Exercises — Birefringence — ordinary and extraordinary rays

• The ray obeying Snell's law is the ordinary ray (o-ray). Its electric field $\vec{E}$ is perpendicular to the principal plane, so $\vec{E}$ never tilts toward the optic axis and always feels the single index $n_o$.
• The ray violating ordinary Snell's law is the extraordinary ray (e-ray). Its $\vec{E}$ lies in the principal plane, so as the direction changes the angle between $\vec{E}$ and the optic axis changes, and its index $n_e(\theta )$ changes with it.

A negative crystal is defined by $n_e<n_o$. Here $1.486<1.658$, so calcite is negative. (Positive would be $n_e>n_o$, e.g. quartz.)

Along the optic axis. There $\theta =0$, and the formula gives $\frac{1}{n_e(0{)}^2}=\frac{{\cos }^{2}0}{n_o^2}+\frac{{\sin }^{2}0}{n_e^2}=\frac{1}{n_o^2}\Rightarrow n_e(0)=n_o.$ Both rays feel the same index, travel at the same speed, and never separate.

Step 1 — write the formula. $\frac{1}{n_e(\theta {)}^2}=\frac{{\cos }^2\theta }{n_o^2}+\frac{{\sin }^2\theta }{n_e^2}$. Step 2 — evaluate the trig. ${\cos }^{2}30^{\circ }=0.75$, ${\sin }^{2}30^{\circ }=0.25$. Step 3 — plug in. $n_o^2=2.749$, $n_e^2=2.208$: $\frac{1}{n_e^2}=\frac{0.75}{2.749}+\frac{0.25}{2.208}=0.27282+0.11322=\mathrm{0.38604.}$ Step 4 — invert. The formula gives us $1/n_e^2$, but we want $n_e$ itself. Why these two operations: to undo "square then reciprocal" we apply the inverse operations in reverse order — first take the reciprocal to get $n_e^2$, then take the square root to get $n_e$. Doing both at once, $n_e(30^{\circ })=1/\sqrt{0.38604}=1.6094$. Sanity: it lies between $n_e=1.486$ and $n_o=1.658$ — as any $n_e(\theta )$ must.

Step 1 — phase formula. $\Delta \varphi =\frac{2\pi }{\lambda }|\Delta n|d$. Why: each ray accumulates phase $\frac{2\pi }{\lambda }nd$; subtracting gives the difference driven by $|\Delta n|$. Step 2 — plug in. $\Delta \varphi =\frac{2\pi }{589\times 10^{-9}}(0.172)(30\times 10^{-6})=\frac{2\pi (0.172)(30\times 10^{-6})}{589\times 10^{-9}}.$ Step 3 — evaluate. Numerator $2\pi (0.172)(30\times 10^{-6})=3.2421\times 10^{-5}$; divide by $589\times 10^{-9}$: gives $\Delta \varphi =55.04$ rad. In wavelengths that is $\Delta \varphi /2\pi =8.76$ full cycles.

Step 1. Set $\frac{2\pi }{\lambda }|\Delta n|d=\pi \Rightarrow d=\frac{\lambda }{2|\Delta n|}$. Step 2. $d=\frac{589\times 10^{-9}}{2(0.172)}=1.712\times 10^{-6}\text{m}=1712$ nm. Note this is exactly twice the quarter-wave thickness (856 nm) from the parent note — as it should be, since $\pi$ is twice $\pi /2$. See Wave plates (quarter and half wave).

Reasoning. Rewrite $\frac{1}{n_e(\theta {)}^2}=\frac{{\cos }^2\theta }{n_o^2}+\frac{{\sin }^2\theta }{n_e^2}$. As $\theta$ grows, ${\cos }^2\theta$ (weight on $1/n_o^2$) falls and ${\sin }^2\theta$ (weight on $1/n_e^2$) rises. For calcite $n_e<n_o$, so $1/n_e^2>1/n_o^2$: we are shifting weight onto the larger term, so $1/n_e(\theta {)}^2$ increases, meaning $n_e(\theta )$ itself decreases. Endpoints. $\theta =0$: $n_e(0)=n_o=1.658$ (largest). $\theta =90^{\circ }$: $n_e(90^{\circ })=n_e=1.486$ (smallest). How to read the figure. The horizontal axis is $\theta$ from $0^{\circ }$ to $90^{\circ }$; the vertical axis is the e-ray index. The solid red curve is $n_e(\theta )$: start at its left end and it sits on the upper blue dashed line at $1.658$ (that dashed line marks $n_o$); trace it rightward and it slides down monotonically until it lands on the lower green dashed line at $1.486$ (that marks $n_e$). The two yellow dots at $30^{\circ }$ and $60^{\circ }$ are the points computed in Problems 2.1 and 3.2 — notice the $60^{\circ }$ dot sits lower than the $30^{\circ }$ dot, confirming the curve keeps falling. Because the red curve never rises anywhere, the answer "decreases" is visible at a glance.

Step 1 — get $n_e(60^{\circ })$. ${\cos }^{2}60^{\circ }=0.25$, ${\sin }^{2}60^{\circ }=0.75$: $\frac{1}{n_e^2}=\frac{0.25}{2.749}+\frac{0.75}{2.208}=0.09094+0.33967=0.43061,n_e(60^{\circ })=\mathrm{1.5240.}$ Step 2 — compare. o-ray index $=n_o=1.658$; e-ray index $=1.524$. Smaller index → higher speed. Since $1.524<1.658$, the e-ray is faster here. Big picture: in a negative crystal the e-ray is faster (lower index) for every $\theta \ne 0$; that is exactly what "negative" encodes.

At $\theta =0$ the two indices coincide ($n_e(0)=n_o$), so $\delta (0)=0$ — no splitting, this direction is the optic axis. As $\theta$ grows, $n_e(\theta )$ marches away from $n_o$ toward $n_e$, and the gap widens monotonically (from Problem 3.1). It is maximal when $n_e(\theta )$ reaches its extreme principal value at $\theta =90^{\circ }$: $\delta (90^{\circ })=|n_o-n_e|=|1.658-1.486|=\mathrm{0.172.}$

Step 1. $\frac{2\pi }{\lambda }|\Delta n|d=\frac{\pi }{2}\Rightarrow d=\frac{\lambda }{4|\Delta n|}$. Step 2. $d=\frac{633\times 10^{-9}}{4(0.172)}=9.20\times 10^{-7}\text{m}=920$ nm. What it does: Linear light at $45^{\circ }$ splits equally into o- and e-components; a $\pi /2$ relative delay turns their sum into circularly polarized light. See Wave plates (quarter and half wave) and Polarization of light.

Step 1 — path difference in wavelengths. $N=\frac{|\Delta n|d}{\lambda }=\frac{0.172\times 2.30\times 10^{-6}}{589\times 10^{-9}}=0.6716$ wavelengths. Step 2 — phase. $\Delta \varphi =2\pi N=4.220$ rad. Reduce mod $2\pi$: $4.220$ rad is already $<2\pi (6.283)$, so effective phase $=4.220$ rad $=0.6716\times 2\pi$. Step 3 — compare — and WHY. Only the phase modulo $2\pi$ changes the output polarization, because adding a whole $2\pi$ (one full wavelength of delay) returns each wave to an identical state. So we ask which "clean" plate the leftover $0.6716\times 2\pi$ resembles. A half-wave plate is $0.5\times 2\pi$; a quarter-wave plate is $0.25\times 2\pi$ or equivalently $0.75\times 2\pi$ (three-quarter delay = quarter-wave behaviour with fast/slow roles swapped, since $0.75\times 2\pi =2\pi /2+2\pi /4$ folds an extra half-wave onto a quarter-wave). Our leftover $0.6716$ is closest to $0.75$, a distance of $0.083$, versus $0.17$ from $0.5$ — so the plate behaves closest to a quarter-wave plate, with an added half-wave folded in.

Step 1. With fast axes aligned, the "fast" component stays fast through both plates and the "slow" stays slow, so the phase differences add: $\Delta {\varphi }_{\text{tot}}=\pi /2+\pi /2=\pi$. Step 2. A $\pi$ phase difference is a half-wave plate. So two aligned quarter-wave plates = one half-wave plate. Total thickness $2\times 856=1712$ nm, matching Problem 2.3. ✔

Step 1 — target. $\frac{1}{n_e(\theta {)}^2}=\frac{1}{1.60^2}=\frac{1}{2.56}=0.390625$. Step 2 — a smart substitution and WHY. The formula has both ${\cos }^2\theta$ and ${\sin }^2\theta$ in it, which look like two unknowns. But they are locked together by the identity ${\cos }^2\theta +{\sin }^2\theta =1$. So set ${\sin }^2\theta =s$; then automatically ${\cos }^2\theta =1-s$. Why this helps: it collapses the two trig terms into one unknown $s$, turning the equation into a plain linear equation in $s$ — much easier than juggling angles directly. $0.390625=\frac{1-s}{2.749}+\frac{s}{2.208}.$ Step 3 — solve the linear equation. Expand each term: $\frac{1-s}{2.749}=0.36377-0.36377s$ and $\frac{s}{2.208}=0.45290s$. Add: $0.390625=0.36377+(0.45290-0.36377)s=0.36377+0.08913s.$ Isolate $s$: $s=\frac{0.390625-0.36377}{0.08913}=\frac{0.026855}{0.08913}=\mathrm{0.30130.}$ Step 4 — back out $\theta$, and handle the two roots / domain. From ${\sin }^2\theta =s=0.30130$ we get $\sin \theta =\pm \sqrt{0.30130}=\pm 0.54891$. Why we keep only one: the propagation angle to the optic axis lives in the physical domain $0^{\circ }\le \theta \le 90^{\circ }$, where $\sin \theta$ is never negative, so the negative root is discarded. Taking the positive root, $\theta =\arcsin (0.54891)=33.29^{\circ }$. (Algebraically ${\sin }^2\theta =s$ also allows $\theta =180^{\circ }-33.29^{\circ }=146.71^{\circ }$, but that too falls outside $0^{\circ }$–$90^{\circ }$, so $33.29^{\circ }$ is the unique physical answer.) How to read the figure. Same axes as Problem 3.1: $\theta$ horizontal, e-ray index vertical, red curve $=n_e(\theta )$. The horizontal yellow dashed line is drawn at the target height $1.60$. Follow it rightward until it meets the red curve — that single meeting point (there is only one inside $0^{\circ }$–$90^{\circ }$, because the red curve is strictly decreasing and therefore crosses any horizontal level at most once) is the solution, marked by the green dot. Drop straight down the vertical blue dotted line to read the axis: it lands at $\theta \approx 33.3^{\circ }$, confirming the algebra. Because $1.60$ sits near the top of the index range (close to $n_o=1.658$), the crossing happens at a small $\theta$ — exactly what the picture shows.

Step 1 — ideal. $d=\frac{\lambda }{4|\Delta n|}=\frac{500\times 10^{-9}}{4(0.172)}=7.2674\times 10^{-7}\text{m}=726.7$ nm. Step 2 — why fractional errors are equal. The phase is $\Delta \varphi =\frac{2\pi }{\lambda }|\Delta n|d=Kd$, where $K=\frac{2\pi }{\lambda }|\Delta n|$ is a constant (fixed wavelength and material). Since $\Delta \varphi$ is directly proportional to $d$, if $d$ changes by $\delta d$ then $\Delta \varphi$ changes by $K\delta d$, and the fractional change is $\frac{\delta (\Delta \varphi )}{\Delta \varphi }=\frac{K\delta d}{Kd}=\frac{\delta d}{d}.$ The constant $K$ cancels — that is why the two fractional errors are identical. Step 3 — evaluate. $\frac{\delta d}{d}=\frac{50}{726.7}=0.0688=6.88\%.$ Interpretation: a $6.9\%$ error in phase means $\Delta \varphi =\pi /2\times (1\pm 0.0688)$, i.e. the plate is a slightly imperfect QWP and produces slightly elliptical rather than perfectly circular light.

Step 1 — target. $\frac{1}{1.572^2}=\frac{1}{2.471184}=0.404663$. Step 2 — same substitution $s={\sin }^2\theta$ (so ${\cos }^2\theta =1-s$, collapsing to one linear unknown): $0.404663=\frac{1-s}{2.749}+\frac{s}{2.208}=0.36377+0.08913s.$ Step 3 — solve for $s$. $s=\frac{0.404663-0.36377}{0.08913}=\frac{0.040893}{0.08913}=\mathrm{0.45881.}$ Step 4 — angle (physical root only). $\sin \theta =+\sqrt{0.45881}=0.67735$ (the negative root is excluded because $0^{\circ }\le \theta \le 90^{\circ }$) $\Rightarrow \theta =\arcsin (0.67735)=42.62^{\circ }.$ Note: the index-average angle ($42.6^{\circ }$) is not $45^{\circ }$, because index combines as $1/n^2$, not linearly — the midpoint in index is not the midpoint in angle.

Can you, from a blank page: state which ray obeys Snell (::: the o-ray) — write the e-index formula (::: $\frac{1}{n_e(\theta {)}^2}=\frac{{\cos }^2\theta }{n_o^2}+\frac{{\sin }^2\theta }{n_e^2}$) — give the QWP thickness formula (::: $d=\frac{\lambda }{4|\Delta n|}$) — say which ray is faster in calcite (::: the e-ray, lower index)? If yes, you have mastered D4.