1. Jul 20, 2011

### Lindsayyyy

1. The problem statement, all variables and given/known data

A laser is focussed on an area A=0.1cm². The radiation pressure is p=10 pascal. Whats the power of the laser when 50% of the light is absorbed?

2. Relevant equations

I'm not sure, I tried it via:

Poynting vector:

$$S= \epsilon_{0}*c*E^{2}$$

$$p=\epsilon_{0}*E^{2}$$

3. The attempt at a solution

Because 50% is absorbed I think the pressure has to be divided by two. I solved the equation of the Poynting vector, so I have E² and put that in the equation for the pressure and I just solved it then, the solution is: 6*10^9 W/m² which I think is pretty high, but to be honest I'm not good in evaluations.

Can anyone help me out? Where are my mistakes? I'm a bit wondered too, because I didn't use the area which was given at all (the exercise has two more tasks, but it doesn't look like I need that A somewhere later).

Thanks for the help

Last edited: Jul 20, 2011
2. Jul 20, 2011

### WatermelonPig

Intensity = sqrt(epsilon-not/mu-not)(Electric field amplitude)^2
Assuming no dielectric.

3. Jul 21, 2011

### I like Serena

Hi Lindsayyyy!

This is what I found on wikipedia:
"Radiation pressure is the pressure exerted upon any surface exposed to electromagnetic radiation. If absorbed, the pressure is the power flux density divided by the speed of light. If the radiation is totally reflected, the radiation pressure is doubled. For example, the radiation of the Sun at the Earth has a power flux density of 1,370 W/m2, so the radiation pressure is 4.6 µPa (absorbed)."