Birefringence — ordinary and extraordinary rays: Worked examples

2.5.18 · D3 · Physics › Optics › Birefringence — ordinary and extraordinary rays

Yahan sab kuch wahi use karta hai jo parent note ne banaya. Ek formula jis par hum baar baar rely karte hain:

Scenario matrix

Cell	Case class	Kya special hai	Example
A	$θ = 0^{\circ}$ (optic axis ke saath)	degenerate: koi splitting nahi, $n_{e} (θ) = n_{o}$	Ex 1
B	$θ = 9 0^{\circ}$ (optic axis ke across)	limiting: maximum splitting, $n_{e} (θ) = n_{e}$	Ex 2
C	general $0 < θ < 9 0^{\circ}$	index $n_{o}$ aur $n_{e}$ ke beech interpolate karta hai	Ex 3
D	negative crystal ( $n_{e} < n_{o}$ )	$n_{e} (θ)$ $θ$ ke saath badhta hai… ruko, girta hai — sign check	Ex 3, Ex 4
E	positive crystal ( $n_{e} > n_{o}$ )	$n_{e} (θ)$ $θ$ ke saath badhta hai; opposite trend	Ex 4
F	degenerate crystal ( $n_{e} = n_{o}$ )	isotropic: birefringence sabhi $θ$ ke liye vanish ho jaata hai	Ex 5
G	phase / wave-plate design ( $Δ ϕ$ )	target retardation ke liye thickness $d$ chuno	Ex 6, Ex 7
H	walk-off angle (e-ray $\hat{k}$ se bend karta hai)	"extraordinary" bending khud	Ex 8
I	real-world word problem	Nicol-prism style separation choice	Ex 9
J	exam twist (solve backwards for $θ$ )	given index se angle nikalo	Ex 10

Matrix ko checklist ki tarah padho. Neeche har example batata hai ki woh kaun se cell(s) clear karta hai.

Worked examples

Worked example Example 1 — Cell A: optic axis

ke saath propagation ( $θ = 0^{\circ}$ ) Calcite: $n_{o} = 1.658$ , $n_{e} = 1.486$ . Jab e-ray seedha optic axis ke saath travel kare, toh woh kaun sa index feel karta hai?

Forecast: Optic axis ke saath crystal mein koi birefringence nahi hoti — yeh definition hai. Toh guess: dono rays ek hi index feel karein. Kaun sa?

Step 1. $θ = 0$ workhorse mein daalo: $\frac{1}{n _{e} ( 0 ) ^{2}} = \frac{c o s ^{2} 0 ^{\circ}}{n _{o}^{2}} + \frac{s i n ^{2} 0 ^{\circ}}{n _{e}^{2}} = \frac{1}{n _{o}^{2}} + 0.$ Yeh step kyun? $cos 0^{\circ} = 1$ aur $sin 0^{\circ} = 0$ — seedha axis ke saath travel karne ka matlab hai "sab along, kuch nahi across", isliye $n_{e}$ wala term completely off ho jaata hai. Step 2. Isliye $n_{e} (0) = n_{o} = 1.658$ . Yeh step kyun? $1/ n_{e} (0)^{2} = 1/ n_{o}^{2}$ ko invert karne par seedha $n_{e} (0) = n_{o}$ milta hai.

Verify: E-ray index, o-ray index ke barabar hai, toh dono rays ek hi speed se travel karti hain aur kabhi separate nahi hoti — yahi optic axis ki definition hai. ✅

Figure — Birefringence — ordinary and extraordinary rays

Worked example Example 3 — Cell C & D: ek

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

Forecast: Answer $1.486$ aur $1.658$ ke beech hona chahiye. Kyunki calcite negative hai, jaise $θ$ $0$ se $9 0^{\circ}$ tak badhta hai, index $1.658$ se neeche $1.486$ ki taraf slide hona chahiye. Toh $3 0^{\circ}$ par ek value expect karo jo range ke upar ke hisse ke kareeb ho.

Step 1. $cos^{2} 3 0^{\circ} = 0.75$ , $sin^{2} 3 0^{\circ} = 0.25$ . Yeh step kyun? $cos 3 0^{\circ} = \frac{3}{2}$ toh $cos^{2} = \frac{3}{4}$ ; dono ka sum $1$ hona chahiye, isliye $sin^{2} = \frac{1}{4}$ milta hai. Step 2. $n_{o}^{2} = 1.65 8^{2} = 2.749$ , $n_{e}^{2} = 1.48 6^{2} = 2.208$ . Step 3. $\frac{1}{n _{e} ( 3 0 ^{\circ} ) ^{2}} = \frac{0.75}{2.749} + \frac{0.25}{2.208} = 0.2728 + 0.1132 = 0.3860.$ Yeh step kyun? Dono ratios ka seedha substitution. Step 4. $n_{e} (3 0^{\circ}) = 1/ 0.3860 = 1.609$ .

Verify: $1.486 < 1.609 < 1.658$ ✅, aur $1.609$ , $4 5^{\circ}$ ke answer ( $1.565$ parent note se) ke comparison mein $1.658$ ke zyada kareeb hai — index sach mein negative crystal mein $θ$ badhne ke saath neeche slide karta hai. ✅

Worked example Example 5 — Cell F: degenerate crystal (

$n_{e} = n_{o}$ ) Maan lo ek crystal mein $n_{o} = n_{e} = 1.60$ hai (ek accidentally isotropic case). Kisi bhi $θ$ ke liye $n_{e} (θ)$ nikalo.

Forecast: Agar dono principal indices equal hain, toh "lopsidedness" khatam ho jaati hai. Guess: har direction ke liye ek hi index, kahin bhi koi splitting nahi.

Step 1. $n_{o} = n_{e} = n$ ke saath: $\frac{1}{n _{e} ( θ ) ^{2}} = \frac{c o s ^{2} θ}{n ^{2}} + \frac{s i n ^{2} θ}{n ^{2}} = \frac{c o s ^{2} θ + s i n ^{2} θ}{n ^{2}} = \frac{1}{n ^{2}} .$ Yeh step kyun? $cos^{2} θ + sin^{2} θ = 1$ hamesha true hota hai (unit circle par Pythagoras), toh $θ$ -dependence collapse ho jaati hai. Step 2. Sabhi $θ$ ke liye $n_{e} (θ) = n = 1.60$ .

Verify: Koi bhi angle splitting nahi deta — yeh bas ordinary glass hai. Formula correctly isotropic case mein reduce ho jaata hai, jo exactly wahi hai jo "birefringence ko do distinct indices chahiye" demand karta tha. ✅ (Dekhein Dielectric tensor and anisotropic media — kyun equal diagonal entries ka matlab isotropy hai.)

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

$λ = 589$ nm par, principal indices use karte hue ( $θ = 9 0^{\circ}$ , toh $∣Δ n ∣ = 0.172$ ), sabse patli thickness $d$ nikalo jo half-wave retardation deta ho, $Δ ϕ = π$ .

Forecast: Half-wave plate polarization flip karta hai; ise quarter-wave plate se double retardation chahiye (parent Ex 3 ne $π /2$ ke liye $856$ nm diya tha). Guess: lagbhag $2 \times 856 \approx 1712$ nm.

Step 1. Thickness $d$ ke baad phase difference $Δ ϕ = \frac{2 π}{λ} ∣Δ n ∣ d$ hai. Yeh step kyun? Dono rays same physical distance $d$ travel karti hain lekin alag speeds par; optical-path gap $∣Δ n ∣ d$ hai, aur gap ka har wavelength $2 π$ phase ka hota hai. Step 2. $Δ ϕ = π$ set karo aur solve karo: $d = \frac{λ}{2∣Δ n ∣}$ . Step 3. $d = \frac{589 \times 1 0 ^{- 9}}{2 ( 0.172 )} = 1.712 \times 1 0^{- 6} m = 1712$ nm.

Verify: Units: $\frac{m}{(dimensionless)} = m$ ✅. Aur yeh exactly quarter-wave value se double hai — Wave plates (quarter and half wave) ke saath consistent. ✅

Worked example Example 7 — Cell G twist: ek

given plate ka retardation Ek calcite plate $d = 2.0 μ m$ moti hai, principal indices, $λ = 589$ nm. $Δ ϕ$ kya hai, aur kya yeh ek "clean" wave plate hai?

Forecast: $2.0 μ$ m half-wave value ( $1712$ nm) se thoda zyada hai. Toh $Δ ϕ$ $π$ se thoda zyada hona chahiye. Guess: lagbhag $1.1 π$ se $1.2 π$ ke beech.

Step 1. $Δ ϕ = \frac{2 π}{λ} ∣Δ n ∣ d = \frac{2 π ( 0.172 ) ( 2.0 \times 1 0 ^{- 6} )}{589 \times 1 0 ^{- 9}}$ . Yeh step kyun? Same phase formula, ab $Δ ϕ$ ke liye forward solve kar rahe hain. Step 2. Andar numerator: $0.172 \times 2.0 \times 1 0^{- 6} = 3.44 \times 1 0^{- 7}$ . $589 \times 1 0^{- 9} = 5.89 \times 1 0^{- 7}$ se divide karo: ratio $= 0.5840$ . Step 3. $Δ ϕ = 2 π \times 0.5840 = 3.669$ rad $= 1.168 π$ .

Verify: $1.168 π$ , $π$ se thoda upar hai — forecast se match karta hai. Yeh na clean half-wave plate ( $π$ ) hai na quarter-wave ( $π /2$ ); extra $0.168 π$ ka matlab hai output elliptically polarized hai. ✅ (Dekhein Polarization of light.)

Worked example Example 8 — Cell H: e-ray

walk-off angle Calcite mein $\hat{k}$ aur optic axis ke beech $θ = 4 5^{\circ}$ par, e-ray ki energy $\hat{k}$ se $α$ angle par travel karti hai, jo is formula se milta hai: $tan (θ + α) = (\frac{n _{o}}{n _{e}})^{2} tan θ .$ $α$ nikalo. (Yahi $α$ literally woh reason hai kyun e-ray "extraordinary" hai.)

Forecast: Kyunki calcite negative hai ( $n_{o} > n_{e}$ ), $(n_{o} / n_{e})^{2} > 1$ , toh ray direction $\hat{k}$ se aage tilt hoti hai — kuch degrees ka ek chota sa positive walk-off expect karo.

Step 1. $(n_{o} / n_{e})^{2} = (1.658/1.486)^{2} = (1.11575)^{2} = 1.24489$ . Yeh step kyun? Yeh ratio "anisotropy strength" hai; $> 1$ ka matlab hai e-ray $\hat{k}$ se door jhukti hai. Step 2. $tan 4 5^{\circ} = 1$ , toh $tan (4 5^{\circ} + α) = 1.24489$ . Step 3. $4 5^{\circ} + α = arctan (1.24489) = 51.22 9^{\circ}$ , isliye $α = 6.22 9^{\circ}$ . Yeh step kyun? $arctan$ jawaab deta hai "kaun sa angle is tangent ko rakhta hai?" — yeh left side ke tan ko undo karta hai.

Verify: $α \approx 6. 2^{\circ}$ , ek chota sa positive tilt — forecast se aur us famous "double image shift" se match karta hai jo tum calcite crystal ke through dekhte ho. Agar hum positive crystal ( $n_{o} < n_{e}$ ) use karte toh ratio $< 1$ hota aur $α$ negative hota (doosri taraf tilt karta). ✅

Worked example Example 9 — Cell I: real-world word problem (kaun sa crystal zyada split karta hai?)

Tumhe ek Nicol prism ke liye $1.0$ mm path par do polarizations ko physically separate karna hai, aur tum calcite ( $Δ n = - 0.172$ ) ya quartz ( $Δ n = + 0.009$ ) choose kar sakte ho. Kaun sa rays ke beech zyada optical-path separation deta hai, aur woh kitna bada hai?

Forecast: Separation $∣Δ n ∣$ ke saath badhti hai. Calcite ka $0.172$ , quartz ke $0.009$ ko bahut peeche chhod deta hai, toh calcite roughly $0.172/0.009 \approx 19 \times$ se jeetta hai. Guess: calcite lagbhag $0.17$ mm ka path gap deta hai.

Step 1. Optical-path gap $= ∣Δ n ∣ d$ . Yeh step kyun? Wave plate jaisa hi optical-path idea, lekin ab hum un dono rays ke physical separation ki parwah karte hain jo accumulate hoti hai. 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 ~19× zyada separate karta hai — exactly isliye real Nicol prisms calcite se kaate jaate hain, quartz se nahi. ✅

Worked example Example 10 — Cell J: exam twist (angle ke liye

backwards solve karo) Calcite mein tum measure karte ho ki e-ray index $n_{e} (θ) = 1.600$ feel kar raha hai. Optic axis se woh kis angle $θ$ par travel kar raha hai?

Forecast: $1.600$ , $1.486$ aur $1.658$ ke beech hai, toh ek valid $θ$ exist karta hai. Yeh $n_{o} = 1.658$ ke kareeb hai, aur negative crystal mein high index ka matlab chota $θ$ hai — $4 5^{\circ}$ se kuch kam expect karo.

Step 1. Workhorse rearrange karo. $cos^{2} θ = 1 - sin^{2} θ$ use karte hue: $\frac{1}{n _{e} ( θ ) ^{2}} = \frac{1}{n _{o}^{2}} + sin^{2} θ (\frac{1}{n _{e}^{2}} - \frac{1}{n _{o}^{2}}) .$ Yeh step kyun? Sab kuch $sin^{2} θ$ par collect karne se hum ek unknown isolate kar sakte hain. Step 2. Numbers plug karo: $\frac{1}{n _{e} ( θ ) ^{2}} = 1/2.56 = 0.390625$ ; $\frac{1}{n _{o}^{2}} = 0.363805$ ; $\frac{1}{n _{e}^{2}} - \frac{1}{n _{o}^{2}} = 0.452899 - 0.363805 = 0.089094$ . Step 3. $sin^{2} θ = \frac{0.390625 - 0.363805}{0.089094} = \frac{0.026820}{0.089094} = 0.30103$ . Step 4. $sin θ = 0.30103 = 0.54866$ , toh $θ = arcsin (0.54866) = 33.2 7^{\circ}$ . Yeh step kyun? $arcsin$ jawaab deta hai "kaun sa angle is sine ko rakhta hai?" — woh inverse jo $sin θ$ isolate karne ke baad chahiye.

Verify: $θ = 33.2 7^{\circ}$ forward formula mein re-substitute karo: $cos^{2} = 0.699$ , $sin^{2} = 0.301$ , jo deta hai $1/ n_{e}^{2} = 0.699/2.749 + 0.301/2.208 = 0.2543 + 0.1363 = 0.3906$ , toh $n_{e} = 1.600$ ✅. Aur $33.2 7^{\circ} < 4 5^{\circ}$ , forecast se match karta hai. ✅

Recall Matrix par quick self-test

$θ = 0$ par e-ray index kiske barabar hota hai? ::: $n_{o}$ (optic axis ke saath koi splitting nahi) Negative crystal mein, $θ$ $0 \to 9 0^{\circ}$ jaane par $n_{e} (θ)$ badhta hai ya girta hai? ::: Girta hai ( $n_{o}$ se neeche $n_{e}$ tak) Agar $n_{o} = n_{e}$ ho, toh $n_{e} (θ)$ kya hai? ::: Sabhi $θ$ ke liye ek constant $n$ — crystal isotropic hai Wave plate kitna mota hona chahiye yeh kya control karta hai? ::: $∣Δ n ∣$ aur target $Δ ϕ$ : $d = λ Δ ϕ / (2 π ∣Δ n ∣)$ E-ray "walk off" kyun karta hai? ::: Iska energy flow $\hat{k}$ se $α$ tilt karta hai kyunki $(n_{o} / n_{e})^{2} \neq = 1$