### Video Transcript

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.