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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_1^+)}
\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).


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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA