A plane EM wave travelling along z-deflection is described by $\overrightarrow{\mathrm{E}}={\mathrm{E}}_0 \sin \left(\mathrm{kz}-\mathrm{\omega t}\right)\hat{\mathrm{i}}$ and $\overrightarrow{\mathrm{B}}={\mathrm{B}}_0 \sin \left(\mathrm{kz}-\mathrm{\omega t}\right)\hat{\mathrm{j}}$ . Then:

Energy density of electric field $\mathrm{E}$ is ${\mathrm{u}}_{\mathrm{E}}=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2$

Energy density due to magnetic field $B$ is ${\mathrm{u}}_{\mathrm{B}}=\frac{1}{2}\frac{{\mathrm{B}}^2}{{\mathrm{\mu }}_0}$

Total energy density of wave $=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2+\frac{1}{2}\frac{{\mathrm{B}}^2}{{\mathrm{\mu }}_0}$

Now, ${\left({\mathrm{u}}_{\mathrm{E}}\right)}_{\mathrm{av}}=\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}{\mathrm{u}}_{\mathrm{E}}\mathrm{dt}=\frac{1}{\mathrm{T}}{\int }_0^{\mathrm{T}}\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2$

                   $$\begin{gathered} =\frac{{\mathrm{\epsilon }}_0}{2\mathrm{T}}{\int }_0^{\mathrm{T}}{{\mathrm{E}}_0}^2{\sin }^2\left(\mathrm{kx}-\mathrm{\omega t}\right)\mathrm{dt} \\ =\frac{{\mathrm{\epsilon }}_0}{2\mathrm{T}}\left({{\mathrm{E}}_0}^2\times \frac{\mathrm{T}}{2}\right)=\frac{1}{4}{\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2 \left(\because {\int }_0^{\mathrm{T}}{\sin }^2\left(\mathrm{kx}-\mathrm{\omega t}\right)\mathrm{dt}=\frac{\mathrm{T}}{2}\right) \end{gathered}$$

Similarly, ${\left({\mathrm{u}}_{\mathrm{B}}\right)}_{\mathrm{av}}=\frac{1}{4}\frac{{{\mathrm{B}}_0}^2}{{\mathrm{\mu }}_0}$

$\therefore {\mathrm{u}}_{\mathrm{av}}={\left({\mathrm{u}}_{\mathrm{E}}\right)}_{\mathrm{av}}+{\left({\mathrm{u}}_{\mathrm{B}}\right)}_{\mathrm{av}}=\frac{1}{4}{\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2+\frac{1}{4}\frac{{{\mathrm{B}}_0}^2}{{\mathrm{\mu }}_0}$

$\because {\mathrm{E}}_0=\mathrm{c} {\mathrm{B}}_0$ and $\mathrm{c}=\frac{1}{{\sqrt{{\mathrm{\mu }}_0\mathrm{\epsilon }}}_0}$

$$\begin{gathered} \therefore \frac{1}{4}\frac{{{\mathrm{B}}_0}^2}{{\mathrm{\mu }}_0}=\frac{1}{4}\frac{{{\mathrm{E}}_0}^2/{\mathrm{c}}^2}{{\mathrm{\mu }}_0}=\frac{{{\mathrm{E}}_0}^2}{4{\mathrm{\mu }}_0}\times {\mathrm{\mu }}_0{\mathrm{\epsilon }}_0=\frac{1}{4}{\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2 \\ {\mathrm{u}}_{\mathrm{av}}=\frac{1}{4}{\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2+\frac{1}{4}\frac{{{\mathrm{B}}_0}^2}{{\mathrm{\mu }}_0}=\frac{1}{4}{\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2+\frac{1}{4}{\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2=\frac{1}{2}{\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2 \end{gathered}$$

Time averaged intensity, ${\mathrm{I}}_{\mathrm{av}}={\mathrm{u}}_{\mathrm{av}}\mathrm{c}=\frac{1}{2}{\mathrm{c\epsilon }}_0{{\mathrm{E}}_0}^2$