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

$Γ_{∥} = 0$ when $η_{2} cos θ_{t} = η_{1} cos θ_{i}$ . Squaring and using Snell's law for $cos^{2} θ_{t} = 1 - (β_{1}^{2} / β_{2}^{2}) sin^{2} θ_{i}$ ,

η_{2}^{2} (1 - \frac{β _{1}^{2}}{β _{2}^{2}} sin^{2} θ_{B ∥}) = η_{1}^{2} (1 - sin^{2} θ_{B ∥}) ⟹ sin^{2} θ_{B ∥} (1 - \frac{η _{2}^{2} β _{1}^{2}}{η _{1}^{2} β _{2}^{2}}) = 1 - \frac{η _{2}^{2}}{η _{1}^{2}},

and with $η_{2}^{2} / η_{1}^{2} = μ_{2} ε_{1} / (μ_{1} ε_{2})$ and $η_{2} β_{1} / (η_{1} β_{2}) = ε_{1} / ε_{2}$ :

Brewster angle, P polarisation

sin^{2} θ_{B ∥} = \frac{1 - μ _{2} ε _{1} / ( μ _{1} ε _{2} )}{1 - ε _{1}^{2} / ε _{2}^{2}} . (6.4.30)

For media with the same permeability (all nonmagnetic media)

sin θ_{B ∥} = \frac{1}{1 + ε _{1} / ε _{2}} ⟺ tan θ_{B ∥} = \frac{ε _{2}}{ε _{1}} = \frac{n _{2}}{n _{1}}, θ_{B ∥} + θ_{t} = 9 0^{\circ} . (6.4.31)

The last relation follows from Snell's law: at $θ_{B ∥}$ , $sin θ_{t} = \frac{n _{1}}{n _{2}} sin θ_{B ∥} = \frac{n _{1}}{n _{2}} \frac{n _{2}}{n _{1}^{2} + n _{2}^{2}} = \frac{n _{1}}{n _{1}^{2} + n _{2}^{2}} = cos θ_{B ∥}$ . The Brewster angle always exists for P polarisation between two dielectrics, from either side: air to glass $θ_{B} = arctan 1.5 = 56. 3^{\circ}$ , glass to air $arctan (1/1.5) = 33. 7^{\circ} = 9 0^{\circ} - 56. 3^{\circ}$ .

In plain words

At the Brewster angle a P-polarised wave crosses the interface without any reflection, and the reflected and the refracted ray would be at right angles to each other. An S wave is always partly reflected. Light reflected at $θ_{B}$ is therefore purely S-polarised: polarisation by reflection, discovered by David Brewster in 1815.

Why nothing is reflected at the Brewster angle. The transmitted field drives the bound dipoles of medium 2, which oscillate along \vec E^t and radiate the reflected wave. At \theta_B the direction of reflection (dashed) is parallel to \vec E^t, since \theta_B+\theta_t=90^\circ, and a dipole does not radiate along its own axis. The dipoles of an S wave point along \vec a_z and radiate in every direction of the plane of incidence. — **Figure 6.4.12.** Why nothing is reflected at the Brewster angle. The transmitted field drives the bound dipoles of medium 2, which oscillate along $E^{t}$ and radiate the reflected wave. At $θ_{B}$ the direction of reflection (dashed) is parallel to $E^{t}$ , since $θ_{B} + θ_{t} = 9 0^{\circ}$ , and a dipole does not radiate along its own axis. The dipoles of an S wave point along $a_{z}$ and radiate in every direction of the plane of incidence.

The dipole picture

The reflected wave is not produced by the surface itself. It is radiated by the bound dipoles of medium 2, which the transmitted field sets oscillating along $E^{t}$ (Figure 6.4.12). An oscillating dipole radiates most strongly at right angles to its axis and not at all along it. At $θ_{B}$ the reflected direction is exactly the axis of the P dipoles, so they cannot send anything back. The S dipoles point out of the plane of incidence and radiate equally in all directions within it, which is why an S wave has no Brewster angle in nonmagnetic media.

Quick check

Sunlight is reflected by a glass window ( $n = 1.5$ ) at the Brewster angle. Which fraction of the S power is reflected?

At $θ_{B} = 56. 3^{\circ}$ , $θ_{t} = 33. 7^{\circ}$ and, in units of $η_{0}$ , $Γ_{⊥} = \frac{c o s θ _{B} /1.5 - c o s θ _{t}}{c o s θ _{B} /1.5 + c o s θ _{t}}$ . With $cos θ_{t} = sin θ_{B}$ and $tan θ_{B} = n$ this simplifies to $Γ_{⊥} = \frac{1 - n ^{2}}{1 + n ^{2}} = - 0.385$ , so $∣ Γ_{⊥} ∣^{2} = 14.8 %$ of the S power. The P power is not reflected at all.

6.4.9 The Brewster angle in P polarisation