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

Let the incident wave be S-polarised. With Eq. (6.4.1) and Eq. (6.4.2) the incident field is

E^{i} (x, y) H^{i} (x, y) = E_{0}^{i} e^{- j β_{1} (x c o s θ_{i} + y s i n θ_{i})} a_{z}, = \frac{E _{0}^{i}}{η _{1}} (sin θ_{i} a_{x} - cos θ_{i} a_{y}) \times e^{- j β_{1} (x c o s θ_{i} + y s i n θ_{i})} . (6.4.3)

The magnetic field follows from $H = \frac{1}{η} a_{n} \times E$ with

(cos θ_{i} a_{x} + sin θ_{i} a_{y}) \times a_{z} = cos θ_{i} (a_{x} \times a_{z}) + sin θ_{i} (a_{y} \times a_{z}) = - cos θ_{i} a_{y} + sin θ_{i} a_{x} .

The reflected wave travels back into medium 1 at an angle $θ_{r}$ which we do not yet know. Its direction is $a_{n}^{r} = - cos θ_{r} a_{x} + sin θ_{r} a_{y}$ : the $x$ -component is reversed, the wave keeps moving up the interface. With $(- cos θ_{r} a_{x} + sin θ_{r} a_{y}) \times a_{z} = sin θ_{r} a_{x} + cos θ_{r} a_{y}$ ,

E^{r} (x, y) H^{r} (x, y) = E_{0}^{r} e^{- j β_{1} (- x c o s θ_{r} + y s i n θ_{r})} a_{z}, = \frac{E _{0}^{r}}{η _{1}} (sin θ_{r} a_{x} + cos θ_{r} a_{y}) \times e^{- j β_{1} (- x c o s θ_{r} + y s i n θ_{r})} . (6.4.4)

The transmitted wave repeats the incident one with the constants of medium 2 and the angle $θ_{t}$ :

E^{t} (x, y) H^{t} (x, y) = E_{0}^{t} e^{- j β_{2} (x c o s θ_{t} + y s i n θ_{t})} a_{z}, = \frac{E _{0}^{t}}{η _{2}} (sin θ_{t} a_{x} - cos θ_{t} a_{y}) \times e^{- j β_{2} (x c o s θ_{t} + y s i n θ_{t})} . (6.4.5)

Figure 6.4.2 shows the reference directions. At normal incidence these three waves reduce to those of Section 6.3, with $E$ along $a_{z}$ instead of $a_{y}$ .

Perpendicular (S) polarisation. All three electric fields point out of the page, along \vec a_z, and the magnetic fields lie in the plane of incidence, each forming a right-handed triad with \vec E and its own direction of travel. The arrows are the reference directions for which E_0^r and E_0^t are positive. — **Figure 6.4.2.** Perpendicular (S) polarisation. All three electric fields point out of the page, along $a_{z}$ , and the magnetic fields lie in the plane of incidence, each forming a right-handed triad with $E$ and its own direction of travel. The arrows are the reference directions for which $E_{0}^{r}$ and $E_{0}^{t}$ are positive.

Compared with normal incidence the problem has more unknowns. Besides the amplitudes $E_{0}^{r}$ and $E_{0}^{t}$ we do not know the angles $θ_{r}$ and $θ_{t}$ . Two boundary conditions cannot fix four numbers, unless they hold in a stronger sense than before. They do: the conditions must be satisfied not at one point but at every point $y$ of the interface, and this is what fixes the angles.

6.4.2 Perpendicular (S) polarisation: the three waves