Video Transcript
What intensity of light would be needed to exert a 1.0-newton force on a polished copper plate with an area of 2.5 meter squared? Assume that all of the light incident on the copper plate is reflected. Use a value of 3.00 times 10 to the eight meters per second for the speed of light in vacuum.
Okay, so this is a question about light reflecting from a surface. That surface is a polished copper plate, which we’re told has an area of 2.5 meters squared. Let’s label this area as 𝐴. We’re asked to work out what intensity this light must have in order to exert a force of 1.0 newtons on this plate. We’ll label this force as 𝐹. Let’s suppose that this here is the copper plate. And we’ll begin by thinking about how it is that light can exert a force on this plate. We can recall that even though light waves don’t have mass, they can still transfer momentum. So if we have a load of light waves like this one colliding with and reflecting off the plate, then those light waves experience a change in momentum.
This momentum change means that there is a force, since whenever something’s momentum changes by an amount Δ𝑝 over a time Δ𝑡, there’s a force 𝐹 equal to Δ𝑝 divided by Δ𝑡. So when these light waves reflect off the copper plate, they must be exerting a force on it. In this case, we know the area 𝐴 of the plate. We can recall that whenever a force acts across the surface with a particular area, we get a pressure on that surface equal to the force divided by the surface area. If we label the pressure as capital 𝑃, then for a force 𝐹 acting across a surface with an area 𝐴, we can write this equation in terms of symbols as 𝑃 is equal to 𝐹 divided by 𝐴. This means that the light waves are exerting a pressure on the copper plate. This is known as radiation pressure.
The question tells us to assume that all of the light which is incident on the copper plate gets reflected. We can recall that when 100 percent of the light incident on a surface gets reflected, the radiation pressure on that surface, capital 𝑃, is equal to two times the intensity of the light 𝐼 divided by the speed of light 𝑐. In our case, the quantity that we’re trying to find is the intensity of the light. So that’s the 𝐼 in this equation. In order to work out this value, we first need to find the value of the radiation pressure capital 𝑃. To do this, we can make use of this equation here, which tells us how to calculate the radiation pressure in terms of two quantities that we already know, the force 𝐹 and the area 𝐴 of the surface. So let’s take our values for 𝐹 and 𝐴 and sub them into this equation.
When we do this, we find that 𝑃 is equal to 1.0 newtons divided by 2.5 meters squared. Then evaluating the expression gives a result for the radiation pressure 𝑃 of 0.4 newtons per meter squared. Looking again at this equation here, we see that we now know the value of the quantity 𝑃. And we’re told in the question to use a value of 3.00 times 10 to the eight meter per second for the speed of light in vacuum. So that’s our value for the quantity 𝑐. This means that we know the values for all the quantities in this equation except the intensity 𝐼, which we’re trying to find. To calculate this value, we need to take the equation and rearrange it to make 𝐼 the subject.
The first step is to multiply both sides of the equation by the speed of light 𝑐. Then on the right-hand side of the equation, we now have a 𝑐 in the numerator which cancels with the 𝑐 in the denominator. From here, we then divide both sides of the equation by two. Again, looking at the right-hand side, we find that the two in the numerator cancels the two in the denominator. So we get this expression here, which we can also write the other way around to say that 𝐼 the intensity of the light is equal to the speed of light 𝑐 multiplied by the radiation pressure 𝑃 divided by two. So if we now take our values for 𝑐 and 𝑃 and sub them into this equation, then we can calculate the value of 𝐼.
This value of 𝑃 that we’ve calculated is the radiation pressure which corresponds to a force of 1.0 newtons acting over the 2.5-meter-squared area of the copper plate. So when we put that value into this equation, the value of 𝐼 that we will find is the intensity of light that would exert a 1.0-newton force on the plate. And this intensity that would be needed to exert this 1.0-newton force is exactly what the question is asking us to find. After subbing the values for 𝑐 and 𝑃, we get that 𝐼 is equal to 3.00 times 10 to the eight meters per second multiplied by 0.4 newtons per meter squared divided by two.
We can notice that the speed of light has units of meters per second, which is the SI base unit for speed, and the radiation pressure has units of newtons per meter squared, the SI base unit for pressure. Since both of these quantities are expressed in their SI base units, then the value for 𝐼 that we’ll calculate will be in the SI base unit for intensity, which is the watt per meter squared. Evaluating the expression gives a result of six times 10 to the seven watts per meter squared.
So our answer to the question is that the intensity of light that would be needed to exert a 1.0-newton force on the polished copper plate is six times 10 to the seven watts per meter squared.
