17-Phys-A7 Optics · May 2015

Question 4 of 6: Fresnel reflection/transmission at a water surface

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2015, 98-Phys-A7 Optics. Three hours, closed book, one approved Casio or Sharp calculator. Question 1 is mandatory; the rubric asks for Question 1 plus any four of Questions 2–6 (only the first four questions appearing in the answer book are marked) — all six are worked here, because the set is a study resource.

Reference texts. E. Hecht, Optics, 5th ed.; F. L. Pedrotti, L. M. Pedrotti and L. S. Pedrotti, Introduction to Optics, 3rd ed.

Question 4: Fresnel reflection/transmission at a water surface (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — normal-incidence Fresnel coefficients. Take the interface at $y=0$ with normal $\hat y$ (the same axes as Question 3), medium 1 ($n_1$) at $y<0$, medium 2 ($n_2$) at $y>0$, all waves normally incident/reflected/transmitted along $\pm\hat y$, $\mathbf E$ along $\hat z$ (choice of polarization is immaterial at normal incidence, since TE and TM coincide there). Write the fields as

$$E_i=E_{0i}\cos(k_1y-\omega t),\quad E_r=E_{0r}\cos(-k_1y-\omega t),\quad E_t=E_{0t}\cos(k_2y-\omega t),$$

with the associated $H$ fields from $H=nE/\eta_0$ (incident and transmitted $H$ along $+\hat x$; reflected $H$ along $-\hat x$, since $\mathbf S=\mathbf E\times\mathbf H$ must reverse with the reversed propagation direction). Maxwell's boundary conditions require the tangential $E$ and tangential $H$ to be continuous at $y=0$ (no free surface charge/current on a dielectric interface):

$$\begin{gathered} E_{0i}+E_{0r}=E_{0t} \qquad \text{(tangential }E\text{ continuity)}\\ n_1(E_{0i}-E_{0r})=n_2E_{0t} \qquad \text{(tangential }H\text{ continuity, using }H=nE/\eta_0\text{)} \end{gathered}$$

Solving this pair of linear equations for $r=E_{0r}/E_{0i}$ and $t=E_{0t}/E_{0i}$:

$$\boxed{r=\dfrac{n_1-n_2}{n_1+n_2}}, \qquad \boxed{t=\dfrac{2n_1}{n_1+n_2}}.$$

($t$ can exceed $1$ — that is normal for the field amplitude; the energy transmittance $T=(n_2/n_1)t^2\le1$ always.)

Part (b) — internal reflection, water→air.

Given. $n_1=1.33$ (water, incidence side), $n_2=1.00$ (air), full Fresnel equations $r_s=\dfrac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}$, $r_p=\dfrac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t}$ with $\sin\theta_t=(n_1/n_2)\sin\theta_i$.

Find. $r_s(\theta_i)$, $r_p(\theta_i)$ (and $t_s,t_p$) with numerical values at the salient features: $\theta_i=0$, the internal Brewster angle, and the critical angle.

Approach. Evaluate the boxed normal-incidence result at $\theta_i=0$; find the internal Brewster angle from $r_p=0$; find the critical angle from $\sin\theta_t=1$; sketch the four curves over $0\le\theta_i<\theta_c$ (beyond $\theta_c$, $\theta_t$ is complex — total internal reflection, $|r|=1$).

1. Normal incidence. $r_0=\dfrac{n_1-n_2}{n_1+n_2}=\dfrac{1.33-1.00}{1.33+1.00}=\boxed{+0.1416}$ (and $t_0=2n_1/(n_1+n_2)=1.1416$).
2. Internal Brewster angle ($r_p=0$). $\theta_{B,\text{int}}=\arctan(n_2/n_1)=\arctan(1/1.33)=\boxed{36.9^\circ}$.
3. Critical angle (onset of total internal reflection, $\sin\theta_t=1$). $\theta_c=\arcsin(n_2/n_1)=\arcsin(1/1.33)=\boxed{48.8^\circ}$; for $\theta_i>\theta_c$, $|r_s|=|r_p|=1$ (all light reflected, with a phase shift).

Part (c) — external reflection, air→water.

Given. $n_1=1.00$ (air, incidence side), $n_2=1.33$ (water); same Fresnel formulas as (b), now with $n_1 \lt n_2$ so no total internal reflection occurs (real $\theta_t$ for all $\theta_i\le90^\circ$).

Find. $r_s(\theta_i)$, $r_p(\theta_i)$ with numerical salient values at $\theta_i=0$, the (external) Brewster angle, and grazing incidence.

1. Normal incidence. $r_0=\dfrac{n_1-n_2}{n_1+n_2}=\dfrac{1.00-1.33}{1.00+1.33}=\boxed{-0.1416}$ ($t_0=2n_1/(n_1+n_2)=0.858$) — same magnitude as (b), opposite sign, as required by the Stokes reversibility relation $r_{12}=-r_{21}$.
2. External Brewster angle ($r_p=0$). $\theta_B=\arctan(n_2/n_1)=\arctan(1.33)=\boxed{53.1^\circ}$.
3. Grazing incidence ($\theta_i\to90^\circ$). $r_s\to-1$, $r_p\to+1$ (total reflection at grazing angle, both polarizations, with opposite sign conventions) — visible on the sketch as both curves converging toward $|r|=1$ at the right edge.

Feature	Internal (water→air)	External (air→water)
$r$ at $\theta_i=0$	$+0.142$	$-0.142$
Brewster angle ($r_p=0$)	$36.9^\circ$	$53.1^\circ$
Critical angle	$48.8^\circ$	— (none)

Part (d) — polarizing sunglasses. Light reflecting off a horizontal water or road surface at close to the external Brewster angle ($53.1^\circ$ for water) has $r_p\approx0$: the reflected glare is almost purely $s$-polarized. For a horizontal reflecting surface the plane of incidence is vertical, so the $s$-polarization direction (perpendicular to the plane of incidence) is horizontal — reflected glare is therefore predominantly horizontally polarized light. Polarizing sunglasses are most effective exactly when the reflecting surface is viewed near its Brewster angle (i.e. looking down at water or a road at a moderate angle, roughly $50$–$55^\circ$ from vertical for water); a shallower or steeper viewing angle admits more of the (still partially polarized) $p$-component and the glare is not fully blocked. The transmission axis of the sunglasses is oriented vertically under normal use, so as to absorb the horizontally-polarized glare while passing the (largely unpolarized) vertically-polarized component of the scene.

---