# 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 $\theta_t$.

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

$\mathbf{p_2}=\mathbf{P_t}e^{j(\omega t - \cos\theta_t K_2 x - \sin\theta_t K_2 y)}$
!Reflection and transmission of obliquely incident planar
wave.\|440x440px
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.

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

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

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

