Question

Show that for a continuous sinusoidal electromagnetic wave, the peak intensity is twice the average intensity $\left(I_0=2I_{\mathrm{ave}}\right)$, using either the fact that ${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\sqrt{\mathbf{2}{\mathbf{E}}_{\mathbf{rms}}}\mathbf{,}\mathbf{or}\mathbf{}{\mathbf{B}}_{\mathbf{0}}\mathbf{=}\sqrt{\mathbf{2}{\mathbf{B}}_{\mathbf{rms}}}$, where rms means average (actually root mean square, a type of average).

Short answer

The solution is${\mathrm{I}}_0=2{\mathrm{I}}_{\mathrm{avg}}$ .

Step-by-step solution

Define electromagnetic waves

Electromagnetic waves, or EM waves, are waves that are formed when an electric field and a magnetic field vibrate together.

Explanation

It is known that a wave's intensity is related to its amplitude squared and is given by

${\mathrm{I}}_0={\mathrm{kE}}_0^2$

Where ${\mathrm{I}}_0$is peak intensity, k is proportionality constant, ${\mathrm{E}}_0$is peak electric field strength.

And the average intensity is calculated as

${\mathrm{I}}_{\mathrm{avg}}={{\mathrm{kE}}_{\mathrm{rms}}}^2$

Where ${\mathrm{I}}_{\mathrm{avg}}$is average intensity, k is proportionality constant, ${\mathrm{E}}_{\mathrm{rms}}$ is average electric field strength.

Therefore,

$\frac{{\mathrm{I}}_0}{{\mathrm{I}}_{\mathrm{avg}}}=\frac{{\mathrm{kE}}_0^2}{{{\mathrm{kE}}_{\mathrm{rms}}}^2}$

Substitute the value of ${\mathrm{E}}_0=\sqrt{2{\mathrm{E}}_{\mathrm{rms}}}$in the above equation,

$$\begin{gathered} \frac{{\mathrm{I}}_0}{{\mathrm{I}}_{\mathrm{avg}}}=\frac{{\mathrm{kE}}_0^2}{{{\mathrm{kE}}_{\mathrm{rms}}}^2} \\ \frac{{\mathrm{I}}_0}{{\mathrm{I}}_{\mathrm{avg}}}=\frac{{\left(\sqrt{2}{\mathrm{E}}_{\mathrm{rms}}\right)}^2}{{{\mathrm{E}}_{\mathrm{rms}}}^2} \\ \frac{{\mathrm{I}}_0}{{\mathrm{I}}_{\mathrm{avg}}}=2 \\ {\mathrm{I}}_0=2{\mathrm{I}}_{\mathrm{avg}} \\ \mathrm{Therefore},{\mathrm{I}}_0=2{\mathrm{I}}_{\mathrm{avg}}. \end{gathered}$$