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Intensity of Electromagnetic Waves

Intensity of Electromagnetic Waves :- The intensity of electromagnetic (EM) waves at a point refers to the average amount of energy crossing per second per unit area normally around that point during the propagation of EM wave. It is a measure of the wave’s power and is proportional to the square of the wave’s amplitude.

Let us consider an imaginary cylinder of cross section area ΔA lying along the direction of propagation of EM wave so that wave crosses the area ΔA normally.

If u_avg. is the average energy density, then the total energy ΔU contained in the imaginary cylinder will be :

$\displaystyle \Delta U={{u}_{avg.}}(c\Delta t)(\Delta A)$

The intensity of electromagnetic waves at any point inside this imaginary cylinder,

$\displaystyle I=\frac{\Delta U}{\Delta A\times \Delta t}={{u}_{avg.}}c$ …..(1)

Now for EM waves,

$\displaystyle {{u}_{avg.}}={{\left( {{u}_{E}}+{{u}_{B}} \right)}_{avg.}}={{\left( \frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}+\frac{1}{2}\frac{{{B}^{2}}}{{{\mu }_{0}}} \right)}_{avg.}}$

$\displaystyle {{u}_{avg.}}={{\left[ \frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}{{\sin }^{2}}(kx-\omega t)+\frac{1}{2}\frac{B_{0}^{2}{{\sin }^{2}}(kx-\omega t)}{{{\mu }_{0}}} \right]}_{avg.}}$

(As $\displaystyle E={{E}_{0}}\sin (kx-\omega t)\text{ }and\text{ }B={{B}_{0}}\sin (kx-\omega t)$ )

$\displaystyle {{u}_{avg.}}=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}{{\left[ {{\sin }^{2}}(kx-\omega t) \right]}_{avg.}}+\frac{1}{2}\frac{B_{0}^{2}}{{{\mu }_{0}}}{{\left[ {{\sin }^{2}}(kx-\omega t) \right]}_{avg.}}$

$\displaystyle {{u}_{avg.}}=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}\times \frac{1}{2}+\frac{1}{2}\frac{B_{0}^{2}}{{{\mu }_{0}}}\times \frac{1}{2}$

$\displaystyle \Rightarrow {{u}_{avg.}}=\frac{1}{2}{{\varepsilon }_{0}}{{\left( \frac{{{E}_{0}}}{\sqrt{2}} \right)}^{2}}+\frac{1}{2{{\mu }_{0}}}{{\left( \frac{{{B}_{0}}}{\sqrt{2}} \right)}^{2}}$

$\displaystyle \Rightarrow {{u}_{avg.}}=\frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2}\frac{B_{rms}^{2}}{{{\mu }_{0}}}$

So intensity,

$\displaystyle I={{u}_{avg.}}c=\left[ \frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2}\frac{B_{rms}^{2}}{{{\mu }_{0}}} \right]\times c$ …..(2)

In terms of electric field,

$\displaystyle I=\left[ \frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2{{\mu }_{0}}}\frac{E_{rms}^{2}}{{{c}^{2}}} \right]\times c=\left[ \frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2{{\mu }_{0}}}\frac{E_{rms}^{2}}{\left( \frac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}} \right)} \right]\times c$

$\displaystyle \Rightarrow I=\frac{1}{2}c{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2}c{{\varepsilon }_{0}}E_{rms}^{2}$

$\displaystyle \therefore I=c{{\varepsilon }_{0}}E_{rms}^{2}=\frac{1}{2}c{{\varepsilon }_{0}}E_{0}^{2}$ …..(3)

In terms of magnetic field,

$\displaystyle I=\left[ \frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2}\frac{B_{rms}^{2}}{{{\mu }_{0}}} \right]\times c=\left[ \frac{1}{2}{{\varepsilon }_{0}}{{c}^{2}}B_{rms}^{2}+\frac{1}{2}\frac{B_{rms}^{2}}{{{\mu }_{0}}} \right]\times c$

$\displaystyle \Rightarrow I=\left[ \frac{1}{2}{{\varepsilon }_{0}}\times \left( \frac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}} \right)\times B_{rms}^{2}+\frac{1}{2}\frac{B_{rms}^{2}}{{{\mu }_{0}}} \right]\times c=\left[ \frac{1}{2}\frac{B_{rms}^{2}}{{{\mu }_{0}}}+\frac{1}{2}\frac{B_{rms}^{2}}{{{\mu }_{0}}} \right]\times c$

$\displaystyle \therefore I=\frac{B_{rms}^{2}}{{{\mu }_{0}}}\times c=\frac{1}{2}\frac{B_{0}^{2}}{{{\mu }_{0}}}\times c$ …..(4)

Intensity of Electromagnetic Waves as an average value of Poynting Vector

The intensity of electromagnetic waves can be defined as the average value of the Poynting vector.

$\displaystyle I=\left\langle \left| {\vec{S}} \right| \right\rangle =\left\langle \frac{EB}{{{\mu }_{0}}} \right\rangle$ …..(5)

$\displaystyle \Rightarrow I=\left\langle \frac{\left[ {{E}_{0}}\sin (kx-\omega t) \right]\left[ {{B}_{0}}\sin (kx-\omega t) \right]}{{{\mu }_{0}}} \right\rangle$

$\displaystyle I=\frac{{{E}_{0}}{{B}_{0}}}{{{\mu }_{0}}}\left\langle {{\sin }^{2}}(kx-\omega t) \right\rangle$

$\displaystyle I=\frac{{{E}_{0}}{{B}_{0}}}{{{\mu }_{0}}}\times \frac{1}{2}$

$\displaystyle I=\frac{1}{{{\mu }_{0}}}\times \frac{{{E}_{0}}}{\sqrt{2}}\times \frac{{{B}_{0}}}{\sqrt{2}}=\frac{1}{{{\mu }_{0}}}\times {{E}_{rms}}\times {{B}_{rms}}$