Calculating electric field from radiant flux density

[eep]

Hi,
The problem is stated as follows:

"A laser provides pulses of EM-radiation in vacuum lasting $10^{-12}$ seconds. If the radiant flux density is $$10^{20} {\frac{W}{m^2}}$$, determine the amplitude of the electric field of the beam."

So far, I figure that the period of one wave is $$10^{-12}$$ seconds. The instantaneous energy flux density is equal to $$\frac{E^2}{{\mu_o}c}}$$. So an intergral of the instantaneous energy flux denisty over a period should equal the radiant flux density, no?

$$\int_{t=0}^{t=10^{-12}} \frac{E^2}{\mu_oc} dt = 10^{20} \frac{W}{m^2}$$

Now, $$E = E_o cos({\omega}t)$$ and $${\omega} * 10^{-12} = 2\pi$$. So I should be able to integrate and solve for $$E_o$$?

 

[CarlB]

No, the $$10^{-12}$$ figure gives the duration of the pulse, not the period. I don't think that you need that number to make the calculation. For that matter, an E&M pulse with a period of 1 picosecond wouldn't be a "laser", but instead would be a short wavelength radar. So my guess is that the $$10^{-12}$$ is included to either confuse you or to make you integrate out to find an average electric field when you compute the wattage.

Admittedly, I did not take undergraduate E&M and I'm a little hazy about units, but your integral looks to me like the thing you're integrating is correct. But my suggestion is that you get rid of the integral and the $$dt$$. Other than that, I believe you can now solve for $$E_0$$, which is the amplitude of the electric field. If you leave the integral in, you will definitely have the wrong units (check em).

By the way, you probably already know that the average value of $$\cos^2$$ is 0.5, which gets back to why they gave such a large number for the pulse duration (that is, so that the average would be correct).

Carl