Problem 6.4.6: Circular polarisation at the Brewster angle

Problem 6.4.6

Circular polarisation at the Brewster angle

A circularly polarised light beam in air strikes a calm water surface at the Brewster angle. At optical frequencies water is a lossless dielectric with $n = 4/3$ ( $μ_{r} = 1$ ). Each of the two orthogonal components of the incident $E$ has the amplitude $E_{0}$ .

Find $θ_{B}$ and the angle of refraction.
Find $Γ_{⊥}$ , $τ_{⊥}$ , $Γ_{∥}$ and $τ_{∥}$ .
Describe the polarisation of the reflected and transmitted waves.
What fraction of the incident power is reflected? Compare with normal incidence.

Adapted from: B. M. Notaroš, Electromagnetics, Example 10.25.

Try it yourself first.

Strategy

A circularly polarised wave is an S part and a P part of equal amplitude $E_{0}$ , $9 0^{\circ}$ out of phase. The interface treats the two parts separately with their own Fresnel coefficients, and the reflected and transmitted parts are recombined afterwards. Each part carries half of the incident power, so $R = \frac{1}{2} (∣ Γ_{⊥} ∣^{2} + ∣ Γ_{∥} ∣^{2})$ .

$tan θ_{B} = n_{2} / n_{1} = 4/3$ : $θ_{B} = 53.1 3^{\circ}$ ( $sin θ_{B} = 0.8$ , $cos θ_{B} = 0.6$ ), and $θ_{t} = 9 0^{\circ} - θ_{B} = 36.8 7^{\circ}$ . Snell checks: $sin θ_{t} = \frac{3}{4} \times 0.8 = 0.6$ .

Divide the numerators and denominators by $η_{0}$ , with $η_{1} / η_{0} = 1$ and $η_{2} / η_{0} = 0.75$ :

Γ_{⊥} τ_{⊥} Γ_{∥} τ_{∥} = \frac{0.75 \times 0.6 - 0.8}{0.75 \times 0.6 + 0.8} = \frac{- 0.35}{1.25} = - 0.28, = 1 + Γ_{⊥} = 0.72, = \frac{0.75 \times 0.8 - 0.6}{0.75 \times 0.8 + 0.6} = 0, = \frac{2 \times 0.75 \times 0.6}{1.2} = 0.75.

Reflected: only the S part survives, with amplitude $0.28 E_{0}$ . The reflected light is linearly polarised, with $E$ parallel to the water surface. Transmitted: the P part $0.75 E_{0}$ and the S part $0.72 E_{0}$ , still $9 0^{\circ}$ apart in time: elliptically polarised, with the axial ratio $0.75/0.72 = 1.04$ , almost circular, and the same sense of rotation.

$R = \frac{1}{2} (0.2 8^{2} + 0) = 0.0392$ . At normal incidence $Γ = \frac{0.75 - 1}{0.75 + 1} = - \frac{1}{7}$ for both parts, $R = 1/49 = 2.0 %$ , and the reflected wave is circular with the opposite handedness (it travels the other way).

Answer

θ_{B} = 53.1 3^{\circ}, θ_{t} = 36.8 7^{\circ}, Γ_{⊥} = - 0.28, τ_{⊥} = 0.72, Γ_{∥} = 0, τ_{∥} = 0.75, R = 3.9 %

Check

$T_{⊥} = \frac{c o s θ _{t}}{0.75 c o s θ _{i}} τ_{⊥}^{2} = \frac{0.8}{0.45} \times 0.7 2^{2} = 0.9216 = 1 - 0.2 8^{2}$ and $T_{∥} = \frac{0.8}{0.45} \times 0.7 5^{2} = 1$ .

(b) Axial ratio of the reflected wave. — (a) Reflected power.

What the graphs say

Only at $θ_{B}$ is the reflection purely S: polarisation by reflection. This is why glare from water is horizontally polarised and why polarising sunglasses, with a vertical transmission axis, remove it. Below $θ_{B}$ the reflection is elliptical with the opposite sense of rotation, above $θ_{B}$ with the same sense, because $Γ_{∥}$ changes sign at $θ_{B}$ .