Engineering Acoustics/Reflection, transmission and refraction of planar waves

Two-dimensional planar waves
Two-dimensional planar pressure waves can be described in Cartesian coordinates by decomposing the wave number into x and  y components,

$$           \mathbf{p}(x,y,t)=\mathbf{P}e^{j(\omega t -K_x x-K_y y)}. $$

Substituting into the general wave equation yields:

$$           \nabla^2 \mathbf{p}- \frac{1}{c_o^2} \frac{\partial^2 \mathbf{p}}{\partial t^2}=0, $$

$$       \mathbf{P}(-K_x^2-K_y^2)+\frac{\omega^2}{c_o^2}\mathbf{P}=0, $$

$$       K=\frac{\omega}{c_o}=\sqrt{K_x^2+K_y^2}. $$

The wave number becomes a vector quantity and may be expressed using the directional cosines,

$$       \vec{K}=K_x \boldsymbol{\hat{\imath}} + K_y \boldsymbol{\hat{\jmath}} = K \cos(\alpha) \boldsymbol{\hat{\imath}} + K \cos(\beta) \boldsymbol{\hat{\jmath}}. $$

Obliquely incident planar waves
Consider an obliquely incident planar wave in medium 1 which approaches the boundary at an angle $$       \theta_i $$ with respect to the normal. Part of the wave is reflected back into medium 1 at an angle $$       \theta_r $$ and the remaining part is transmitted to medium 2 at an angle $$       \theta_t $$.

$$               \mathbf{p_1}=\mathbf{P_i}e^{j(\omega t - \cos\theta_i K_1 x - \sin\theta_i K_1 y)} + \mathbf{P_r}e^{j(\omega t + \cos\theta_r K_1 x - \sin\theta_r K_1 y)}

$$

$$               \mathbf{p_2}=\mathbf{P_t}e^{j(\omega t - \cos\theta_t K_2 x - \sin\theta_t K_2 y)}

$$ Notice that the wave frequency does not change across the boundary, however the specific acoustic impedance does change from medium 1 to medium 2. The propagation speed  is different in each medium, so the wave number changes across the boundary. There are two boundary conditions to be satisfied.


 * 1) The acoustic pressure must be continuous at the boundary.
 * 2) The particle velocity component normal to the boundary must be continuous at the boundary.

Imposition of the first boundary condition yields

$$           \mathbf{p_1}(x=0)=\mathbf{p_2}(x=0),

$$

$$           \mathbf{P_i}e^{-j \sin\theta_i K_1 y} + \mathbf{P_r} e^{-j \sin\theta_r K_1 y}= \mathbf{P_t} e^{-j \sin\theta_t K_2 y}. $$

For continuity to hold, the exponents must be all equal to each other

$$           K_1 \sin\theta_i =K_1 \sin\theta_r=K_2 \sin\theta_t.

$$

This has two implications. First, the angle of incident waves is equal to the angle of reflected waves,

$$           \sin\theta_i = \sin\theta_r

$$

and second, Snell's law is recovered,

$$           \frac{\sin\theta_i}{c_1}=\frac{\sin\theta_t}{c_2}.

$$

The first boundary condition can be expressed using the pressure reflection and transmission coefficients

$$           1+\mathbf{R}=\mathbf{T}.

$$

Imposition of the second boundary condition yields

$$           \mathbf{u_{1x}}(x=0)=\mathbf{u_{2x}}(x=0), $$

$$           \mathbf{u_i}\cos\theta_i+ \mathbf{u_r}\cos\theta_r= \mathbf{u_t}\cos\theta_t. $$

Using the specific acoustic impedance definition yields

$$           \frac{\mathbf{P_i}}{r_1}\cos\theta_i- \frac{\mathbf{P_r}}{r_1}\cos\theta_r= \frac{\mathbf{P_t}}{r_2}\cos\theta_t. $$

Using the reflection coefficient, the transmission coefficient and the acoustic impedance ratio leads to

$$           1- \mathbf{R}= \frac{\cos\theta_t}{\cos\theta_i}\frac{\mathbf{T}}{\zeta}. $$

Solving for the pressure reflection coefficient yields:

$$           \mathbf{R}=\mathbf{T}-1=\frac{\frac{\cos\theta_i}{\cos\theta_t}\zeta-1}{\frac{\cos\theta_i}{\cos\theta_t}\zeta+1}=\frac{\frac{r_2}{\cos\theta_t}-\frac{r_1}{\cos\theta_i}}{\frac{r_2}{\cos\theta_t}+\frac{r_1}{\cos\theta_i}}.

$$

Solving for the pressure transmission coefficient yields:

$$           \mathbf{T}=\mathbf{R}+1=\frac{2 \frac{\cos\theta_i}{\cos\theta_t}\zeta}{\frac{\cos\theta_i}{\cos\theta_t}\zeta +1}=\frac{2\frac{r_2}{\cos\theta_t}}{\frac{r_2}{\cos\theta_t}+\frac{r_1}{\cos\theta_i}}.

$$

Solving for the specific acoustic impedance ratio yields

$$           \zeta = \frac{\cos\theta_t}{\cos\theta_i}\Big(\frac{1+\mathbf{R}}{1-\mathbf{R}}\Big) = \frac{\cos\theta_t}{\cos\theta_i}\Big(\frac{\mathbf{T}}{2-\mathbf{T}}\Big). $$

Rayleigh reflection coefficient
The Rayleigh reflection coefficient relates the angle of incidence from Snell's law to the angle of transmission in the equations for $$       \mathbf{R} $$, $$       \mathbf{T} $$ and $$       \zeta $$. From the trigonometric identity,

$$           \cos^2\theta_t+\sin^2\theta_t=1

$$

and using Snell's law,

$$           \cos\theta_t=\sqrt{1-\Big( \frac{c_2}{c_1}\sin\theta_i \Big)^2}.

$$

Notice that for the angle of transmission to be real,

$$           c_2<\frac{c_1}{\sin\theta_i}

$$

must be met. Thus, there is a critical angle of incidence such that

$$           \sin{\theta_c}=\frac{c_1}{c_2}.

$$

The Rayleigh reflection coefficient are substituted back into the equations for $$       \mathbf{R} $$, $$       \mathbf{T} $$ and $$       \zeta $$ to obtain expression only in term of impedance and angle of incidence.

$$           \mathbf{R}==\frac{\cos\theta_i\zeta-\sqrt{1-\Big( \frac{c_2}{c_1}\sin\theta_i \Big)^2}}{\cos\theta_i\zeta+\sqrt{1-\Big( \frac{c_2}{c_1}\sin\theta_i \Big)^2}}

$$

$$           \mathbf{T}=\frac{2 \cos\theta_i\zeta}{\cos\theta_i\zeta +\sqrt{1-\Big( \frac{c_2}{c_1}\sin\theta_i \Big)^2}}

$$

$$           \zeta = \frac{\sqrt{1-\Big( \frac{c_2}{c_1}\sin\theta_i \Big)^2}}{\cos\theta_i}\Big(\frac{1+\mathbf{R}}{1-\mathbf{R}}\Big) = \frac{\sqrt{1-\Big( \frac{c_2}{c_1}\sin\theta_i \Big)^2}}{\cos\theta_i}\Big(\frac{\mathbf{T}}{2-\mathbf{T}}\Big).

$$