### 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 ๐ก. The intensity of the light, 1.00 kilowatts per meter squared, weโll call ๐ผ. And the area of the mirror, 2.00 meters squared, weโll call ๐ด. We want to know three things involving this scenario: the energy of the light reflected from the mirror, which weโll call ๐ธ, the momentum imparted to the mirror, ๐, and the force exerted on the mirror, ๐น.

Letโs start out by drawing a sketch. We have a mirror of area ๐ด. And onto the mirror light of intensity ๐ผ 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 letโs recall the connections between intensity, power, and energy. Intensity, ๐ผ, 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 ๐ผ equals ๐ธ over ๐ก times ๐ด or energy ๐ธ equals ๐ผ times ๐ก times ๐ด. 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 ๐. 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 ๐, where ๐ the speed of light weโll treat as exactly 3.00 times 10 to the eighth meters per second. When we plug in for ๐ธ and ๐, being careful to use units of joules for the energy, and calculate ๐, we find that its equal to 1.33 times 10 to the negative fifth kilograms meters per second. Thatโs 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, letโs start by recalling the impulse momentum theorem. This states that impulse, the quantity ๐น times the change in time, is equal to an objectโs change in momentum ๐. Written in terms of our variables in this problem, ๐น times ๐ก equals ๐ or ๐น equals momentum over time.

Weโre given a value of time in the problem statement. And weโve solved for the momentum in a previous part of the problem. So weโre ready to solve for ๐น. When we enter in the values for ๐ and ๐ก, since time is simply one second the math is simpler, and we find that ๐น is 1.33 times 10 to the negative fifth newtons. This is the amount of force exerted on the mirror over one second.