Video Transcript
The diagram shows a 100 percent reflective surface that is parallel to light rays with a constant intensity of 40 watts per meter squared. The thickness of the surface is negligible. What is the radiation pressure on the surface due to the light?
Our problem statement tells us the intensity of these incoming rays of light. If we had a surface with an area of one square meter, then 40 watts of power, or 40 joules of energy every second, would fall on that surface. Notice though that this would only happen if the surface were oriented perpendicularly to the rays of light, that is, if the surface looked like this from a side-on view. We can see that the reason the surface needs to be oriented this way so that a full 40 watts of power falls on every square meter is because it’s in this orientation of the surface that it’s fully exposed to the rays of light.
If the surface were instead tilted at some angle smaller than 90 degrees, then it would receive fewer light rays. If it was at an even smaller angle, it would receive fewer rays still. And as an extreme example of this, if the surface were at zero degrees, that is, parallel to our incoming rays of light, then none of those light rays would land on the surface. Therefore, no particles of light, photons, would land on the surface. And so there would be no radiation pressure.
Recalling that pressure is defined as a force spread out over an area, we know that radiation pressure will have units of force, newtons, say, divided by units of area, say, square meters. For a surface like the one we have here that is parallel to the incoming rays of light, since none of those rays land on the surface, the surface experiences a radiation pressure of zero newtons per meter squared.
