Exercises — Birefringence — ordinary and extraordinary rays

• Jo ray Snell's law follow karti hai woh ordinary ray (o-ray) hai. Uski electric field $\vec{E}$ principal plane ke perpendicular hai, isliye $\vec{E}$ kabhi optic axis ki taraf nahi jhukti aur hamesha single index $n_o$ feel karti hai.
• Jo ray ordinary Snell's law violate karti hai woh extraordinary ray (e-ray) hai. Uski $\vec{E}$ principal plane mein hoti hai, isliye jab direction badlti hai toh $\vec{E}$ aur optic axis ke beech ka angle badlta hai, aur uska index $n_e(\theta )$ uske saath badlta rehta hai.

Negative crystal ki defining condition hai $n_e<n_o$. Yahan $1.486<1.658$ hai, isliye calcite negative hai. (Positive hota agar $n_e>n_o$, jaise quartz mein hota hai.)

Optic axis ke along. Wahan $\theta =0$ hai, aur formula deta hai $\frac{1}{n_e(0{)}^2}=\frac{{\cos }^{2}0}{n_o^2}+\frac{{\sin }^{2}0}{n_e^2}=\frac{1}{n_o^2}\Rightarrow n_e(0)=n_o.$ Dono rays same index feel karti hain, same speed se travel karti hain, aur kabhi alag nahi hoti.

Step 1 — formula likho. $\frac{1}{n_e(\theta {)}^2}=\frac{{\cos }^2\theta }{n_o^2}+\frac{{\sin }^2\theta }{n_e^2}$. Step 2 — trig evaluate karo. ${\cos }^{2}30^{\circ }=0.75$, ${\sin }^{2}30^{\circ }=0.25$. Step 3 — plug in karo. $n_o^2=2.749$, $n_e^2=2.208$: $\frac{1}{n_e^2}=\frac{0.75}{2.749}+\frac{0.25}{2.208}=0.27282+0.11322=\mathrm{0.38604.}$ Step 4 — invert karo. Formula humein $1/n_e^2$ deta hai, lekin hum $n_e$ chahte hain. Yeh do operations kyun: "square then reciprocal" ko undo karne ke liye hum inverse operations ko ulte order mein apply karte hain — pehle reciprocal lete hain taaki $n_e^2$ mile, phir square root lete hain taaki $n_e$ mile. Dono ek saath karte hue, $n_e(30^{\circ })=1/\sqrt{0.38604}=1.6094$. Sanity check: yeh $n_e=1.486$ aur $n_o=1.658$ ke beech mein hai — jaisa ki kisi bhi $n_e(\theta )$ ka hona zaroori hai.

Step 1 — phase formula. $\Delta \varphi =\frac{2\pi }{\lambda }|\Delta n|d$. Kyun: har ray phase accumulate karti hai $\frac{2\pi }{\lambda }nd$; subtract karne par $|\Delta n|$ se driven difference milta hai. Step 2 — plug in karo. $\Delta \varphi =\frac{2\pi }{589\times 10^{-9}}(0.172)(30\times 10^{-6})=\frac{2\pi (0.172)(30\times 10^{-6})}{589\times 10^{-9}}.$ Step 3 — evaluate karo. Numerator $2\pi (0.172)(30\times 10^{-6})=3.2421\times 10^{-5}$; divide by $589\times 10^{-9}$: milta hai $\Delta \varphi =55.04$ rad. Wavelengths mein yeh hai $\Delta \varphi /2\pi =8.76$ full cycles.

Step 1. $\frac{2\pi }{\lambda }|\Delta n|d=\pi \Rightarrow d=\frac{\lambda }{2|\Delta n|}$ set karo. Step 2. $d=\frac{589\times 10^{-9}}{2(0.172)}=1.712\times 10^{-6}\text{m}=1712$ nm. Note karo ki yeh parent note mein quarter-wave thickness (856 nm) ka bilkul do guna hai — jaisa hona chahiye, kyunki $\pi$ exactly $\pi /2$ ka do guna hai. Dekho Wave plates (quarter and half wave).

Reasoning. Rewrite karo $\frac{1}{n_e(\theta {)}^2}=\frac{{\cos }^2\theta }{n_o^2}+\frac{{\sin }^2\theta }{n_e^2}$. Jab $\theta$ badhta hai, ${\cos }^2\theta$ ($1/n_o^2$ par weight) girta hai aur ${\sin }^2\theta$ ($1/n_e^2$ par weight) badhta hai. Calcite ke liye $n_e<n_o$, isliye $1/n_e^2>1/n_o^2$: hum weight ko bade term par shift kar rahe hain, isliye $1/n_e(\theta {)}^2$ increase hoga, matlab $n_e(\theta )$ itself decrease karega. Endpoints. $\theta =0$: $n_e(0)=n_o=1.658$ (sabse bada). $\theta =90^{\circ }$: $n_e(90^{\circ })=n_e=1.486$ (sabse chhota). Figure kaise padhein. Horizontal axis $\theta$ hai $0^{\circ }$ se $90^{\circ }$ tak; vertical axis e-ray index hai. Solid red curve $n_e(\theta )$ hai: iski left end se shuru karo aur yeh upper blue dashed line par $1.658$ par baith ti hai (woh dashed line $n_o$ mark karti hai); ise rightward trace karo aur yeh monotonically slide karte hue lower green dashed line par $1.486$ par land karti hai (woh $n_e$ mark karti hai). $30^{\circ }$ aur $60^{\circ }$ par yellow dots Problems 2.1 aur 3.2 mein compute kiye gaye points hain — notice karo ki $60^{\circ }$ wala dot $30^{\circ }$ wale se neeche hai, jo confirm karta hai ki curve girti rehti hai. Kyunki red curve kabhi kahin rise nahi karti, "decreases" ka jawab ek nazar mein dikhta hai.

Step 1 — $n_e(60^{\circ })$ nikalo. ${\cos }^{2}60^{\circ }=0.25$, ${\sin }^{2}60^{\circ }=0.75$: $\frac{1}{n_e^2}=\frac{0.25}{2.749}+\frac{0.75}{2.208}=0.09094+0.33967=0.43061,n_e(60^{\circ })=\mathrm{1.5240.}$ Step 2 — compare karo. O-ray index $=n_o=1.658$; e-ray index $=1.524$. Chhota index → zyada speed. Kyunki $1.524<1.658$ hai, e-ray faster hai yahan. Big picture: negative crystal mein e-ray har $\theta \ne 0$ ke liye faster hota hai (lower index); exactly yahi "negative" encode karta hai.

$\theta =0$ par dono indices coincide karte hain ($n_e(0)=n_o$), isliye $\delta (0)=0$ — koi splitting nahi, yeh direction optic axis hi hai. Jab $\theta$ badhta hai, $n_e(\theta )$ $n_o$ se door $n_e$ ki taraf jaata hai, aur gap monotonically badhta hai (Problem 3.1 se). Yeh tab maximum hota hai jab $n_e(\theta )$ apni extreme principal value tak pahunch jaati hai $\theta =90^{\circ }$ par: $\delta (90^{\circ })=|n_o-n_e|=|1.658-1.486|=\mathrm{0.172.}$

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. Yeh kya karta hai: $45^{\circ }$ par linear light o- aur e-components mein equally split hoti hai; $\pi /2$ relative delay unka sum circularly polarized light mein badal deta hai. Dekho Wave plates (quarter and half wave) aur Polarization of light.

Step 1 — wavelengths mein path difference. $N=\frac{|\Delta n|d}{\lambda }=\frac{0.172\times 2.30\times 10^{-6}}{589\times 10^{-9}}=0.6716$ wavelengths. Step 2 — phase. $\Delta \varphi =2\pi N=4.220$ rad. Reduce mod $2\pi$: $4.220$ rad already $<2\pi (6.283)$ hai, isliye effective phase $=4.220$ rad $=0.6716\times 2\pi$. Step 3 — compare — aur KYUN. Sirf modulo $2\pi$ wala phase output polarization ko change karta hai, kyunki pura $2\pi$ (ek full wavelength delay) add karne par har wave identical state mein wapas aa jaati hai. Isliye hum poochhte hain ki bacha hua $0.6716\times 2\pi$ kis "clean" plate se milta hai. Half-wave plate $0.5\times 2\pi$ hai; quarter-wave plate $0.25\times 2\pi$ ya equivalently $0.75\times 2\pi$ hai (three-quarter delay = quarter-wave behaviour with fast/slow roles swapped, kyunki $0.75\times 2\pi =2\pi /2+2\pi /4$ ek extra half-wave ko quarter-wave par fold karta hai). Humara bacha hua $0.6716$ sabse close $0.75$ ke hai, distance $0.083$ hai, jabki $0.5$ se $0.17$ hai — isliye plate sabse zyada quarter-wave plate ki tarah behave karti hai, ek added half-wave ke saath.

Step 1. Fast axes aligned hone par, "fast" component dono plates mein fast rehta hai aur "slow" slow rehta hai, isliye phase differences add hote hain: $\Delta {\varphi }_{\text{tot}}=\pi /2+\pi /2=\pi$. Step 2. $\pi$ phase difference ek half-wave plate hai. Isliye do aligned quarter-wave plates = ek half-wave plate. Total thickness $2\times 856=1712$ nm, jo Problem 2.3 se match karta hai. ✔

Step 1 — target. $\frac{1}{n_e(\theta {)}^2}=\frac{1}{1.60^2}=\frac{1}{2.56}=0.390625$. Step 2 — ek smart substitution aur KYUN. Formula mein ${\cos }^2\theta$ aur ${\sin }^2\theta$ dono hain, jo do unknowns ki tarah dikhte hain. Lekin yeh identity ${\cos }^2\theta +{\sin }^2\theta =1$ se lock hain. Isliye set karo ${\sin }^2\theta =s$; toh automatically ${\cos }^2\theta =1-s$. Yeh kyun help karta hai: yeh do trig terms ko ek unknown $s$ mein collapse kar deta hai, equation ko $s$ mein ek plain linear equation bana deta hai — angles directly juggle karne se kahin zyada easy. $0.390625=\frac{1-s}{2.749}+\frac{s}{2.208}.$ Step 3 — linear equation solve karo. Har term expand karo: $\frac{1-s}{2.749}=0.36377-0.36377s$ aur $\frac{s}{2.208}=0.45290s$. Add karo: $0.390625=0.36377+(0.45290-0.36377)s=0.36377+0.08913s.$ $s$ isolate karo: $s=\frac{0.390625-0.36377}{0.08913}=\frac{0.026855}{0.08913}=\mathrm{0.30130.}$ Step 4 — $\theta$ back out karo, aur do roots / domain handle karo. ${\sin }^2\theta =s=0.30130$ se milta hai $\sin \theta =\pm \sqrt{0.30130}=\pm 0.54891$. Kyun hum sirf ek rakhte hain: optic axis se propagation angle physical domain $0^{\circ }\le \theta \le 90^{\circ }$ mein hota hai, jahan $\sin \theta$ kabhi negative nahi hota, isliye negative root discard ho jaata hai. Positive root lete hue, $\theta =\arcsin (0.54891)=33.29^{\circ }$. (Algebraically ${\sin }^2\theta =s$ se $\theta =180^{\circ }-33.29^{\circ }=146.71^{\circ }$ bhi possible hai, lekin woh bhi $0^{\circ }$–$90^{\circ }$ ke bahar hai, isliye $33.29^{\circ }$ unique physical answer hai.) Figure kaise padhein. Same axes jaise Problem 3.1 mein: $\theta$ horizontal, e-ray index vertical, red curve $=n_e(\theta )$. Horizontal yellow dashed line target height $1.60$ par draw ki gayi hai. Ise rightward follow karo jab tak yeh red curve se nahi milti — woh single meeting point (sirf ek hai $0^{\circ }$–$90^{\circ }$ ke andar, kyunki red curve strictly decreasing hai aur isliye kisi bhi horizontal level ko zyada se zyada ek baar cross karti hai) solution hai, green dot se mark kiya gaya hai. Seedha neeche vertical blue dotted line se axis tak jaao: yeh $\theta \approx 33.3^{\circ }$ par land karta hai, algebra confirm karta hai. Kyunki $1.60$ index range ke top ke paas hai (close to $n_o=1.658$), crossing chhote $\theta$ par hoti hai — exactly yahi picture dikhati hai.

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 — kyun fractional errors equal hain. Phase hai $\Delta \varphi =\frac{2\pi }{\lambda }|\Delta n|d=Kd$, jahan $K=\frac{2\pi }{\lambda }|\Delta n|$ ek constant hai (fixed wavelength aur material). Kyunki $\Delta \varphi$ directly $d$ ke proportional hai, agar $d$ mein $\delta d$ ka change ho toh $\Delta \varphi$ mein $K\delta d$ ka change hoga, aur fractional change hai $\frac{\delta (\Delta \varphi )}{\Delta \varphi }=\frac{K\delta d}{Kd}=\frac{\delta d}{d}.$ Constant $K$ cancel ho jaata hai — yahi wajah hai ki dono fractional errors identical hain. Step 3 — evaluate karo. $\frac{\delta d}{d}=\frac{50}{726.7}=0.0688=6.88\%.$ Interpretation: Phase mein $6.9\%$ error ka matlab hai $\Delta \varphi =\pi /2\times (1\pm 0.0688)$, yaani plate thodi imperfect QWP hai aur perfectly circular ki jagah thodi elliptical light produce karti hai.

Step 1 — target. $\frac{1}{1.572^2}=\frac{1}{2.471184}=0.404663$. Step 2 — same substitution $s={\sin }^2\theta$ (toh ${\cos }^2\theta =1-s$, ek linear unknown mein collapse hota hai): $0.404663=\frac{1-s}{2.749}+\frac{s}{2.208}=0.36377+0.08913s.$ Step 3 — $s$ solve karo. $s=\frac{0.404663-0.36377}{0.08913}=\frac{0.040893}{0.08913}=\mathrm{0.45881.}$ Step 4 — angle (sirf physical root). $\sin \theta =+\sqrt{0.45881}=0.67735$ (negative root exclude hai kyunki $0^{\circ }\le \theta \le 90^{\circ }$) $\Rightarrow \theta =\arcsin (0.67735)=42.62^{\circ }.$ Note: Index-average angle ($42.6^{\circ }$) $45^{\circ }$ nahi hai, kyunki index $1/n^2$ ke through combine hota hai, linearly nahi — index mein midpoint angle mein midpoint nahi hota.

Kya tum, ek blank page se: bata sakte ho kaun sa ray Snell follow karta hai (::: o-ray) — e-index formula likh sakte ho (::: $\frac{1}{n_e(\theta {)}^2}=\frac{{\cos }^2\theta }{n_o^2}+\frac{{\sin }^2\theta }{n_e^2}$) — QWP thickness formula de sakte ho (::: $d=\frac{\lambda }{4|\Delta n|}$) — bata sakte ho calcite mein kaun sa ray faster hai (::: e-ray, lower index)? Agar haan, toh tumne D4 master kar liya hai.