Home exercise 6.2.1: Ground-penetrating radar in wet soil

$\omega \epsilon =2\pi \times 10^8\times 16\times 10^{-9}/36\pi =0.0889S/m$, so $p=0.02/0.0889=0.225$: neither class. Use the exact formulas.

$\omega \sqrt{\mu \epsilon }=2\pi f\sqrt{16}/c=8.378rad/m$, $\omega \sqrt{\mu \epsilon /2}=5.924rad/m$ and $\sqrt{1+p^2}=1.025$:

$\alpha \approx \frac{\sigma }{2}\sqrt{{\mu }_0/16{\epsilon }_0}=0.01\times 30\pi =0.942Np/m$, only $0.6\%$ high: the low-loss formula is still usable at $p=0.225$.

Down and back the path is $2d$, so $2d\times 8.14=40\mathrm{dB}$ gives $d=40/(2\times 8.14)=2.46\mathrm{m}$.

${\beta }^2-{\alpha }^2=71.06-0.88=70.2=(\omega \sqrt{\mu \epsilon }{)}^2$ and $2\alpha \beta =15.8=\omega {\mu }_0\sigma$, the two relations of Eq. (6.2.9). The wavelength in the soil, $0.75\mathrm{m}$, is practically ${\lambda }_0/\sqrt{{\epsilon }_r}=0.75\mathrm{m}$: the loss hardly changes the phase. Drier soil (smaller $\sigma$) lets the same radar see much deeper, while a clay layer with $\sigma \sim 0.1S/m$ stops it within a metre.