 Lesson Explainer: Radiation Pressure on Perfectly Reflective Surfaces | Nagwa Lesson Explainer: Radiation Pressure on Perfectly Reflective Surfaces | Nagwa

# Lesson Explainer: Radiation Pressure on Perfectly Reflective Surfaces Physics

In this explainer, we will learn how to calculate the pressure exerted on an object by light that is reflected by it.

We can recall that light, or electromagnetic radiation in general, may be modeled as a wave. It turns out that electromagnetic waves transfer momentum through space. This phenomenon was first predicted by the physicist James Clerk Maxwell back in 1862. It was confirmed experimentally almost 40 years later in 1900 by Russian physicist Pyotr Lebedev.

We can recall that the usual definition of momentum is where is the mass of an object and is its velocity.

This definition works well when we are considering physical objects. However, electromagnetic waves do not have any mass. So we might wonder what is going on—how can they have a momentum?

Well, the formula only applies for objects that have a mass. Electromagnetic waves do have a momentum, but it is calculated in a different way.

For our purposes, it does not matter precisely how this momentum is defined. The important thing is simply that they do have momentum, because it is this fact that means they must exert a pressure on surfaces that they are incident on.

To understand how this works, we will consider an example of an object, such as a ball, colliding with a wall. We will imagine that the ball travels directly toward the wall.

The ball has some mass and has a velocity directed toward the wall. This means that it has some momentum, which we will call , directed toward the wall. This is shown in the diagram below:

After colliding with the wall, the ball moves in the opposite direction, away from the wall. So, after the collision, its momentum, which we will call , is directed away from the wall:

Evidently, the momentum of the ball changes in the collision. This means that the ball must experience a force. That force is provided by the wall.

We can recall that Newton’s third law tells us that every action has an equal and opposite reaction. In this case, that means that since the wall exerts a force on the ball, then the ball also exerts a force on the wall, as shown in the diagram below:

Now imagine that we have many balls colliding with the wall in the same way.

We now have a force acting all over the wall’s surface, as a result of all of the collisions. Then, we can calculate the pressure on the wall. This pressure, , is given by where is the force and is the area over which the force is acting.

What we have seen from this situation is the following. Whenever an object with momentum collides with something such as a wall, this object experiences a change in momentum. This change in momentum means that a force from the wall is acting on the object. Newton’s third law means that the object exerts an equal and oppositely directed force on the wall. This force on the wall means the object is exerting a pressure on it.

Just as a wall effectively acts to reflect a ball, a mirror acts to reflect an electromagnetic wave. Because electromagnetic waves also have momentum, we can follow a similar line of reasoning to the case of the ball.

Let’s consider an electromagnetic wave reflecting off a perfectly reflective mirror. The wave travels at the speed of light, , toward the mirror and has a certain amount of momentum. After reflecting off the mirror, it is traveling in the opposite direction at this same speed . At this point, it has the same magnitude of momentum as before the reflection, but now this momentum is in the opposite direction.

Following the same reasoning as before, we can conclude that if electromagnetic radiation is incident on a surface then it will exert a force and hence a pressure on that surface. We call this pressure radiation pressure.

When talking about radiation pressure, we do not tend to talk about it in terms of momentum. Instead, we express it in terms of quantities that are more convenient when talking about radiation. We can define radiation pressure as follows

When electromagnetic radiation of an intensity is incident on a perfectly reflective surface, the radiation pressure exerted on the surface is given by where is the speed of light.

In our equation for radiation pressure, the intensity, , of electromagnetic radiation is the power transmitted by that radiation per unit area. Mathematically, we can write this as

Note that we have chosen to keep this as a word equation. Our reason for this is that power is often represented by the letter , which we are already using to represent pressure. Keeping our intensity equation in terms of words avoids any potential for confusion.

The power of electromagnetic radiation is the rate of energy transfer, in other words, the energy transferred per unit time. Mathematically, we can write this as

Since intensity is the power per area, it has the unit watts per square metre (W/m2). Then, the radiation pressure, , has the unit of divided by the unit of , the speed of light. So the unit of must be . We can recognize that and that we can also write . Then, we can write the unit of as . Finally, we can recognize that , where the pascal is the unit of pressure. So our radiation pressure does indeed have units of pressure.

Let’s have a look at an example problem where we need to calculate radiation pressure on a surface.

### Example 1: Calculating Radiation Pressure for a Perfectly Reflective Surface

Light with an intensity of 60 W/m2 is directed at a reflective surface. What is the pressure exerted by the light on the surface? Use a value of m/s for the speed of light in vacuum.

We are told that we have light with an intensity of directed at a surface. We are asked to work out the pressure exerted on the surface.

The question tells us that the surface is perfectly reflective. This means we can use our formula for the radiation pressure.

We are told to use a value of , so substituting and our given value of into our radiation pressure equation gives us

It is important to remember that our formula for the radiation pressure only applies for perfectly reflective surfaces, that is, for surfaces that reflect of the incident electromagnetic radiation.

It turns out that, for surfaces that absorb some of the radiation, the radiation pressure is lower than in the case of a perfect reflector.

To see why this is the case, we will go back to our analogy of a ball colliding with a wall.

When we first considered this, we imagined that the ball bounced off the wall with the same magnitude of momentum as it initially had when traveling toward the wall.

Let’s now imagine instead that somehow the ball gets absorbed by the wall, or that it sticks to it, rather than rebounding. Perhaps this seems a little unphysical. To make it a more reasonable scenario, perhaps we could imagine some very strong double-sided tape on the ball that sticks it onto the wall where it collides.

However we picture this, the net result is that the ball ends up with zero momentum, since it is stationary at the location of the wall.

We will suppose that the ball travels toward the wall with some initial momentum , and we know it ends up with a final momentum . So in this case, the change in momentum of the ball is clearly less than the case where the ball bounced off the wall and ended up with in the opposite direction to its initial momentum .

Since the momentum change of the ball is lower, this means that there is a smaller force acting on it. Then, by Newton’s third law, the force on the wall must also be smaller.

Finally, we can recall that pressure is force per area. So, if the force acting on a given wall as a result of collisions from balls is reduced, then the pressure on the wall must also be reduced.

Exactly the same thing happens with electromagnetic radiation.

If all of the radiation incident on a surface were to be absorbed, then the radiation would experience half of the momentum change relative to the case where it gets completely reflected. And so, for the case of a perfectly absorbing surface, the radiation pressure would be half of that in the perfectly reflective case:

In reality, most surfaces fall somewhere between these two extremes. That is, when electromagnetic radiation is incident on them, some gets absorbed, while some gets reflected. Then, the radiation pressure also falls between the two extreme cases. The more of the radiation that gets reflected, the greater the radiation pressure.

In this explainer, we only consider examples in which the surface is perfectly reflective; in this case, we have seen that the radiation pressure is given by the maximum value:

So far, we have been considering the radiation pressure. This is a quantity that is independent of the size of the surface that the electromagnetic radiation is incident on.

In some situations, we have a surface of a known size and we are interested in the total force on that surface as a result of incident electromagnetic radiation. In this case, we may recall that pressure is the force per area. In other words, for a force acting over an area , we have

We can rearrange this to make the force the subject by multiplying both sides by the area :

We can then substitute into this our expression for the radiation pressure on a perfectly reflective surface, . This gives us an expression for the force acting on a perfectly reflective surface as a result of incident radiation.

### Equation: Force Arising from Radiation Pressure

The force acting on a perfectly reflective surface of area as a result of incident radiation of intensity is given by where is the speed of light.

Let’s now look at an example where we are asked to calculate the force exerted on a surface by electromagnetic radiation.

### Example 2: Calculating the Force Exerted by Light on a Perfectly Reflective Surface

Light with an intensity of 50.0 W/m2 is directed at a reflective surface. The surface has an area of 2.25 m2. What force is exerted by the light on the surface? Use a value of m/s for the speed of light in a vacuum.

We are told that we have light with an intensity of directed at a surface with an area of . We are asked to work out the force exerted on the surface by the light.

The question tells us that the surface is perfectly reflective. This means we can use our formula for the force exerted by the light.

We are told to use a value of , so substituting and our given values of and into our force equation gives us

Let’s have a look at the units on the right-hand side of this expression. We have units of . We can write this more simply as W⋅s/m.

We can recognize that , which means that we can write the unit as joules per metre (J/m).

Finally, we can also write or, equivalently, . And so the units on the right-hand side of our expression for the force are equivalent to units of newtons.

Then, evaluating this expression for the force gives us our answer: the force exerted by the light on the surface is

We will finish up by looking at one more example.

### Example 3: Calculating the Intensity of Light Required to Exert a Certain Force

A large asteroid is on a collision course with Earth. The asteroid has a mass of 300‎ ‎000 kg. The asteroid can be deflected if a 0.010 N force can be applied to it for s. If a powerful laser is used to shine light on a 12 m2 area of the asteroid, what is the intensity of light needed in order to deflect the asteroid using radiation pressure? Assume that the surface of the asteroid is reflective. Use a value of m/s for the speed of light in vacuum. Give your answer in scientific notation to one decimal place.

This question is asking us to calculate the intensity of light required in order to use a laser to deflect an asteroid. We will label this intensity we are trying to calculate as .

We are told that a force of must be applied for s in order to successfully deflect it and that the laser will be shone over an area of .

The question also tells us that we can assume the surface of the asteroid is reflective. This means that we can use our formula: where is the speed of light in vacuum, for which we are told to use a value of .

We are looking to calculate the intensity , so we need to rearrange the equation to make the subject. To do this, we multiply both sides by and divide both sides by . This gives us:

We have been given the values of , , and in the question, so all that remains is to substitute them into this equation.

It is worth noticing that we were also given some additional information in the question—namely the length of time that the force must be applied for to deflect the asteroid, and the mass of the asteroid. As we have seen, we do not actually need to use those values in order to calculate the intensity of the laser light that is required.

Substituting our values of , , and into our equation for the intensity, , we get

Evaluating this expression, we get that the intensity, , is given by

Finally, we note that the question asked us to give this answer in scientific notation to one decimal place. We already have the value in scientific notation, so we just need to round it to one decimal place of precision. Doing this, we get our answer to the question: to one decimal place, the intensity of light required to deflect the asteroid is given by

### Key Points

• When electromagnetic radiation is incident on a surface, it exerts a pressure on that surface. This pressure is known as radiation pressure.
• For a perfectly reflective surface, the radiation pressure exerted by electromagnetic radiation of intensity is given by , where is the speed of light.
• If the surface absorbs some of the light rather than reflecting of it, then the radiation pressure will be reduced from this value.
• The force exerted by light of intensity incident on a surface of area is given by .