Reflection and Refraction

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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 denote the phase velocity in wave impedance :

$\displaystyle c_i = \frac{\omega}{k_i}, \quad i=1,2$

In impedance

, 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,

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

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