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

Now let $H$ be perpendicular to the plane of incidence and $E$ lie in it (Figure 6.4.10). The incident wave travels along $a_{n}^{i} = cos θ_{i} a_{x} + sin θ_{i} a_{y}$ , and its electric field is perpendicular to that direction within the $x y$ -plane:

E^{i} (x, y) H^{i} (x, y) = E_{0}^{i} (- 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})}, = \frac{E _{0}^{i}}{η _{1}} e^{- j β_{1} (x c o s θ_{i} + y s i n θ_{i})} a_{z}, (6.4.24)

E^{r} (x, y) H^{r} (x, y) = E_{0}^{r} (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})}, = - \frac{E _{0}^{r}}{η _{1}} e^{- j β_{1} (- x c o s θ_{r} + y s i n θ_{r})} a_{z}, (6.4.25)

E^{t} (x, y) H^{t} (x, y) = E_{0}^{t} (- 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})}, = \frac{E _{0}^{t}}{η _{2}} e^{- j β_{2} (x c o s θ_{t} + y s i n θ_{t})} a_{z} . (6.4.26)

The reference directions of the electric fields are chosen so that the magnetic fields point along $\pm a_{z}$ and each wave obeys $a_{n} \times E = η H$ . For the incident wave $(cos θ_{i} a_{x} + sin θ_{i} a_{y}) \times (- sin θ_{i} a_{x} + cos θ_{i} a_{y}) = (cos^{2} θ_{i} + sin^{2} θ_{i}) a_{z} = a_{z}$ . For the reflected wave $(- cos θ_{r} a_{x} + sin θ_{r} a_{y}) \times (sin θ_{r} a_{x} + cos θ_{r} a_{y}) = - a_{z}$ , which explains the minus sign in $H^{r}$ . Both reference vectors of $E^{i}$ and $E^{r}$ have the same tangential component $+ cos θ a_{y}$ , and at normal incidence both become $a_{y}$ .

Parallel (P) polarisation. The magnetic fields point out of (\odot) or into (\otimes) the page and the electric fields lie in the plane of incidence. With these reference directions the tangential parts of \vec E^i and \vec E^r point the same way, along +\vec a_y. — **Figure 6.4.10.** Parallel (P) polarisation. The magnetic fields point out of ( $⊙$ ) or into ( $\otimes$ ) the page and the electric fields lie in the plane of incidence. With these reference directions the tangential parts of $E^{i}$ and $E^{r}$ point the same way, along $+ a_{y}$ .

Common mistake · Comparing

Γ_{∥}

between books

The reference direction of $E^{r}$ in P polarisation is a convention. With the choice made here, $Γ_{∥} = Γ_{⊥}$ at normal incidence. Some books reverse $E^{r}$ (so that $H^{r}$ points along $+ a_{z}$ ), and their $Γ_{∥}$ has the opposite sign. Both are correct with their own figures. Before taking a formula or a numerical value of $Γ_{∥}$ from another source, check its reference directions. Exercise 6.4.5 and Exercise 6.4.6 show where the sign matters.

Boundary conditions

The tangential components at $x = 0$ are $E_{y}$ and $H_{z}$ . Phase matching works exactly as for S polarisation, so Snell's law Eq. (6.4.10) holds and the common factor $e^{- j β_{1} y s i n θ_{i}}$ cancels. With $θ_{r} = θ_{i}$ ,

E_{y} : H_{z} : E_{0}^{i} cos θ_{i} + E_{0}^{r} cos θ_{i} = E_{0}^{t} cos θ_{t}, \frac{E _{0}^{i}}{η _{1}} - \frac{E _{0}^{r}}{η _{1}} = \frac{E _{0}^{t}}{η _{2}} . (6.4.27)

With $Γ_{∥} = E_{0}^{r} / E_{0}^{i}$ and $τ_{∥} = E_{0}^{t} / E_{0}^{i}$ these read

(1 + Γ_{∥}) cos θ_{i} 1 - Γ_{∥} = τ_{∥} cos θ_{t}, = \frac{η _{1}}{η _{2}} τ_{∥} . (6.4.28)

Note that $1 + Γ_{∥} \neq = τ_{∥}$ : the electric field is not entirely tangential, and only its $y$ -part, $cos θ$ times the amplitude, has to be continuous. From the second equation $τ_{∥} = (η_{2} / η_{1}) (1 - Γ_{∥})$ . Substituting into the first, $(1 + Γ_{∥}) η_{1} cos θ_{i} = (1 - Γ_{∥}) η_{2} cos θ_{t}$ , and solving for $Γ_{∥}$ :

Fresnel coefficients, parallel (P) polarisation

Γ_{∥} τ_{∥} = \frac{η _{2} cos θ _{t} - η _{1} cos θ _{i}}{η _{2} cos θ _{t} + η _{1} cos θ _{i}}, = \frac{2 η _{2} cos θ _{i}}{η _{2} cos θ _{t} + η _{1} cos θ _{i}}, (6.4.29)

with the reference directions of Figure 6.4.10. The power fractions are given by Eq. (6.4.15), $R_{∥} = ∣ Γ_{∥} ∣^{2}$ and $T_{∥} = \frac{η _{1} c o s θ _{t}}{η _{2} c o s θ _{i}} ∣ τ_{∥} ∣^{2}$ .

In plain words

For P polarisation each medium presents the impedance $η cos θ$ , the ratio of the tangential $E$ to the tangential $H$ , and the formula is again the normal-incidence one written with these impedances. Multiplying the two equations Eq. (6.4.28) gives $(1 - Γ_{∥}^{2}) cos θ_{i} = (η_{1} / η_{2}) τ_{∥}^{2} cos θ_{t}$ , that is $R + T = 1$ .

If medium 2 is a perfect conductor, $η_{2} = 0$ and the coefficients become $Γ_{∥} = - 1$ and $τ_{∥} = 0$ , as for S. At normal incidence $Γ_{∥} = Γ_{⊥}$ . Snell's law, total reflection and the evanescent wave apply to P polarisation unchanged. Beyond $θ_{c}$ , with $cos θ_{t} = - j ∣ cos θ_{t} ∣$ ,

Γ_{∥} ∣ Γ_{∥} ∣ = - \frac{η _{1} cos θ _{i} + j η _{2} ∣ cos θ _{t} ∣}{η _{1} cos θ _{i} - j η _{2} ∣ cos θ _{t} ∣}, = 1.

Figure 6.4.11 shows the P coefficients for air on glass. $Γ_{∥}$ starts at $- 0.2$ , as $Γ_{⊥}$ does, but rises to zero at $56. 3^{\circ}$ , changes sign and tends to $+ 1$ at grazing incidence. The angle at which it vanishes is the subject of the next subsection.

The P coefficients for air on glass (n_1=1, n_2=1.5), from Eq. (6.4.29). \Gamma_\parallel passes through zero at the Brewster angle 56.3^\circ, where \tau_\parallel=\eta_2/\eta_1=1/1.5. — **Figure 6.4.11.** The P coefficients for air on glass ( $n_{1} = 1$ , $n_{2} = 1.5$ ), from Eq. (6.4.29). $Γ_{∥}$ passes through zero at the Brewster angle $56. 3^{\circ}$ , where $τ_{∥} = η_{2} / η_{1} = 1/1.5$ .

6.4.8 Parallel (P) polarisation

Boundary conditions