91Ó°ÊÓ

Intensity of sunlight $I=10\frac{\mathrm{μ°Â}}{{\mathrm{m}}^2}$

Distance$r=10\mathrm{km}$

Consider the intensity relation of wave, square both the sides, and solve for electric field component. Finally, substitutingtheelectric field component value inthemagnetic field component relation, find its amplitude. Using intensity and power relation, and find the transmission power.

Formulae:

$c=\frac{E_m}{B_m}$

$I=\frac{E_m^2}{2{\mathrm{μ}}_{0}c}$

$P=2\mathrm{Ï€}{r}^{2}I$

The time averaged rate per unit area at which energy is transported, $S_{avg},$is called the intensity $I$ of the wave.

$$\begin{gathered} I=\frac{E_m^2}{{24}_{n}c} \\ {E}_{rms}=\frac{E_m}{\sqrt{2}} \\ I=\frac{E_m^2}{2{\mathrm{μ}}_{0}c} \end{gathered}$$

Substitute the values in the equation and solve as:

role="math" localid="1662974648726" $\begin{array}{rcl}{E}_m& =& \sqrt{2{\mathrm{μ}}_{0}Ic}\\ & =& \sqrt{2\left(4\mathrm{π}×{10}^{-7}\frac{\mathrm{T}·\mathrm{m}}{\mathrm{A}}\right)\left(10×{10}^{-6}\frac{\mathrm{W}}{{\mathrm{m}}^2}\right)\left(3×{10}^8\frac{\mathrm{m}}{\mathrm{s}}\right)}\\ & =& 8.7×{10}^{-2}\frac{\text{V}}{\mathrm{m}}\end{array}$

The amplitude of the electric field component at the airplane is$8.7×{10}^{-2}\frac{\mathrm{V}}{\mathrm{m}}$

The wave speed cand amplitudes of electric and magnetic fields are related as

$$\begin{gathered} c=\frac{E}{B} \\ c=\frac{E_m}{B_m} \end{gathered}$$

The amplitude of magnetic field component in the wave is given by

$\begin{array}{rcl}{B}_m& =& \frac{E_m}{c}\\ & =& \frac{8.7×{10}^{-2}\frac{\mathrm{V}}{\mathrm{m}}}{3×{10}^8\frac{\mathrm{m}}{\mathrm{s}}}\\ & =& 2.9×{10}^{-10}\mathrm{T}\end{array}$

The amplitude of the magnetic field component at the airplane is$2.9×{10}^{-10}\text{T}$

The intensity I and power P are related as:

$$\begin{gathered} I=\frac{P}{4\mathrm{Ï€}{r}^2} \\ P=4\mathrm{Ï€}{r}^{2}I \end{gathered}$$

The intensity oftheemerging polarized light is half of the original intensity of light.

role="math" localid="1662975216508" $\begin{array}{rcl}P& =& \frac{4\mathrm{π}{r}^{2}I}{2}\\ & =& 2\mathrm{π}{r}^2\\ & =& 2\mathrm{π}{\left(10×{10}^3\mathrm{m}\right)}^2\left(10×{10}^{-6}\mathrm{W}/{\mathrm{m}}^2\right)\\ & =& 6.3×{10}^3\mathrm{W}\end{array}$

The transmission power is $6.3×{10}^3\text{W}$.