Show that average value of radiant flux density $'S'$ over a single period $'T'$ is given by $S = \frac{1}{{2c{\mu _0}}}E_0^2$.

Radiant flux density,

$\mathrm{S}=\frac{1}{\mu_{0}}(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}})$

$\therefore \mathrm{S}=c^{2} \in_{0}(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}) \ldots \text { (1) }\left[\because c=\frac{1}{\sqrt{\mu_{0} \in_{0}}}\right]$

Let electromagnetic waves propagate in $x$-direction. Electric field vector in $y$-direction and mag netic field vector in $z$-direction. Hence,

$\overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{E}_{0}} \cos (k x-\omega t)$

$\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}_{0}} \cos (k x-\omega t)$

$\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}=\left(\overrightarrow{\mathrm{E}_{0}} \times \overrightarrow{\mathrm{B}_{0}}\right) \cos ^{2}(k x-\omega t)$

$\mathrm{S}=c^{2} \in_{0}(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}) \quad[\because \text { From equation (1)] }$

$c^{2} \in_{0}\left(\overrightarrow{\mathrm{E}_{0}} \times \overrightarrow{\mathrm{B}_{0}}\right) \cos ^{2}(k x-\omega t)$

$\left[\because\left|\overrightarrow{\mathrm{E}}_{0} \times \overrightarrow{\mathrm{B}}_{0}\right|=\mathrm{E}_{0} \mathrm{~B}_{0} \sin 90^{\circ}=\mathrm{E}_{0} \mathrm{~B}_{0}\right]$

and $\int_{0}^{\mathrm{T}} \cos ^{2}(k x-\omega t) d t=\frac{\mathrm{T}}{2}$