Project Seafarer was an ambitious program to construct an enormous antenna, buried underground on a site about $\mathbf{10000}\mathbf{}\mathbf{}{\mathbf{km}}^{\mathbf{2}}$in area. Its purpose was to transmit signals to submarines while they were deeply submerged. If the effective wavelength were$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}$Earth radii, what would be the (a) frequency and(b) period of the radiations emitted? Ordinarily, electromagnetic radiations do not penetrate very far into conductors such as seawater, and so normal signals cannot reach the submarines.

Area, $A=10000{\mathrm{km}}^2$.

The wavelength,$\lambda =1.0\times {10}^4$ earth radii.

Earth radii is,$6.37\times {10}^6\mathrm{m}$ .

Speed of light,$c=3\times {10}^8\text{m/s}$ .

The transmission signals from a submarine deeply submerged underground have wavelength $1.0\times {10}^4$earth radii. Their frequency and time period can be calculated using the general formula for frequency. Earth radii can be considered as the average distance from the physical center to the surface, having value $6.37\times {10}^6\mathrm{m}$. Also, the time period of the radiation is the inverse of the frequency.

Formula:

Frequency is,$f=\frac{c}{\lambda }$ .

The time period is, $T=\frac{1}{f}$.

Frequency can be calculated as,

$f=\frac{c}{\lambda }$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}f& =& \frac{3\times {10}^8}{1.0\times {10}^4\times 6.37\times {10}^6}\\ & =& 4.7\times {10}^{-3}\mathrm{Hz}\\ & & \end{array}$

Thus, the frequency of the emitted radiations is $4.7\times {10}^{-3}\mathrm{Hz}$

The time period can be calculated as,

$T=\frac{1}{f}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}T& =& \frac{1}{4.7\times {10}^{-3}}\\ & =& 212\mathrm{seconds}\\ & & \end{array}$

Thus, the period of the emitted radiations is $212\mathrm{seconds}$