A 150-watt light bulb emits 5.00 percent of its energy as electromagnetic radiation. The bulb is at the center of a perfectly light absorbing sphere of radius 10.00 meters. What is the radiation pressure on the sphere’s interior?
We’re told in the problem statement that the light bulb has a power rating of 150 watts. We’ll call that 𝑃. We’re also told that the efficiency of the light bulb in generating light is 5.00 percent. We’ll call that 𝐸. The bulb sits at the center of a sphere of radius 10.00 meters which we’ll refer to as 𝑟. We want to know the radiation pressure on the sphere’s interior due to the light from the light bulb. We’ll call this radiation pressure 𝑅𝑃.
Let’s begin by drawing a diagram of the situation. We have a light bulb that sits at the center of a sphere of radius 𝑟 equals 10.00 meters. As the light shines, it emits photons, discrete units of light, which run into and absorbed by the interior of the sphere. Because photons have momentum, all these many photons hitting the inside of the sphere creates pressure in the same way as a force over an area creates pressure. To figure out what 𝑅𝑃 — the radiation pressure on the inside of the sphere created by the light from the bulb — is, let’s recall a relationship for radiation pressure. Radiation pressure is equal to source power 𝑃 divided by the speed of light 𝑐 times the area over which that power is distributed 𝐴. When we apply this relationship to our situation, we see that we are given the power 𝑃; 𝑐 we assume to be exactly 3.00 times 10 to the eighth meters per second, the speed of light; and 𝐴 is the area of the sphere surrounding the bulb. We can recall that the area of a sphere is given by four times 𝜋 times the sphere radius squared.
So we can replace 𝐴 in our equation, for radiation pressure, with four times 𝜋 times 𝑟 squared. Now we’re ready to plug in for 𝑃, 𝑐, and 𝑟. As we do, we’ll be careful to use the true power actually emitted in the form of electromagnetic radiation by the bulb. That means that in our numerator, instead of simply having 150 watts, we include the efficiency of the bulb. Five percent as a decimal is 0.05. When we enter 3.00 times 10 to the eighth meters per second for 𝑐 and 10.00 for 𝑟 and enter these values on our calculator, we find that — to three significant figures — the radiation pressure on the inside of the absorbing sphere is 1.99 times 10 to the negative 11th newtons per meters squared. That’s the pressure created by the light from this light bulb.