Birefringence — ordinary and extraordinary rays: Worked examples

Everything here uses only what the parent built. The one formula we lean on constantly:

The scenario matrix

Cell	Case class	What is special	Example
A	$θ = 0^{\circ}$ (along optic axis)	degenerate: no splitting, $n_{e} (θ) = n_{o}$	Ex 1
B	$θ = 9 0^{\circ}$ (across optic axis)	limiting: maximum splitting, $n_{e} (θ) = n_{e}$	Ex 2
C	general $0 < θ < 9 0^{\circ}$	index interpolates between $n_{o}$ and $n_{e}$	Ex 3
D	negative crystal ( $n_{e} < n_{o}$ )	$n_{e} (θ)$ rises with $θ$ … wait, falls — sign check	Ex 3, Ex 4
E	positive crystal ( $n_{e} > n_{o}$ )	$n_{e} (θ)$ increases with $θ$ ; opposite trend	Ex 4
F	degenerate crystal ( $n_{e} = n_{o}$ )	isotropic: birefringence vanishes for all $θ$	Ex 5
G	phase / wave-plate design ( $Δ ϕ$ )	pick thickness $d$ for a target retardation	Ex 6, Ex 7
H	walk-off angle (e-ray bends off $\hat{k}$ )	the "extraordinary" bending itself	Ex 8
I	real-world word problem	Nicol-prism style separation choice	Ex 9
J	exam twist (solve backwards for $θ$ )	given the index, find the angle	Ex 10

Read the matrix as a checklist. Each example below names the cell(s) it clears.

Worked examples

Worked example Example 1 — Cell A: propagation

along the optic axis ( $θ = 0^{\circ}$ ) Calcite: $n_{o} = 1.658$ , $n_{e} = 1.486$ . What index does the e-ray feel when it travels straight down the optic axis?

Forecast: Along the optic axis the crystal is defined to have no birefringence. So guess: both rays feel the same index. Which one?

Step 1. Put $θ = 0$ into the workhorse: $\frac{1}{n _{e} ( 0 ) ^{2}} = \frac{c o s ^{2} 0 ^{\circ}}{n _{o}^{2}} + \frac{s i n ^{2} 0 ^{\circ}}{n _{e}^{2}} = \frac{1}{n _{o}^{2}} + 0.$ Why this step? $cos 0^{\circ} = 1$ and $sin 0^{\circ} = 0$ — travelling straight down the axis means "all along, none across", so the $n_{e}$ term switches completely off. Step 2. Therefore $n_{e} (0) = n_{o} = 1.658$ . Why this step? Inverting $1/ n_{e} (0)^{2} = 1/ n_{o}^{2}$ gives $n_{e} (0) = n_{o}$ directly.

Verify: The e-ray index equals the o-ray index, so the two rays travel at the same speed and never separate — precisely the definition of the optic axis. ✅

Figure — Birefringence — ordinary and extraordinary rays

Worked example Example 3 — Cell C & D: a general angle in a

negative crystal ( $θ = 3 0^{\circ}$ ) Calcite ( $n_{e} < n_{o}$ , negative). Find $n_{e} (3 0^{\circ})$ .

Forecast: The answer must sit between $1.486$ and $1.658$ . Because calcite is negative, as $θ$ grows from $0$ to $9 0^{\circ}$ the index should slide down from $1.658$ toward $1.486$ . So at $3 0^{\circ}$ expect a value nearer the top of that range.

Step 1. $cos^{2} 3 0^{\circ} = 0.75$ , $sin^{2} 3 0^{\circ} = 0.25$ . Why this step? $cos 3 0^{\circ} = \frac{3}{2}$ so $cos^{2} = \frac{3}{4}$ ; the two must add to $1$ , giving $sin^{2} = \frac{1}{4}$ . Step 2. $n_{o}^{2} = 1.65 8^{2} = 2.749$ , $n_{e}^{2} = 1.48 6^{2} = 2.208$ . Step 3. $\frac{1}{n _{e} ( 3 0 ^{\circ} ) ^{2}} = \frac{0.75}{2.749} + \frac{0.25}{2.208} = 0.2728 + 0.1132 = 0.3860.$ Why this step? Straight substitution of the two ratios. Step 4. $n_{e} (3 0^{\circ}) = 1/ 0.3860 = 1.609$ .

Verify: $1.486 < 1.609 < 1.658$ ✅, and $1.609$ is closer to $1.658$ than the $4 5^{\circ}$ answer ( $1.565$ from the parent note) — the index really does slide downward as $θ$ increases in a negative crystal. ✅

Worked example Example 5 — Cell F: the degenerate crystal (

$n_{e} = n_{o}$ ) Suppose a crystal has $n_{o} = n_{e} = 1.60$ (an accidentally isotropic case). Find $n_{e} (θ)$ for any $θ$ .

Forecast: If both principal indices are equal, the "lopsidedness" is gone. Guess: one index for every direction, no splitting anywhere.

Step 1. With $n_{o} = n_{e} = n$ : $\frac{1}{n _{e} ( θ ) ^{2}} = \frac{c o s ^{2} θ}{n ^{2}} + \frac{s i n ^{2} θ}{n ^{2}} = \frac{c o s ^{2} θ + s i n ^{2} θ}{n ^{2}} = \frac{1}{n ^{2}} .$ Why this step? $cos^{2} θ + sin^{2} θ = 1$ is always true (Pythagoras on the unit circle), so the $θ$ -dependence collapses. Step 2. $n_{e} (θ) = n = 1.60$ for all $θ$ .

Verify: No angle gives splitting — this is just ordinary glass. The formula correctly reduces to the isotropic case, which is exactly what "birefringence needs two distinct indices" demanded. ✅ (See Dielectric tensor and anisotropic media for why equal diagonal entries mean isotropy.)

Worked example Example 6 — Cell G: half-wave plate thickness for calcite

At $λ = 589$ nm, using principal indices ( $θ = 9 0^{\circ}$ , so $∣Δ n ∣ = 0.172$ ), find the thinnest thickness $d$ giving a half-wave retardation, $Δ ϕ = π$ .

Forecast: A half-wave plate flips polarization; it needs twice the retardation of a quarter-wave plate (parent Ex 3 gave $856$ nm for $π /2$ ). Guess: about $2 \times 856 \approx 1712$ nm.

Step 1. The phase difference after thickness $d$ is $Δ ϕ = \frac{2 π}{λ} ∣Δ n ∣ d$ . Why this step? The two rays travel the same physical distance $d$ but at different speeds; the optical-path gap is $∣Δ n ∣ d$ , and each wavelength of gap is $2 π$ of phase. Step 2. Set $Δ ϕ = π$ and solve: $d = \frac{λ}{2∣Δ n ∣}$ . Step 3. $d = \frac{589 \times 1 0 ^{- 9}}{2 ( 0.172 )} = 1.712 \times 1 0^{- 6} m = 1712$ nm.

Verify: Units: $\frac{m}{(dimensionless)} = m$ ✅. And it is exactly twice the quarter-wave value — consistent with Wave plates (quarter and half wave). ✅

Worked example Example 7 — Cell G twist: retardation of a

given plate A calcite plate is $d = 2.0 μ m$ thick, principal indices, $λ = 589$ nm. What is $Δ ϕ$ , and is it a "clean" wave plate?

Forecast: $2.0 μ$ m is a bit more than the half-wave value ( $1712$ nm). So $Δ ϕ$ should be a bit more than $π$ . Guess: around $1.1 π$ to $1.2 π$ .

Step 1. $Δ ϕ = \frac{2 π}{λ} ∣Δ n ∣ d = \frac{2 π ( 0.172 ) ( 2.0 \times 1 0 ^{- 6} )}{589 \times 1 0 ^{- 9}}$ . Why this step? Same phase formula, now solving forward for $Δ ϕ$ . Step 2. Numerator inside: $0.172 \times 2.0 \times 1 0^{- 6} = 3.44 \times 1 0^{- 7}$ . Divide by $589 \times 1 0^{- 9} = 5.89 \times 1 0^{- 7}$ : ratio $= 0.5840$ . Step 3. $Δ ϕ = 2 π \times 0.5840 = 3.669$ rad $= 1.168 π$ .

Verify: $1.168 π$ is just above $π$ — matches the forecast. It is not a clean half-wave plate ( $π$ ) nor quarter-wave ( $π /2$ ); the extra $0.168 π$ means the output is elliptically polarized. ✅ (See Polarization of light.)

Worked example Example 8 — Cell H: the e-ray

walk-off angle For calcite at $θ = 4 5^{\circ}$ between $\hat{k}$ and the optic axis, the e-ray's energy travels at an angle $α$ off $\hat{k}$ given by $tan (θ + α) = (\frac{n _{o}}{n _{e}})^{2} tan θ .$ Find $α$ . (This $α$ is literally why the e-ray is "extraordinary".)

Forecast: Since calcite is negative ( $n_{o} > n_{e}$ ), $(n_{o} / n_{e})^{2} > 1$ , so the ray direction tilts further than $\hat{k}$ — expect a small positive walk-off of a few degrees.

Step 1. $(n_{o} / n_{e})^{2} = (1.658/1.486)^{2} = (1.11575)^{2} = 1.24489$ . Why this step? This ratio is the "anisotropy strength"; $> 1$ means the e-ray leans away from $\hat{k}$ . Step 2. $tan 4 5^{\circ} = 1$ , so $tan (4 5^{\circ} + α) = 1.24489$ . Step 3. $4 5^{\circ} + α = arctan (1.24489) = 51.22 9^{\circ}$ , hence $α = 6.22 9^{\circ}$ . Why this step? $arctan$ answers "which angle has this tangent?" — it undoes the tan on the left.

Verify: $α \approx 6. 2^{\circ}$ , a small positive tilt — matches the forecast and the famous "double image shift" you see through a calcite crystal. If we had used a positive crystal ( $n_{o} < n_{e}$ ) the ratio would be $< 1$ and $α$ would be negative (tilts the other way). ✅

Worked example Example 9 — Cell I: real-world word problem (which crystal splits more?)

You need to physically separate two polarizations for a Nicol prism over a $1.0$ mm path, and you may pick calcite ( $Δ n = - 0.172$ ) or quartz ( $Δ n = + 0.009$ ). Which gives the larger optical-path separation between the rays, and how big is it?

Forecast: Separation grows with $∣Δ n ∣$ . Calcite's $0.172$ dwarfs quartz's $0.009$ , so calcite wins by roughly $0.172/0.009 \approx 19 \times$ . Guess: calcite gives a path gap of order $0.17$ mm.

Step 1. Optical-path gap $= ∣Δ n ∣ d$ . Why this step? Same optical-path idea as the wave plate, but now we care about the physical separation the two rays accumulate. Step 2. Calcite: $0.172 \times 1.0 \times 1 0^{- 3} = 1.72 \times 1 0^{- 4}$ m $= 0.172$ mm. Step 3. Quartz: $0.009 \times 1.0 \times 1 0^{- 3} = 9.0 \times 1 0^{- 6}$ m $= 0.009$ mm.

Verify: Ratio $= 0.172/0.009 = 19.1$ ✅. Calcite separates ~19× more — exactly why real Nicol prisms are cut from calcite, not quartz. ✅

Worked example Example 10 — Cell J: exam twist (solve

backwards for the angle) In calcite you measure the e-ray feeling index $n_{e} (θ) = 1.600$ . At what angle $θ$ to the optic axis is it travelling?

Forecast: $1.600$ sits between $1.486$ and $1.658$ , so a valid $θ$ exists. It is closer to $n_{o} = 1.658$ , and in a negative crystal high index means small $θ$ — expect something under $4 5^{\circ}$ .

Step 1. Rearrange the workhorse. Using $cos^{2} θ = 1 - sin^{2} θ$ : $\frac{1}{n _{e} ( θ ) ^{2}} = \frac{1}{n _{o}^{2}} + sin^{2} θ (\frac{1}{n _{e}^{2}} - \frac{1}{n _{o}^{2}}) .$ Why this step? Collecting everything onto $sin^{2} θ$ lets us isolate the one unknown. Step 2. Plug numbers: $\frac{1}{n _{e} ( θ ) ^{2}} = 1/2.56 = 0.390625$ ; $\frac{1}{n _{o}^{2}} = 0.363805$ ; $\frac{1}{n _{e}^{2}} - \frac{1}{n _{o}^{2}} = 0.452899 - 0.363805 = 0.089094$ . Step 3. $sin^{2} θ = \frac{0.390625 - 0.363805}{0.089094} = \frac{0.026820}{0.089094} = 0.30103$ . Step 4. $sin θ = 0.30103 = 0.54866$ , so $θ = arcsin (0.54866) = 33.2 7^{\circ}$ . Why this step? $arcsin$ answers "which angle has this sine?" — the inverse we need after isolating $sin θ$ .

Verify: Re-substitute $θ = 33.2 7^{\circ}$ into the forward formula: $cos^{2} = 0.699$ , $sin^{2} = 0.301$ , giving $1/ n_{e}^{2} = 0.699/2.749 + 0.301/2.208 = 0.2543 + 0.1363 = 0.3906$ , so $n_{e} = 1.600$ ✅. And $33.2 7^{\circ} < 4 5^{\circ}$ , matching the forecast. ✅

Recall Quick self-test on the matrix

At $θ = 0$ the e-ray index equals what? ::: $n_{o}$ (no splitting along the optic axis) In a negative crystal, does $n_{e} (θ)$ rise or fall as $θ$ goes $0 \to 9 0^{\circ}$ ? ::: It falls (from $n_{o}$ down to $n_{e}$ ) If $n_{o} = n_{e}$ , what is $n_{e} (θ)$ ? ::: A single constant $n$ for all $θ$ — the crystal is isotropic What controls how thick a wave plate must be? ::: $∣Δ n ∣$ and the target $Δ ϕ$ : $d = λ Δ ϕ / (2 π ∣Δ n ∣)$ Why does the e-ray "walk off"? ::: Its energy flow tilts by $α$ from $\hat{k}$ because $(n_{o} / n_{e})^{2} \neq = 1$