Plane Waves at an Impedance Discontinuity

By continuity, waves must agree on boundary plane:

$$\left<\underline{k}_1^+,\underline{r}\right> = \left<\underline{k}_1^-,\underline{r}\right> = \left<\underline{k}_2^+,\underline{r}\right>$$

where $\underline{r}=(0,y,z)$ denotes any vector in the boundary plane. Thus, at $x=0$ we have

$$k_{1y}^+\,y + k_{1z}^+\,z = k_{1y}^-\,y + k_{1z}^-\,z = k_{2y}^+\,y + k_{2z}^+\,z$$

If the incident wave is constant along $z$, then $k_{1z}^+=0$, requiring $k_{1z}^- = k_{2z}^+ = 0$, leaving

$$k_{1y}^+\,y = k_{1y}^-\,y =k_{2y}^+\,y$$

or

$$\zbox{k_1\sin(\theta_1^+) =k_1\sin(\theta_1^-) =k_2\sin(\theta_2^+)}$$