Plane-Wave Scattering at an Angle

Figure C.18 shows the more general situation (as compared to Fig.C.15) of a sinusoidal traveling plane wave encountering an impedance discontinuity at some arbitrary angle of incidence, as indicated by the vector wavenumber $ \underline{k}_1^+$. The mathematical details of general sinusoidal plane waves in air and vector wavenumber are reviewed in §B.8.1.

Figure C.18: Sinusoidal plane wave scattering at an impedance discontinuity--oblique angle of incidence $ \theta _1^+$.
\includegraphics{eps/planewavescatangle}

At the boundary between impedance $ R_1$ and $ R_2$, we have, by continuity of pressure,

\begin{eqnarray*}
k_1\sin(\theta_1^+)
&=&k_1\sin(\theta_1^-)
\;=\;k_2\sin(\theta_2^+)
\end{eqnarray*}

as we will now derive.

Let the impedance change be in the $ \underline{x}=(0,y,z)$ plane. Thus, the impedance is $ R_1$ for $ x\le0$ and $ R_2$ for $ x>0$. There are three plane waves to consider:

  • The incident plane wave with wave vector $ \underline{k}_1^+$
  • The reflected plane wave with wave vector $ \underline{k}_1^-$
  • The transmitted plane wave with wave vector $ \underline{k}_2^+$
By continuity, the waves must agree on boundary plane:

$\displaystyle \left<\underline{k}_1^+,\underline{r}\right> = \left<\underline{k}_1^-,\underline{r}\right> = \left<\underline{k}_2^+,\underline{r}\right>
$

where $ \underline{r}=(0,y,z)$ denotes any vector in the boundary plane. Thus, at $ x=0$ we have

$\displaystyle k_{1y}^+\,y + k_{1z}^+\,z
= k_{1y}^-\,y + k_{1z}^-\,z = k_{2y}^+\,y + k_{2z}^+\,z.
$

If the incident wave is constant along $ z$, then $ k_{1z}^+=0$, requiring $ k_{1z}^- = k_{2z}^+ = 0$, leaving

$\displaystyle k_{1y}^+\,y = k_{1y}^-\,y =k_{2y}^+\,y
$

or

$\displaystyle \zbox {k_1\sin(\theta_1^+) =k_1\sin(\theta_1^-) =k_2\sin(\theta_2^+)} \protect$ (C.56)

where $ \theta$ is defined as zero when traveling in the direction of positive $ x$ for the incident ( $ \underline{k}_1^+$) and transmitted ( $ \underline{k}_2^+$) wave vector, and along negative $ x$ for the reflected ( $ \underline{k}_1^-$) wave vector (see Fig.C.18).

Reflection and Refraction

The first equality in Eq.$ \,$(C.56) implies that the angle of incidence equals angle of reflection:

$\displaystyle \zbox {\theta_1^+=\theta_1^-} % \isdef\theta_1}
$

We now wish to find the wavenumber in medium 2. Let $ c_i$ denote the phase velocity in wave impedance $ R_i$:

$\displaystyle c_i = \frac{\omega}{k_i}, \quad i=1,2
$

In impedance $ R_2$, we have in particular

$\displaystyle \omega^2 \eqsp c_2^2 k_2^2 \eqsp c_2^2 \left[(k^+_{2x})^2 + (k^+_{2y})^2\right].
$

Solving for $ k^+_{2x}$ gives

$\displaystyle k^+_{2x} \eqsp \sqrt{\frac{\omega^2}{c_2^2} - (k^+_{2y})^2}
\eqsp \sqrt{\frac{\omega^2}{c_2^2} - k_2^2\sin^2(\theta_2^+)}.
$

Since $ k_1\sin(\theta_1^+)=k_2\sin(\theta_2^+)$ from above,

$\displaystyle k^+_{2x}
\eqsp \sqrt{\frac{\omega^2}{c_2^2} - k_1^2\sin^2(\theta...
...\eqsp
\sqrt{\frac{\omega^2}{c_2^2}-\frac{\omega^2}{c_1^2}\sin^2(\theta_1^+)}.
$

We have thus derived

$\displaystyle \zbox {k^+_{2x}
\eqsp \frac{\omega}{c_2}\sqrt{1 - \frac{c_2^2}{c_1^2}\sin^2(\theta_1^+)}.}
$

We earlier established $ k^+_{2y} = k^+_{1y}$ (for agreement along the boundary, by pressure continuity). This describes the refraction of the plane wave as it passes through the impedance-change boundary. The refraction angle depends on ratio of phase velocities $ c_2/c_1$. This ratio is often called the index of refraction:

$\displaystyle n \isdef \frac{c_2}{c_1}
$

and the relation $ k_1\sin(\theta_1^+)=k_2\sin(\theta_2^+)$ is called Snell's Law (of refraction).


Evanescent Wave due to Total Internal Reflection

Note that if $ c_1 < c_2 \vert\sin(\theta_1^+)\vert$, the horizontal component of the wavenumber in medium 2 becomes imaginary. In this case, the wave in medium 2 is said to be evanescent, and the wave in medium 1 undergoes total internal reflection (no power travels from medium 1 to medium 2). The evanescent-wave amplitude decays exponentially to the right and oscillates ``in place'' (like a standing wave). ``Tunneling'' is possible given a medium 3 beyond medium 2 in which wave propagation resumes.

To show explicitly the exponential decay and in-place oscillation in an evanescent wave, express the imaginary wavenumber as $ k_x\isdef
-j\kappa_x$. Then we have

\begin{eqnarray*}
p(t,\underline{x}) &=&
\cos\left(\omega t - \underline{k}^T\...
...-k_x x}\right\}}}\\ [5pt]
&=& e^{-k_x x} \cos(\omega t - k_y y).
\end{eqnarray*}

Thus, an imaginary wavenumber corresponds to an exponentially decaying evanescent wave. Note that the time dependence (cosine term) applies to all points to the right of the boundary. Since evanescent waves do not really ``propagate,'' it is perhaps better to speak of an ``evanescent acoustic field'' or ``evanescent standing wave'' instead of ``evanescent waves''.

For more on the physics of evanescent waves and tunneling, see [295].


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Longitudinal Waves in Rods
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Plane-Wave Scattering