Video Transcript
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.
