Reflection and Refraction

Above we derived

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

The first equality implies

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

(Angle of incidence equals angle of reflection)

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

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

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

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

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

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

We have derived

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

We earlier established $k^+_{2y} = k^+_{1y}$.

• This describes the refraction of the plane wave as it passes through the impedance-change boundary.
• Refraction angle depends on ratio of phase velocities $c_2/c_1$.
• This ratio is often called the index of refraction:
  $$n \mathrel{\stackrel{\mathrm{\Delta}}{=}}\frac{c_2}{c_1}$$
  and the relation $k_1\sin(\theta_1^+)=k_2\sin(\theta_2^+)$ is called Snell's Law (of refraction).