Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

There are no surface currents on the interface between two dielectrics, so the tangential components of $E$ and of $H$ are continuous at $x = 0$ . For the S waves the tangential components are $E_{z}$ and $H_{y}$ :

E_{z}^{i} (0, y) + E_{z}^{r} (0, y) H_{y}^{i} (0, y) + H_{y}^{r} (0, y) = E_{z}^{t} (0, y), = H_{y}^{t} (0, y) for all y . (6.4.6)

Inserting the waves Eq. (6.4.3) to Eq. (6.4.5) at $x = 0$ gives

E_{0}^{i} e^{- j β_{1} y s i n θ_{i}} + E_{0}^{r} e^{- j β_{1} y s i n θ_{r}} = E_{0}^{t} e^{- j β_{2} y s i n θ_{t}}, (6.4.7)

- \frac{E _{0}^{i}}{η _{1}} cos θ_{i} e^{- j β_{1} y s i n θ_{i}} + \frac{E _{0}^{r}}{η _{1}} cos θ_{r} e^{- j β_{1} y s i n θ_{r}} = - \frac{E _{0}^{t}}{η _{2}} cos θ_{t} e^{- j β_{2} y s i n θ_{t}} . (6.4.8)

Each equation is a sum of three exponentials in $y$ . If their exponents were different, the three terms would oscillate along the interface at different rates and could cancel at a few isolated points at best, never at every $y$ . The equations can hold for all $y$ only if the three exponents are equal:

β_{1} y sin θ_{i} = β_{1} y sin θ_{r} = β_{2} y sin θ_{t} . (6.4.9)

This is called phase matching: the three waves must have the same phase variation along the interface.

Laws of reflection and refraction (Snell's law)

θ_{r} = θ_{i}, β_{1} sin θ_{i} = β_{2} sin θ_{t} ⟺ \frac{sin θ _{t}}{sin θ _{i}} = \frac{β _{1}}{β _{2}} . (6.4.10)

With the refractive index $n = c μ ε = μ_{r} ε_{r}$ , so that $β = n ω / c$ , Snell's law reads

n_{1} sin θ_{i} = n_{2} sin θ_{t}, (6.4.11)

and for nonmagnetic media $n = ε_{r}$ . Both laws hold for any polarisation.

In plain words

The angle of reflection equals the angle of incidence. A wave that enters a denser medium ( $n_{2} > n_{1}$ ) is bent towards the normal, a wave that enters a less dense medium is bent away from it. Both statements follow from one requirement: the crests of all three waves must meet along the interface with the same spacing.

Figure 6.4.3 shows phase matching. Successive crests of the incident wave are a wavelength $λ_{1}$ apart along its ray, and they cut the interface at points spaced $λ_{1} / sin θ_{i}$ apart. The crests of the transmitted wave must start from the same points, so their spacing along the interface, $λ_{2} / sin θ_{t}$ , is the same. Since $λ = 2 π / β$ , this is exactly Eq. (6.4.10). The spacing $λ_{1} / sin θ_{i}$ is called the trace wavelength of the wave on the interface. It is always longer than both $λ_{1}$ and $λ_{2}$ .

Phase matching. The lines are the crests (planes of constant phase) of the incident and the transmitted wave, drawn for \lambda_2=0.625\,\lambda_1 and \theta_i=40^\circ, which give \theta_t=23.7^\circ. Each incident crest continues as a transmitted crest, so both sets cut the interface with the same spacing, the trace wavelength \lambda_1/\sin\theta_i=\lambda_2/\sin\theta_t. — **Figure 6.4.3.** Phase matching. The lines are the crests (planes of constant phase) of the incident and the transmitted wave, drawn for $λ_{2} = 0.625 λ_{1}$ and $θ_{i} = 4 0^{\circ}$ , which give $θ_{t} = 23. 7^{\circ}$ . Each incident crest continues as a transmitted crest, so both sets cut the interface with the same spacing, the trace wavelength $λ_{1} / sin θ_{i} = λ_{2} / sin θ_{t}$ .

Quick check

Light in air meets a calm water surface ( $n = 1.33$ ) at $3 0^{\circ}$ from the normal. At which angle does it travel in the water, and at which angle does the reflected light leave?

$sin θ_{t} = sin 3 0^{\circ} /1.33 = 0.376$ , so $θ_{t} = 22. 1^{\circ}$ : the light is bent towards the normal. The reflected light leaves at $θ_{r} = θ_{i} = 3 0^{\circ}$ on the other side of the normal.

Common mistake · Measuring the angles from the interface

Snell's law $n_{1} sin θ_{i} = n_{2} sin θ_{t}$ uses angles measured from the normal. With angles measured from the interface (the grazing angle $ψ = 9 0^{\circ} - θ_{i}$ of radio engineers, or the angle $ϕ$ between a ray and the axis of an optical fibre), the sines become cosines. Problem 6.4.3 and Exercise 6.4.6 use both angles: convert first, then apply the formulas.

6.4.3 The boundary conditions and Snell's law