck, I'm here to help you. Let's just go ahead and get started. We'll work it out together. So we want to use a laser beam to make an object float in the air. We're told that the laser beam is pointing straight upwards, and it exerts radiation pressure on a horizontal disc, and we know the mass and radius of that disc. This problem is very visual, so I actually want to draw this out here so it kind of makes sense. We've got this disc here, and we're told that the radius of this disc, which I'm going to call r_disc = 100 mm (100 millimeters). I'm also told that the mass of the disc is 7.8 grams. The disc, which completely absorbs light, is 2.75 meters from the laser, and the radius of the beam is 50 millimeters. Basically, you have this horizontal disc here, and the laser beam is pointing straight upwards, and it's going to try to make this object float by using radiation pressure. So, the distance from the disc to the laser is y = 2.75 m, and the radius of the beam is r_laser = 50 mm. We want to figure out what the laser's power is needed to balance this disc. There are a lot of variables, but we really just want to figure out p, the power of the laser.

First, we need to understand what 'float' even means. The object wants to float in the air, which means the disc is balanced and suspended in the air. When objects float, it means that the forces cancel out; basically, the sum of all forces is zero. There is a force produced from radiation pressure and a downward force because of gravity, f_g(force of gravity). When these two are equal to each other, the disc will float. Float really means that f_radiation = f_g. They are equal and opposite. We know the disc completely absorbs light, so we use the absorption equations for radiation pressure, not the reflecting ones. When f_absorb = f_g, the object floats.

To balance the equations, we start with the absorbed force equation, which involves I a / c (without a factor of 2 for absorption). This equals the force of gravity, which is just mg (mass of the disk times the gravitational constant). We know the mass, gravitational constant 9.8, and the speed of light c are constants. What we need is the intensity and the area. However, our target variable is power p. To link intensity and power, we use I = p / a. Substituting intensity with power over area, we get different areas for disk and laser, but we focus on the area hit by the laser beam for calculating force. These areas cancel in the equation p a / c = mg. What remains is p = mgc.

Upon plugging in the numbers: m = 0.0078 kg, g = 9.8 ms^-2, c = 3 ✕ 10⁸ ms^-1. Solving this, you would need about 2.29 ✕ 10⁷ watts (22.9 megawatts) to make a tiny disk suspended in the air just from radiation pressure. Unfortunately, there's no laser currently available that can provide this power but that's the theoretical requirement to make this disk float solely by radiation pressure.

Let me know if you have any questions.

Example 2

Welcome back, everyone. Hopefully, you had a chance to work this problem out on your own. So we're told in this problem that a spacecraft with reflective sail-like material may eventually be used for low-cost space travel. This is a real technology that's actually being used, and the science is sound. We're told that a 400-kilogram satellite near Earth is equipped with these super reflective sails.

I want to draw this out here. So what happens is those sails are going to capture sunlight from the sun, and it's going to generate a radiation pressure. So you have the sun like this, and then at some distance where the Earth is, you're going to have this satellite over here with big reflective sails like this. The sunlight that travels is going to hit those reflective sails, and it's eventually going to push them to the right. This is going to be a force from radiation pressure.

Eventually, what happens is that the satellite is going to pick up speed, and that's actually one way that we could generate large amounts of velocity for a spacecraft with very little fuel. You're only using the light from the sun.

The area of each of these reflective sails is 5,000 meters squared. So, in other words, the area is equal to 5,000, but what happens is the total amount of area is going to be twice of 5,000 because there are 2 of them, both completely reflective. So this total area is going to equal 10,000.

We are using our reflective radiation pressure equations and not our absorbing radiation equations, and the intensity of sunlight is approximately 1350. So let's figure out what the force that's exerted on the satellites is. We use the equation: F r e f l e c t e d = 2 × I A c where I is 1350, A is 10,000, and c is 3 × 10&sup8;. You should get F reflect ≈ 0.09 Newtons.

For the second problem, assuming that the satellite starts from rest, how fast is it moving after 1 year? We use the kinematic equation: V f i n a l = V i n i t i a l + a × t Given that V initial is 0, we find acceleration using F = ma, thus a = F reflect / m = 0.09 / 400 kg, giving us an acceleration of 2.25 × 10⊃-4 m/s². For t = 1 year × 365 days × 24 hours × 60 minutes × 60 seconds, calculate V final ≈ 7,100 m/s.

Even though the force is very small, after acting for a long time, it can produce significant velocity changes. This technology could send the satellite out, and years later, it would be traveling extremely fast out of the solar system. That's it for this one, folks. Let me know if you have any questions. I'll see you in the next video.

Here’s what students ask on this topic:

What is radiation pressure and how is it calculated?

Radiation pressure is the pressure exerted by electromagnetic waves on a surface due to the transfer of momentum. It can be calculated for two scenarios: absorption and reflection. For absorbed light, the force is given by $f=\frac{I}{c}$, where $I$ is the intensity of the light and $c$ is the speed of light. For reflected light, the force is $f=\frac{2I}{c}$. The corresponding pressure equations are $p=\frac{I}{c}$ for absorbed light and $p=\frac{2I}{c}$ for reflected light.

How does radiation pressure affect space technology?

Radiation pressure plays a significant role in space technology, particularly in the design and operation of solar sails. Solar sails use the momentum of photons from the Sun to propel spacecraft without the need for traditional fuel. The force exerted by radiation pressure can be harnessed to accelerate, decelerate, or change the direction of a spacecraft. This technology offers a sustainable and efficient method for long-duration space missions, reducing the dependency on chemical propulsion systems.

Why does reflected light exert more force than absorbed light?

Reflected light exerts more force than absorbed light because the change in momentum is greater. When light is absorbed, it transfers its momentum to the object, resulting in a force given by $f=\frac{I}{c}$. However, when light is reflected, it reverses direction, effectively doubling the change in momentum. This results in a force of $f=\frac{2I}{c}$, making the force exerted by reflected light twice as large as that of absorbed light.

What are the equations for radiation pressure in terms of intensity and speed of light?

The equations for radiation pressure depend on whether the light is absorbed or reflected. For absorbed light, the pressure is given by $p=\frac{I}{c}$, where $I$ is the intensity of the light and $c$ is the speed of light. For reflected light, the pressure is $p=\frac{2I}{c}$. These equations highlight the relationship between the intensity of the electromagnetic waves and the resulting pressure exerted on a surface.

How do you calculate the force exerted by radiation pressure on an object?

To calculate the force exerted by radiation pressure on an object, you need to know the intensity of the light and the area over which it is distributed. For absorbed light, the force is given by $f=\frac{IA}{c}$, where $I$ is the intensity, $A$ is the area, and $c$ is the speed of light. For reflected light, the force is $f=\frac{2IA}{c}$. These equations allow you to determine the force based on the light's intensity and the surface area it impacts.