Reflection and Refraction

The first equality in Eq.(C.56) implies that the angle of incidence equals angle of reflection:

$$\zbox {\theta_1^+=\theta_1^-} % \isdef\theta_1}$$

We now wish to find the wavenumber in medium 2. Let $c_i$ denote the phase velocity in wave impedance $R_i$ :

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

In impedance $R_2$ , we have in particular

$$\omega^2 \eqsp c_2^2 k_2^2 \eqsp c_2^2 \left[(k^+_{2x})^2 + (k^+_{2y})^2\right].$$

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

$$k^+_{2x} \eqsp \sqrt{\frac{\omega^2}{c_2^2} - (k^+_{2y})^2} \eqsp \sqrt{\frac{\omega^2}{c_2^2} - k_2^2\sin^2(\theta_2^+)}.$$

Since $k_1\sin(\theta_1^+)=k_2\sin(\theta_2^+)$ from above,

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

We have thus derived

$$\zbox {k^+_{2x} \eqsp \frac{\omega}{c_2}\sqrt{1 - \frac{c_2^2}{c_1^2}\sin^2(\theta_1^+)}.}$$

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:

$$n \isdef \frac{c_2}{c_1}$$

and the relation $k_1\sin(\theta_1^+)=k_2\sin(\theta_2^+)$ is called Snell's Law (of refraction).