Physics3 question on light, radiation pressure

In summary, a student is struggling with a physics problem involving using radiation pressure from a beam of light to suspend a piece of paper. They have attempted the problem multiple times with different approaches and have not been successful. After receiving feedback and further thought, they realize that the area of the paper does not affect the ability of the beam to hold up the paper, leading to the correct solution.
  • #1
Yroyathon
42
0
Hi folks, first time poster. I'm a math undergrad taking my final physics3 requirement. I'm doing ok in the class, but about 2 out of 8 of the homework problems confound me. I removed the numbers below and replaced them with variables (does that matter on this site?). Mostly I'm interested in getting the concepts right, which approach to use, when to use certain equations.

Below is one problem I've tried to answer 4 times already, incorrectly.

Homework Statement


Suppose that you want to use the radiation pressure from a beam of light to suspend a piece of paper in a horizontal position; the paper has an area of A_1 cm2 and a mass of m_1 grams. Assume that there is no problem with balance, that the paper is dark and absorbs the beam fully, and that the entire beam can be used to hold the paper against the pull of gravity. How many watts must the light produce?


Homework Equations


For radiation pressure, I've used u=F/A, where F is the gravitational force, F=m_1 * (9.8). I wasn't sure if the A here should be the area of the paper A_1, or if we should consider the paper as a point mass and use unit area, A_1 = 1?... so either u=F=m_1*(9.8) or u=F/A=m_1*(9.8)/A_1.

To calculate power, I've been using Power=c*u*A, where c is the speed of light (3*10^8), u is the radiation pressure, and A is the area of the paper A_1. So Power=(3*10^8)*m_1*(9.8)*A_1.

But this hasn't worked. I tried using Power=c*u, and saying that A = 1 there, and using u = F/A = m_1*(9.8)/A_1, so that Power=c*u = (3*10^8)*m_1*(9.8)/A_1, but this didn't work either.

i'm kind of guessing at the approach, and was hoping someone here could explain the parts I'm not getting.

thanks in advance!
,Yroyathon
 
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  • #2
Welcome to PF :smile:

To calculate power, I've been using Power=c*u*A, where c is the speed of light (3*10^8), u is the radiation pressure, and A is the area of the paper A_1. So Power=(3*10^8)*m_1*(9.8)*A_1.

Including A is the best approach.

What does u, the radiation pressure, need to be here?

In the above, you are equating u with m_1*(9.8 m/s^2), which is a force and therefore must be wrong because a pressure cannot equal a force.
 
  • #3
yes, you are right.

after another hour or so of thought, I manage to convince myself that the area in this problem does not matter.

u=F/A=m*g/A, and Power=u*c*A, so Power=m*g*c. Using this the answer turned out to be correct.

The way I managed to convince myself is that if the beam was spread out to cover the entire piece of paper equally, or if the beam was focused on the exact center of the paper, nothing would change in the beam's ability to push-up/levitate the paper.

Thanks a lot for reading and replying. Posting the problem here was helpful because I had to focus even harder on what I knew and didn't know so that I could write the post and describe my attempt at the problem.
 
  • #4
Glad it worked out. Good observation about the area, by the way.
 

Related to Physics3 question on light, radiation pressure

What is light and how does it travel?

Light is a form of electromagnetic radiation that is visible to the human eye. It travels in the form of waves and does not require a medium to travel through. Light is made up of particles called photons.

What is radiation pressure and how does it affect objects?

Radiation pressure is the pressure exerted by electromagnetic radiation on objects. This pressure is caused by the transfer of momentum from the photons in the radiation to the object. The intensity and color of the radiation can affect the amount of radiation pressure on an object.

How is light used in everyday life?

Light has many practical applications in our daily lives. It is used in lighting our homes, powering electronic devices, and in communication technologies such as fiber optics. It is also used in medical imaging and therapy, as well as in various industrial processes.

What is the relationship between light and energy?

Light is a form of energy, specifically electromagnetic energy. It can be described as both a wave and a particle. The energy of light is directly proportional to its frequency, meaning that higher frequency light (such as ultraviolet or x-rays) carries more energy than lower frequency light (such as radio waves).

How does light interact with matter?

When light encounters matter, it can be absorbed, reflected, or transmitted. The way light interacts with matter depends on the properties of both the light and the matter. For example, different materials will absorb or reflect different wavelengths of light, which is why objects appear to have different colors.

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