Video Transcript
Light is directed at a 100 percent reflective surface. The surface has an area of 4.00 meters squared. The light exerts a force of 2.00 times 10 to the negative six newtons on the surface. What is the intensity of the light? Use a value of 3.00 times 10 to the eight meters per second for the speed of light in vacuum.
This question is asking us about light reflecting from a surface. We’re told that the area of this surface is 4.00 meters squared. Let’s label this area as 𝐴. We’re also told that the force exerted by the light on the surface is 2.00 times 10 to the negative six newtons. We’ll label this force as 𝐹. We’re being asked to work out the intensity of the light. But first, let’s begin by thinking about why it is that the light exerts this force on the surface. We can recall that even though light waves don’t have mass, they can still transfer momentum. This means that if 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.
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. What this means is that when the light waves reflect, they must be exerting a force on the surface. In this question, we’ve got light waves exerting a force 𝐹 over a surface with an area 𝐴. Now, whenever there’s a force acting over a surface with a particular area, this causes a pressure on the surface equal to the force divided by the area. If we label the pressure as capital 𝑃, then for a force 𝐹 acting over a surface with an area 𝐴, then we can write this equation in terms of symbols as capital 𝑃 is equal to 𝐹 divided by 𝐴. This means that when the light waves reflect, they’re exerting a pressure on the surface. This is known as radiation pressure.
The question tells us that this surface is 100 percent reflective. We can recall that for a perfectly reflective surface, the radiation pressure capital 𝑃 is equal to two times the intensity of the light 𝐼 divided by the speed of light 𝑐. In this question, we’re trying to calculate the intensity 𝐼 of the light. And we have an equation which relates this intensity to the radiation pressure exerted on the surface. At the moment, we don’t know what that radiation pressure is. But we also have a second equation which tells us how to calculate the radiation pressure using the force and the surface area. Now, the force and the surface area are two quantities that we do know the values of. So let’s take these values for the force 𝐹 and the area 𝐴 and sub them into this equation.
Doing these substitutions gives us this expression here for the radiation pressure. Then, evaluating the expression gives a result of five times 10 to the negative seven newtons per meter squared. So, if we look back at this equation, we see that we now know the value of the radiation pressure capital 𝑃. And we can also notice that we’re told in the question to use a value of 3.00 times 10 to the eight meters per second for the speed of light in vacuum. So that’s our value for the quantity 𝑐. Since we’re trying to calculate the intensity of the light, we need to take this equation and rearrange it to make the intensity 𝐼 the subject.
To do this, we begin by multiplying both sides of the equation by the speed of light 𝑐. Then, on the right-hand side of the equation, we have a 𝑐 in the numerator which cancels with the 𝑐 in the denominator. Next, we divide both sides of the equation by two. Now, on the right-hand side, the two in the numerator cancels with the two in the denominator. Finally, we notice that we can also write this expression the other way around, which gives us an equation that says the intensity 𝐼 is equal to the speed of light 𝑐 multiplied by the radiation pressure 𝑃 divided by two. In this case, we have values for both of the quantities 𝑐 and 𝑃.
If we now sub these values into this equation, we can use them to calculate the intensity 𝐼 of the light. If we take a look at the units in this expression, we can see that the speed of light is expressed in meters per second, the SI base unit for speed, and the radiation pressure is in newtons per meter squared, the SI base unit for pressure. Since both of the quantities on the right-hand side are expressed in their SI base units, this means that the light intensity 𝐼 that we’re going to calculate will be expressed in the SI base unit for intensity, which is the watt per meter squared. When we evaluate this expression, it gives us a result for 𝐼 of 75 watts per meter squared. And so our answer to this question is that the intensity of the light is equal to 75 watts per meter squared.
