# 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.
* The acoustic pressure must be continuous at the boundary.
* 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.
$\mathbf {R} =={\frac {\cos \theta _{i}\zeta -{\sqrt {1-{\Big{\frac {c_{2}}{c_{1}}}\sin \theta _{i}{\Big )}^{2}}}}{\cos \theta _{i}\zeta +{\sqrt {1-{\Big{\frac {c_{2}}{c_{1}}}\sin \theta _{i}{\Big )}^{2}}}}}$
$\mathbf {T} ={\frac {2\cos \theta _{i}\zeta }{\cos \theta _{i}\zeta +{\sqrt {1-{\Big{\frac {c_{2}}{c_{1}}}\sin \theta _{i}{\Big )}^{2}}}}}$
$\zeta ={\frac {\sqrt {1-{\Big{\frac {c_{2}}{c_{1}}}\sin \theta _{i}{\Big )}^{2}}}{\cos \theta _{i}}}{\Big{\frac {1+\mathbf {R} }{1-\mathbf {R} }}{\Big )}={\frac {\sqrt {1-{\Big{\frac {c_{2}}{c_{1}}}\sin \theta _{i}{\Big )}^{2}}}{\cos \theta _{i}}}{\Big{\frac {\mathbf {T} }{2-\mathbf {T} }}{\Big )}.$