### Video Transcript

In this video, we’ll be discussing
radiation pressure, which is pressure exerted by electromagnetic radiation such as
light. It might seem surprising that light
can exert pressure, but we’ll see how this is possible. We’ll be focusing specifically on
light falling on perfectly reflective surfaces. And we’ll be calculating exactly
how much pressure light can exert on a surface like this.

So, let’s begin by recalling that
electromagnetic waves, such as the light coming from this flashlight, transfer
energy through space. Well, it turns out that
electromagnetic waves don’t only transfer energy through space. They also transfer linear
momentum. This was first predicted by a
Victorian physicist, James Clerk Maxwell, as part of his study on
electromagnetism. It was later confirmed with a more
intuitive explanation by the theory of quantum mechanics. But Maxwell’s original prediction
that electromagnetic waves transfer momentum through space was still true.

Now, at this point, we may recall
that the momentum of an object is usually defined as the mass of an object
multiplied by its velocity. But electromagnetic waves don’t
have any mass. So what’s going on? Well, it turns out that this
equation is only true for objects that have mass. Electromagnetic waves can also have
momentum, but their momentum has nothing to do with their mass or lack of mass. This means that we calculate the
momentum of electromagnetic radiation in a different way. But it’s the fact that
electromagnetic waves have momentum that means they must exert a pressure on
surfaces that they’re incident on.

To understand why this is, let’s
imagine an object colliding with a wall. This is an object with mass, and
it’s also moving, which means it has momentum according to the equation 𝑝 equals
𝑚𝑣. Let’s say that the initial momentum
of this object — that is, its momentum before it collides with the wall — is 𝑝
i. Then, after it’s collided with the
wall, let’s say that it bounces away with the same amount of momentum but acting in
the opposite direction.

Since momentum is a vector
quantity, this means it would have the same magnitude but the opposite sign, giving
it a final momentum of negative 𝑝 i. Note that here we’re choosing a
convention where forces or momentum that are acting to the right are positive and
those acting to the left are negative.

Now let’s think more carefully
about this collision. We know that in order for an
object’s momentum to change, it has to experience a force. And this force is given by Δ𝑝, the
change in momentum of the object, over Δ𝑡, the time over which the change in
momentum occurs. This equation is effectively a
statement of Newton’s second law. The change in momentum of this
object is given by the final momentum minus the initial momentum, which here is
negative 𝑝 i minus 𝑝 i.

Evaluating the top of this
fraction, we get negative two 𝑝 i divided by Δ𝑡. This expression represents the size
of the force that the particle must’ve experienced when it collided with the
wall. In other words, we can see that the
wall has exerted a force on the particle. Now, Newton’s third law tells us
that every action has an equal and opposite reaction. This means that since our wall
exerted a force on our object, then our object must’ve exerted an equal and opposite
force on the wall.

So, in this case, we could say that
the force exerted on the wall was two 𝑝 i over Δ𝑡. It’s the same size as the force
that the wall exerted on the object, only positive, signifying that it’s acting in
the opposite direction. Now, if we had lots of similar
objects colliding with the wall, then we’d have a force acting all over the wall’s
surface, which means we’d be able to calculate the pressure. We represent pressure with a
capital 𝑃, and it’s given by force divided by area.

So why are we talking about an
object with mass in a lesson about radiation pressure? Well, what we’ve seen is that when
an object with momentum collides with another object, for example, a wall, it
experiences a change in momentum. Experiencing a change in momentum
must mean it’s experiencing a force. And if it experiences a force, then
it must also exert a force on the wall. And finally, if it’s exerting a
force on the wall, then it must be exerting a pressure on the wall.

It turns out that because
electromagnetic waves also have momentum, we can follow a similar line of
reasoning. Let’s replace our object with an
electromagnetic wave, and we’ll replace our wall with a mirror. Well, we know that waves have
momentum. And we also know that any
electromagnetic wave will be reflected off a perfectly reflective mirror. Importantly, we also know that
electromagnetic waves travel at a fixed speed, the speed of light.

So when a wave is incident on a
mirror, we know that it approaches with a certain amount of momentum. And once it’s been reflected, it
will travel back in the opposite direction with the same magnitude of momentum. So just like our object from
before, if this wave had an initial momentum of 𝑝 i, then it has a final momentum
of negative 𝑝 i. So we know that it’s experienced a
change in momentum. And following the same reasoning as
before, we can conclude that if lots of electromagnetic radiation is incident on a
surface, then it will exert a pressure on that surface. We call this pressure radiation
pressure.

But instead of expressing it in
terms of the momentum of waves, we express it in terms of quantities that are more
convenient when we’re talking about radiation. Specifically, we say that this
pressure, denoted by a capital 𝑃, is equal to two times the intensity of the
radiation, represented by capital 𝐼, divided by the speed of light 𝑐.

When we talk about the intensity of
light, we mean the power transmitted by electromagnetic radiation per unit area. So once again, if we imagine lots
of light waves incident on our mirror, then the intensity measured on the mirror’s
surface would be given by the power of this beam of light divided by the area over
which we’re measuring. This is given by the equation
intensity equals power over area.

It’s important to note here that
power is often represented by a capital 𝑃. However, in this video, we’re
talking about pressure, which we’ve already represented with a capital 𝑃. So for now, we’ll just keep this as
a word equation. Let’s also recall that power is the
rate of energy transfer over time. So as a word equation, we can say
that power equals energy over time. So, this equation shows us how we
can quantify the radiation pressure due to electromagnetic radiation incident on a
perfectly reflective surface. And we can use these two equations
to help us work out the intensity of the radiation.

Now, it’s really important to
remember that this equation only works for perfectly reflective surfaces. That is, it only works for surfaces
that reflect 100 percent of the radiation that’s incident on them. What we actually find is that for
surfaces that are not perfectly reflective, that is, they absorb some radiation, the
amount of radiation pressure experienced is actually less than this.

To see why this is, let’s once
again consider our object with a mass colliding with a wall. Now, when we thought about this
previously, we described the object as bouncing back with the same momentum acting
in the opposite direction. The reason we did this is because
it’s similar to how a wave reflects off a surface. It arrives at the speed of light
and then leaves at the speed of light in the opposite direction, which meant that
its momentum change was twice its initial momentum.

So now, let’s consider what would
happen if the wall could absorb the object on impact. Of course, this isn’t realistic,
but it will help us describe what happens when an object absorbs electromagnetic
radiation. So, let’s say that once again, the
initial momentum of the object is 𝑝 i. But this time, when the object
collides with the wall, it somehow gets absorbed. Well, in this case, we could say
that its final momentum is zero.

Now, let’s follow through the same
reasoning as before and see what happens to our result. This time, when we calculate the
change in momentum of our object by subtracting the initial momentum from the final
momentum, we now have zero minus 𝑝 i over Δ𝑡, which of course is simply equal to
negative 𝑝 i over Δ𝑡. In other words, the force
experienced by this object has halved, as the object has effectively only been
stopped rather than being pushed away in the opposite direction as well.

This means that the force exerted
on the wall is also halved. And if the force is halved, this
means that pressure is halved as well. Even though we’re looking at an
unrealistic situation where our wall absorbs an object, this line of reasoning
actually does help us explain the radiation pressure that would be experienced by a
surface that absorbs radiation. And in fact, we do find that the
radiation pressure on a perfectly absorbing surface is exactly half the radiation
pressure that would be experienced by a perfectly reflective surface.

Now, most real materials are
neither perfectly reflective nor perfectly absorbing, which means that the radiation
pressure they experience would be somewhere between two 𝐼 over 𝑐 and 𝐼 over
𝑐. But of course, in this video, we’re
only focusing on perfect reflectors. So the final thing to mention is
one more really useful equation that we can work out. We now know that the radiation
pressure experienced by a perfectly reflective surface is given by 𝑝 equals two 𝐼
over 𝑐. That is, it’s twice the intensity
divided by the speed of light. And we also know that pressure is
defined as force over area. This means that we can equate these
two expressions.

We can use this to come up with a
formula for the force that’s experienced by a perfectly reflective surface due to
radiation pressure. All we need to do is rearrange for
𝐹. And we can do this by multiplying
both sides of the equation by 𝐴, which gives us 𝐹 equals two 𝐴𝐼 over 𝑐. That is, the force experienced by a
perfectly reflective surface due to radiation pressure is two times the area of the
surface multiplied by the intensity of the radiation divided by the speed of
light.

Okay, so now we’re armed with these
equations, let’s have a go at some practice questions.

Light with an intensity of 60 watts
per meter squared is directed at a 100 percent reflective surface. What is the pressure exerted by the
light on the surface? Use a value of three times 10 to
the eight meters per second for the speed of light in vacuum.

Okay, so in this question, we’re
told that light with a certain intensity is directed at a reflective surface and
that the light exerts pressure on this surface. So we could imagine a flashlight
shining a beam of light onto a mirror. Since the surface is reflected,
this beam of light would be reflected back. Let’s quickly recall that even
though light waves don’t have mass, they can still transfer momentum. And in fact, the process of light
waves colliding with a mirror and then being reflected back in the opposite
direction is in many ways similar to a number of particles colliding with a wall and
bouncing off.

In both of these cases, whether
we’re considering waves or particles, the things colliding with the wall or mirror
undergo a momentum change. And since a change in momentum is
related to force according to the equation 𝐹 equals Δ𝑝 over Δ𝑡 — where Δ𝑝 is a
change in momentum, Δ𝑡 is a change in time, and 𝐹 is a force — we can therefore
conclude that when the waves and particles experience a change in momentum, they
must be exerting a force on the mirror and the wall. Exerting a force over the surface
of a mirror or the surface of a wall means, therefore, that a pressure is
exerted.

And in the case of the light waves,
this is known as radiation pressure. We can recall that the formula for
quantifying the radiation pressure on a perfectly reflective surface is 𝑝 equals
two 𝐼 over 𝑐, where capital 𝑃 is the radiation pressure expressed in newtons per
meter squared. Capital 𝐼 is the intensity of the
radiation, which we usually express in watts per meter squared, and 𝑐 is the speed
of light.

In this question, we’re told that
light with an intensity of 60 watts per meter squared is directed at a surface and
that this surface is 100 percent reflective. We’ve been asked to calculate the
pressure that’s being exerted. This means that we can use this
formula to tell us the answer, since it describes the radiation pressure on a
perfectly reflective surface. All we need to do is substitute in
the intensity of the radiation and the speed of light.

We’re told in the question that the
intensity is 60 watts per meter squared. And we’ve also been told to use a
value of three times 10 to the power of eight for the speed of light in a
vacuum. Plugging all of this into our
calculator, we obtain a value of four times 10 to the power of negative seven
newtons per meter squared. And this is the answer to our
question. If light with an intensity of 60
watts per meter squared is directed at a 100 percent reflective surface, the
pressure exerted by the light on the surface will be four times 10 to the power of
negative seven newtons per meter squared.

So with that answered, let’s try
out another question.

Light with an intensity of 50.0
watts per meter squared is directed at a 100 percent reflective surface. The surface has an area of 2.25
meters squared. What force is exerted by the light
on the surface? Use a value of 3.00 times 10 to the
eight meters per second for the speed of light in a vacuum.

So in this question, we’re told
that light with a certain intensity is incident on a surface and that it exerts a
force on that surface. So let’s think of a flashlight
shining a beam of light on this surface here. In this question, we’re told that
this surface has an area of 2.25 meters squared. We’re also told that the surface is
100 percent reflective, which means that all the radiation that reaches it will be
reflected back.

We can recall that when radiation
is incident on a surface, it applies a pressure on that surface, known as radiation
pressure. Let’s also recall that for a
perfectly reflective surface, such as the one described in this question, the
radiation pressure experienced by the surface is given by the formula 𝑝 equals two
𝐼 over 𝑐. In this formula, 𝑝 is the
radiation pressure, which we generally measure in newtons per meter squared. 𝐼 is the intensity of the
radiation, measured in watts per meter squared. And 𝑐 is the speed of light, which
we’ve been told in the question is 3.00 times 10 to the power of eight meters per
second.

Let’s remind ourselves that the
intensity of radiation measured at a surface is the total power of that radiation
divided by the area that it’s covering. In this question, we’re told that
the light has an intensity of 50.0 watts per meter squared. This means that every square meter
of surface illuminated by the light will receive a total of 50 watts of power. This also means that the total
amount of power received by the entire surface depends on how much of the surface is
illuminated.

If the illuminated area is small,
then only that small area will be receiving the intensity of 50 watts per meter
squared, which means the total amount of received power will be small. If we double the size of the
illuminated area, then we double the total power that the surface receives. In this question, the size of the
illuminated area hasn’t actually been specified, so we’ll assume that the light
covers the whole surface.

Now, this question is asking us to
find the force that’s exerted on the surface. But this formula only tells us the
pressure. However, it’s possible to calculate
the total force by remembering that pressure is defined as force divided by
area. So since the radiation pressure in
this question is equal to two 𝐼 over 𝑐 and equal to 𝐹 over 𝐴, we can equate
these two expressions. That is, we can say that the force
divided by the area is given by two times the intensity divided by the speed of
light.

Since we’re interested in finding
the force, we just need to rearrange this equation to make 𝐹 the subject. We can do this by multiplying both
sides of the equation by 𝐴, giving us the equation 𝐹 equals two 𝐴𝐼 over 𝑐. In other words, the total force
that a perfectly reflective surface experiences due to radiation pressure is equal
to two times the area that the light is incident on multiplied by the intensity of
the light divided by the speed of light.

So to answer this question, all we
need to do is substitute in these values. Once again, we’re assuming that the
area covered by the light is the entire area of the surface, which means 𝐴 has a
value of 2.25 meters squared. We’re told that the intensity of
the light is 50.0 watts per meter squared. And all of this is divided by the
speed of light 𝑐, which we’re told takes a value of 3.00 times 10 to the power of
eight meters per second. Plugging all of this into our
calculator, it gives us a value of 7.5 times 10 to the power of negative seven
newtons.

And this is the answer to our
question. When light with an intensity of
50.0 watts per meter squared is directed at a 100 percent reflective surface with an
area of 2.25 meters squared, the force it will experience due to radiation pressure
is 7.5 times 10 to the power of negative seven newtons.

Now, let’s summarize the key points
that we’ve looked at in this lesson. Electromagnetic waves have
momentum. And we’ve seen how this explains
the fact that they exert pressure on surfaces. This pressure is called radiation
pressure. We’ve also seen that when light
waves are reflected by a surface, the momentum change they experience is greater
than when they’re absorbed, which means the pressure they exert on the surface is
greater. In other words, the more light is
reflected by a surface, the more radiation pressure that surface experiences.

We’ve also seen that for perfectly
reflective surfaces, that is, surfaces that reflect 100 percent of the radiation
incident on them, the equation 𝑝 equals two 𝐼 over 𝑐 gives the relationship
between the radiation pressure experienced by the surface, the intensity of the
radiation, and the speed of light. And the equation 𝐹 equals two 𝐴𝐼
over 𝑐 gives the relationship between the total force experienced by that surface,
the area of the surface, the intensity of the light, and the speed of light.