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Reflection and Refraction

The first equality in Eq.$ \,$(H.58) 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 = c_2^2 k_2^2 = c_2^2 \left[(k^+_{2x})^2 + (k^+_{2y})^2\right]
$

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

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

We have derived

$\displaystyle \zbox {k^+_{2x}
= \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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Previous: Plane-Wave Scattering at an Angle
Next: Evanescent Wave due to Total Internal Reflection

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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