# Video: Calculating the Force Light Exerts on a Reflective Surface

Light with an intensity of 50.0 W/m² is directed at a 100% reflective surface. The surface has an area of 2.25 m². What force is exerted by the light on the surface? Use a value of 3.00 × 10⁸ m/s for the speed of light in a vacuum.

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### 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.