A plane wave is propagating in a uniform medium with propagation constant \((0.6\pi + j\,0.4\pi)\) its wavelength is
5 m
Meaning of the propagation constant. A plane wave in a lossy medium varies as
\(E(z)=E_0 e^{-\gamma z}=E_0 e^{-\alpha z}e^{-j\beta z}\)
where the complex propagation constant is \(\gamma=\alpha+j\beta\). The real part α is the attenuation constant (Np/m), which controls how fast the amplitude decays; the imaginary part β is the phase constant (rad/m), which controls how fast the phase advances with distance — and therefore the wavelength.
Step 1 — pick out β. Comparing with the given \(\gamma = 0.6\pi + j\,0.4\pi\):
\(\alpha = 0.6\pi\ \text{Np/m}, \qquad \beta = 0.4\pi\ \text{rad/m}\)
Step 2 — convert β to wavelength. One wavelength is the distance over which the phase advances by 2π radians:
\(\lambda = \dfrac{2\pi}{\beta}\)
Step 3 — substitute.
\(\lambda = \dfrac{2\pi}{0.4\pi}=\dfrac{2}{0.4}=5\ \text{m}\)
Why α plays no part. The attenuation constant tells you the wave falls to \(e^{-0.6\pi}\approx 0.15\) of its amplitude per metre (about 16 dB/m), but it does not change the spatial periodicity. Only β sets the wavelength. Using 0.6π by mistake would give λ = 3.33 m, which is not among the options — a useful sign that the imaginary part is the one to use.
Related quantities. From β you also get the phase velocity \(v_p=\omega/\beta=f\lambda\); and since \(\alpha\) here is comparable to β, the medium is strongly lossy (in a good dielectric α ≪ β, while in a good conductor α = β).
Hence, the wavelength of the wave is 5 m.
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