The average energy density of an electromagnetic wave in vacuum is $u_{avg}=1.06\times 10^{-11}{\text{ J/m}}^3$. What is the intensity $I$ of the wave?

Solution

1. The intensity is given by $I=u_{avg}c$.
2. Given $u_{avg}=1.06\times 10^{-11}{\text{ J/m}}^3$ and $c=3\times 10^8\text{ m/s}$.
3. Calculating $I=(1.06\times 10^{-11})\times (3\times 10^8)$.
4. This results in $$\begin{array}{rl}I& =3.18\times 10^{-3}{\text{ W/m}}^2\\ & =0.00318{\text{ W/m}}^2\end{array}$$.
5. Hence the answer is (A).