Video Transcript
Light is directed at a 100 percent
reflective surface. The light exerts a pressure of 3.50
times 10 to the negative six newtons per meter squared on the surface. What is the intensity of the
light? Use a value of 3.00 times 10 to the
8 meters per second for the speed of light in vacuum.
This question is about light
reflecting from a surface. We’re told that the light exerts a
pressure on the surface of 3.50 times 10 to the negative six newtons per meter
squared. Let’s label this pressure as
capital 𝑃. In order to understand why the
light exerts a pressure like this, let’s recall that even though light waves don’t
have mass, they can still transfer momentum.
Imagine that we have a load of
light waves like this one colliding with and reflecting off a surface. Then those light waves experience a
change in momentum. A momentum change △𝑃 that occurs
during a time interval △𝑡 means that there is a force 𝐹 equal to △𝑃 divided by
△𝑡. So when the light waves reflect,
they must be exerting a force on the surface. Whenever a force is exerted across
a surface with a particular area, this results in a pressure on the surface equal to
the force divided by the area.
So in this way the light waves
exerted pressure on the surface. This is known as a radiation
pressure. For a perfectly reflective surface,
that is, a surface that reflects 100 percent of the light incident on it, the
radiation pressure 𝑃 exerted on the surface by light with an intensity of 𝐼 is
equal to two times 𝐼 divided by 𝑐, the speed of light. In this question, we’re told that
the light is directed at a 100 percent reflective surface. This means that this equation for
the radiation pressure does apply in this case.
We’re asked to work out the
intensity of the light. We know the value of the radiation
pressure, and we’re told to take the speed of light as 3.00 times 10 to the 8 meters
per second. So if we rearrange this equation to
make 𝐼 the subject, we can then sub in our values for the pressure 𝑃 and the speed
of light 𝑐 to calculate the light’s intensity. In order to make 𝐼 the subject of
the equation, we first need to multiply both sides of the equation by the speed of
light 𝑐.
Then on the right-hand side of the
equation, the 𝑐 in the numerator cancels with the 𝑐 in the denominator. This gives us an equation that says
two times 𝐼 is equal to 𝑐 times 𝑃. Then we divide both sides of the
equation by two. On the left-hand side, the two in
the numerator cancels with the two in the denominator. This gives us an equation that says
intensity 𝐼 is equal to the speed of light 𝑐 multiplied by the radiation pressure
𝑃 divided by two.
We can now take our values for 𝑐
and 𝑃 and sub them into this equation. Doing that gives us this expression
for the intensity 𝐼. We can notice that the speed of
light is given in units of meters per second, which is the SI base unit for
speed. And the radiation pressure has
units of newtons per meter squared, which is the SI base unit for pressure. This means that the light intensity
we’re going to calculate will be in the SI base unit for intensity. This is the watt per meter
squared.
When we evaluate this expression
for 𝐼, it gives us our answer to the question that the intensity of the light is
equal to 525 watts per meter squared.
