Fresnel Equations

When light strikes a planar interface between media with refractive indices $n_1$ and $n_2$The Amplitudes of the reflected and transmitted waves depend on the polarisation.

For an incident wave with amplitude $E_i$The reflection and transmission coefficients are:

s-polarisation (perpendicular to the plane of incidence):

$r_s = \frac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}, \quad t_s = \frac{2n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t}$

p-polarisation (parallel to the plane of incidence):

$r_p = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t}, \quad t_p = \frac{2n_1\cos\theta_i}{n_2\cos\theta_i + n_1\cos\theta_t}$

Reflectance and transmittance (energy fractions):

$R = |r|^2, \quad T = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t|^2$

With $R + T = 1$ (energy conservation).

At the Brewster angle $\theta_B$The reflected beam for p-polarised light has zero amplitude: $r_p = 0$:

$\tan\theta_B = \frac{n_2}{n_1}$

Proof. Setting $r_p = 0$: $n_2\cos\theta_i = n_1\cos\theta_t$. Using Snell’s law $n_1\sin\theta_i = n_2\sin\theta_t$:

$\frac{\cos\theta_i}{\sin\theta_i} = \frac{\cos\theta_t}{\sin\theta_t}$

$\cot\theta_i = \cot\theta_t \implies \theta_i + \theta_t = 90^\circ$

So $\tan\theta_i = \tan\theta_B = n_2/n_1$. $\blacksquare$

At Brewster’s angle, the reflected and refracted beams are perpendicular. This is why polarising Filters work at specific angles for reflected glare.

When $n_1 \gt n_2$ and $\theta_i \gt \theta_c = \arcsin(n_2/n_1)$, $\sin\theta_t \gt 1$So $\cos\theta_t = i\sqrt{\sin^2\theta_t - 1}$ becomes imaginary.

The transmitted field becomes an evanescent wave:

$E_t \propto e^{-\kappa x}\, e^{i(k_z z - \omega t)}$

Where $\kappa = k_0\sqrt{n_1^2\sin^2\theta_i - n_2^2}$ and $k_z = k_0 n_1\sin\theta_i$.

The field decays exponentially with penetration depth $\delta = 1/\kappa$ but propagates along the Interface. No energy is transported into the second medium: $R = 1$.

Frustrated total internal reflection. If a third medium is brought within a few wavelengths of The interface, energy can tunnel across the gap (analogous to quantum tunnelling).

The Fresnel coefficients are real for $\theta_i < \theta_c$ (normal incidence/transmission) and may be positive or negative, indicating phase shifts:

• External reflection ($n_1 < n_2$): $r_s < 0$ for all $\theta_i$ (phase shift of $\pi$ for s-polarisation). $r_p$ changes sign at Brewster’s angle.
• Internal reflection ($n_1 > n_2$): For $\theta_i < \theta_c$, both $r_s$ and $r_p$ are positive at normal incidence. $r_p$ changes sign at Brewster’s angle.

At normal incidence ($\theta_i = 0$):

$r_s = r_p = \frac{n_1 - n_2}{n_1 + n_2}$

The reflection coefficient is negative when $n_1 < n_2$, corresponding to a $\pi$ phase shift. For $n_1 > n_2$, the reflection coefficient is positive (no phase shift).

The reflectance $R$ varies with angle of incidence:

• For s-polarisation, $R_s$ increases monotonically from $((n_1 - n_2)/(n_1 + n_2))^2$ at normal incidence to $1$ at grazing incidence.
• For p-polarisation, $R_p$ drops to $0$ at Brewster’s angle, then increases to $1$ at grazing incidence.

Anti-reflection coatings use destructive interference between reflections from two interfaces. For a single-layer coating of index $n_c$ and thickness $\lambda/4$ on glass ($n_g$), the reflectance at wavelength $\lambda$ is:

$R = \left(\frac{n_c^2 - n_g}{n_c^2 + n_g}\right)^2$

The reflectance is zero when $n_c = \sqrt{n_g}$. For crown glass ($n_g = 1.52$), the optimal coating index is $n_c \approx 1.23$, approximated by magnesium fluoride ($n \approx 1.38$), giving $R \approx 1\%$ per surface.

In total internal reflection, the reflected beam is laterally shifted relative to the geometrically predicted path. This Goos-Hanchen shift arises because the evanescent wave penetrates the second medium before being reflected:

$D = \frac{\lambda}{\pi} \frac{\sin\theta_i}{\sqrt{\sin^2\theta_i - (n_2/n_1)^2}}$

The shift is of order one wavelength for angles near the critical angle and decreases as $\theta_i$ increases beyond $\theta_c$.

Problem. Calculate the reflectance of uncoated glass ($n_g = 1.52$) at normal incidence in air.

Problem. Find the phase difference between s- and p-polarised components after total internal reflection in glass ($n_1 = 1.5$) at $\theta_i = 60^\circ$ with $n_2 = 1$.

At $\theta_i = \theta_c = \arcsin(n_2/n_1)$, we have $\theta_t = 90^\circ$ and $\cos\theta_t = 0$. The Fresnel coefficients become:

$r_s = \frac{n_1\cos\theta_c - 0}{n_1\cos\theta_c + 0} = 1$

$r_p = \frac{n_2\cos\theta_c - 0}{n_2\cos\theta_c + 0} = 1$

Both polarisations have $R = 1$ at the critical angle, and the transmitted wave propagates exactly along the interface with no energy flow into the second medium.