Answer:
z = 0.8 (approx)
Explanation:
given,
Amplitude of 1 GHz incident wave in air = 20 V/m
Water has,
μr = 1
at 1 GHz, r = 80 and σ = 1 S/m.
depth of water when amplitude is down to 1 μV/m
Intrinsic impedance of air = 120 π Ω
Intrinsic impedance of water = [tex]\dfrac{120\pi}{\epsilon_r}[/tex]
Using equation to solve the problem
[tex]E(z) = E_0 e^{-\alpha\ z}[/tex]
E(z) is the amplitude under water at z depth
E_o is the amplitude of wave on the surface of water
z is the depth under water
[tex]\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}[/tex]
[tex]\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}[/tex]
[tex]\alpha =21.07\ Np/m[/tex]
now ,
[tex]1 \times 10^{-6} = 20 e^{-21.07\times z}[/tex]
[tex]e^{21.07\times z}= 20\times 10^{6}[/tex]
taking ln both side
21.07 x z = 16.81
z = 0.797
z = 0.8 (approx)