# Question Video: Finding the Momentum Transfer and Force on an Object due to Radiation Pressure

During a time interval of 1.00 s, light with an intensity of 1.00 kW/m² is incident on a mirror with an area of 2.00 m² and is completely reflected. What is the energy of the light reflected from the mirror? How much momentum is imparted to the mirror? How much force is exerted on the mirror?

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### Video Transcript

During a time interval of 1.00 seconds, light with an intensity of 1.00 kilowatts per meter squared is incident on a mirror with an area of 2.00 meters squared and is completely reflected. What is the energy of the light reflected from the mirror? How much momentum is imparted to the mirror? How much force is exerted on the mirror?

We can call the time interval of 1.00 seconds 𝑡. The intensity of the light, 1.00 kilowatts per meter squared, we’ll call 𝐼. And the area of the mirror, 2.00 meters squared, we’ll call 𝐴. We want to know three things involving this scenario: the energy of the light reflected from the mirror, which we’ll call 𝐸, the momentum imparted to the mirror, 𝑃, and the force exerted on the mirror, 𝐹.

Let’s start out by drawing a sketch. We have a mirror of area 𝐴. And onto the mirror light of intensity 𝐼 is shined for a period of time of 1.00 seconds. Over this time, we want to know how much energy is reflected from the mirror, how much momentum is imparted to it, and how much force is exerted on it.

To solve for the energy of the reflected light, well let’s recall the connections between intensity, power, and energy. Intensity, 𝐼, is power per unit area. And because power is equal to energy per unit time, intensity is also equal to energy per time per area. So 𝐼 equals 𝐸 over 𝑡 times 𝐴 or energy 𝐸 equals 𝐼 times 𝑡 times 𝐴. Each of these three values is given in our problem statement. So we can plug in and solve for energy.

When we do, being careful to use units of watts per meter squared for our intensity, we find that the energy of the reflected light from the mirror over one second is equal to 2.00 kilojoules. Next, you want to turn and consider the moment imparted to the mirror due to this light.

We can recall that the momentum of light is equal to its energy divided by 𝑐. When we apply this relationship to our case, we include a factor of two in the equation. The reason is because the light incident on our surface is reflected back since the surface is a mirror. So the change in momentum of each photon incident on the mirror is twice what it would be if the photons were simply absorbed.

So the momentum imparted to the mirror is two times the energy of the incident light over 𝑐, where 𝑐 the speed of light we’ll treat as exactly 3.00 times 10 to the eighth meters per second. When we plug in for 𝐸 and 𝑐, being careful to use units of joules for the energy, and calculate 𝑃, we find that its equal to 1.33 times 10 to the negative fifth kilograms meters per second. That’s the momentum this light imparts to the mirror.

Lastly, we want to solve for the force that this light exerts on the mirror. To do that, let’s start by recalling the impulse momentum theorem. This states that impulse, the quantity 𝐹 times the change in time, is equal to an object’s change in momentum 𝑝. Written in terms of our variables in this problem, 𝐹 times 𝑡 equals 𝑝 or 𝐹 equals momentum over time.

We’re given a value of time in the problem statement. And we’ve solved for the momentum in a previous part of the problem. So we’re ready to solve for 𝐹. When we enter in the values for 𝑝 and 𝑡, since time is simply one second the math is simpler, and we find that 𝐹 is 1.33 times 10 to the negative fifth newtons. This is the amount of force exerted on the mirror over one second.