Problem 6.4.1: Oblique incidence on a dielectric, both polarisations

Problem 6.4.1

Oblique incidence on a dielectric, both polarisations

A uniform plane wave with electric-field amplitude $E_{0} = 10 V/m$ travels in air and is incident at $θ_{i} = 6 0^{\circ}$ on the plane surface of a lossless nonmagnetic dielectric with $ε_{r} = 3$ .

Find the angle of refraction $θ_{t}$ .
For perpendicular (S) polarisation find $Γ_{⊥}$ , $τ_{⊥}$ and the fractions of the incident power reflected and transmitted.
Repeat (b) for parallel (P) polarisation.
The incident wave is linearly polarised at $4 5^{\circ}$ to the plane of incidence. Describe the polarisation of the reflected and transmitted waves and find the fraction of the power reflected.

Try it yourself first.

Strategy

Snell's law first. Then divide the numerator and the denominator of each Fresnel coefficient by $η_{1}$ , so that only $η_{2} / η_{1} = 1/ 3$ and the two cosines appear. At oblique incidence the transmitted power fraction is the ratio of the normal components of the Poynting vectors, $T = \frac{η _{1} c o s θ _{t}}{η _{2} c o s θ _{i}} ∣ τ ∣^{2}$ , Eq. (6.4.15). In (d) the S and P parts of the field are treated separately and recombined.

$n_{1} = 1$ , $n_{2} = 3$ :

sin θ_{t} = \frac{n _{1}}{n _{2}} sin θ_{i} = \frac{sin 6 0 ^{\circ}}{3} = \frac{1}{2}, θ_{t} = 3 0^{\circ} .

With $cos θ_{i} = \frac{1}{2}$ , $cos θ_{t} = \frac{3}{2}$ and $η_{2} / η_{1} = 1/ 3$ :

Γ_{⊥} τ_{⊥} = \frac{\frac{1}{3} \cdot \frac{1}{2} - \frac{3}{2}}{\frac{1}{3} \cdot \frac{1}{2} + \frac{3}{2}} = \frac{0.2887 - 0.8660}{1.1547} = - 0.5, = 1 + Γ_{⊥} = 0.5,

R_{⊥} T_{⊥} = ∣ Γ_{⊥} ∣^{2} = 0.25, = \frac{η _{1} cos θ _{t}}{η _{2} cos θ _{i}} τ_{⊥}^{2} = 3 \cdot \frac{3 /2}{1/2} \cdot 0.25 = 0.75.

Γ_{∥} τ_{∥} R_{∥} = \frac{\frac{1}{3} \cdot \frac{3}{2} - \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{2} + \frac{1}{2}} = 0, = \frac{2 \cdot \frac{1}{3} \cdot \frac{1}{2}}{1} = \frac{1}{3} = 0.577, = 0, T_{∥} = 3 \cdot \frac{1}{3} = 1.

$6 0^{\circ}$ is the Brewster angle: $tan θ_{B} = n_{2} / n_{1} = 3$ , and indeed $θ_{B} + θ_{t} = 9 0^{\circ}$ .

At $4 5^{\circ}$ to the plane of incidence $E_{⊥} = E_{∥} = 10/ 2 = 7.07 V/m$ , each carrying half of the incident power. The reflected wave has $∣ E_{⊥}^{r} ∣ = 0.5 \times 7.07 = 3.54 V/m$ and $E_{∥}^{r} = 0$ : it is linearly polarised perpendicular to the plane of incidence. The reflected fraction is $\frac{1}{2} R_{⊥} + \frac{1}{2} R_{∥} = \frac{1}{2} \times 0.25 = 12.5 %$ . The transmitted wave has $E_{⊥}^{t} = 0.5 \times 7.07 = 3.54 V/m$ and $E_{∥}^{t} = 0.577 \times 7.07 = 4.08 V/m$ , both real and in phase: it is linearly polarised at $arctan (3.54/4.08) = 40. 9^{\circ}$ to the plane of incidence.

Answer

θ_{t} = 3 0^{\circ}, Γ_{⊥} = - 0.5, τ_{⊥} = 0.5, R_{⊥} = 25 %, Γ_{∥} = 0, τ_{∥} = 0.577, R_{∥} = 0, (d) 12.5 % reflected, pure S

Check

$R + T = 1$ for each polarisation, and the transmitted fraction in (d), $\frac{1}{2} (0.75 + 1) = 87.5 %$ , completes the reflected $12.5 %$ . The angle forms of Home exercise 6.4.2 give the same numbers: $Γ_{⊥} = - sin 3 0^{\circ} / sin 9 0^{\circ} = - 0.5$ and $Γ_{∥} = - tan 3 0^{\circ} / tan 9 0^{\circ} = 0$ .

(b) Electric-field vectors, to scale. — (a) Reflectances of the surface.

What the graphs say

At $θ_{B}$ any incident polarisation leaves the surface as pure S: polarisation by reflection. The transmitted field is tilted towards P ( $40. 9^{\circ}$ instead of $4 5^{\circ}$ ) because all of the P power but only $75 %$ of the S power gets through.