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

At normal incidence the fraction of the incident power that is reflected is $∣Γ ∣^{2}$ and the fraction that is transmitted is $(η_{1} / η_{2}) ∣ τ ∣^{2}$ (Section 6.3). At oblique incidence a factor appears that has to do with geometry, not with the waves. Consider a patch of area $A$ on the interface, lit by a beam (Figure 6.4.5). The incident beam that covers the patch has the cross-section $A cos θ_{i}$ , the reflected beam the same, and the transmitted beam the cross-section $A cos θ_{t}$ . Each beam carries the time-average power density $P = ∣ E_{0} ∣^{2} / (2 η)$ through its own cross-section. The power balance must be written for the power that crosses the patch, that is for the normal components of the Poynting vectors:

\frac{∣ E _{0}^{i} ∣ ^{2}}{2 η _{1}} cos θ_{i} = \frac{∣ E _{0}^{r} ∣ ^{2}}{2 η _{1}} cos θ_{i} + \frac{∣ E _{0}^{t} ∣ ^{2}}{2 η _{2}} cos θ_{t} . (6.4.14)

Reflectance and transmittance at oblique incidence (lossless media)

R T R + T = \frac{P ^{r} cos θ _{i}}{P ^{i} cos θ _{i}} = ∣Γ ∣^{2}, = \frac{P ^{t} cos θ _{t}}{P ^{i} cos θ _{i}} = \frac{η _{1} cos θ _{t}}{η _{2} cos θ _{i}} ∣ τ ∣^{2}, = 1, (6.4.15)

for either polarisation, with $Γ, τ = Γ_{⊥}, τ_{⊥}$ or $Γ_{∥}, τ_{∥}$ . They are also called the reflectivity and the transmissivity of the interface.

In plain words

Whatever is not reflected is transmitted. The reflected fraction is simply $∣Γ ∣^{2}$ , because the reflected beam is a mirror image of the incident one. The transmitted fraction carries the extra factor $cos θ_{t} / cos θ_{i}$ , because the transmitted beam has a different width.

That $R + T = 1$ is not an extra assumption: it follows from the boundary conditions. For S polarisation multiply the two equations Eq. (6.4.12): $(1 + Γ_{⊥}) (1 - Γ_{⊥}) = 1 - Γ_{⊥}^{2} = \frac{η _{1} c o s θ _{t}}{η _{2} c o s θ _{i}} τ_{⊥}^{2}$ , which is $R + T = 1$ . The tangential $E$ and $H$ are continuous, so the normal component of $\frac{1}{2} Re {E \times H^{*}}$ , which is built from the tangential components only, is continuous too.

$The three beams share one footprint A on the interface but have the cross-sections A\cos\theta_i (incident and reflected) and A\cos\theta_t (transmitted), drawn for air on polystyrene at 50^\circ (Problem 6.4.5). A beam that is bent towards the normal becomes *wider*, so its power density drops although it carries the same power.$

Figure 6.4.5. The three beams share one footprint

A

on the interface but have the cross-sections

A cos θ_{i}

(incident and reflected) and

A cos θ_{t}

(transmitted), drawn for air on polystyrene at

5 0^{\circ}

(Problem 6.4.5). A beam that is bent towards the normal becomes *wider*, so its power density drops although it carries the same power.

Common mistake · Forgetting the cosines

Using the normal-incidence formula $T = (η_{1} / η_{2}) ∣ τ ∣^{2}$ at an oblique angle gives $R + T \neq = 1$ . The power densities $P^{i}$ , $P^{r}$ , $P^{t}$ refer to three different cross-sections. Only their normal components $P cos θ$ , the power per unit area of interface, add up. A quick test of any oblique calculation is to check that $R + T = 1$ .

6.4.5 Where the power goes