Reflection and Refraction — Question 10

Question 10

Boundary phase matching fixes frequency and tangential wave number

Two plane waves meet a stationary flat interface at y=0. Their phases along the surface are $k_{x1}x-\omega_1t+\phi_1$ and $k_{x2}x-\omega_2t+\phi_2$. Assume a uniform interface and a nonzero transmitted wave.

1. Explain mathematically why equality of boundary phase dependence for all x and t requires equal frequencies and tangential wave numbers.

2. Use the result and k=n omega/c to derive Snells law and the wavelength ratio.

Question 10 - Solution

Boundary phase matching fixes frequency and tangential wave number

(a) Method and calculation

$$\partial_t(\phi_1^{\mathrm{surf}}-\phi_2^{\mathrm{surf}})=0\Rightarrow\boxed{\omega_1=\omega_2}.$$$$\partial_x(\phi_1^{\mathrm{surf}}-\phi_2^{\mathrm{surf}})=0\Rightarrow\boxed{k_{x1}=k_{x2}}.$$

(b) Interpretation and further calculation

$$k_1\sin\theta_1=k_2\sin\theta_2\Rightarrow\boxed{n_1\sin\theta_1=n_2\sin\theta_2}.$$$$\lambda_j=2\pi/k_j\Rightarrow\boxed{\lambda_2/\lambda_1=n_1/n_2}.$$

Physical check

A constant boundary phase offset can be absorbed into an amplitude coefficient; different x or t dependence cannot match a uniform stationary boundary.