Visual walkthrough — Birefringence — ordinary and extraordinary rays

2.5.18 · D2 · Physics › Optics › Birefringence — ordinary and extraordinary rays

Yeh page Birefringence — ordinary and extraordinary rays ka visual companion hai. Yeh Dielectric tensor and anisotropic media pe lean karta hai — kyun crystal ke paas do indices hain — aur Wave plates (quarter and half wave) mein pay off hota hai.

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Step 0 — Woh teen words jo hum use kar sakte hain

Kisi bhi equation se pehle, aao har symbol ko plain language mein pin down karein aur har ek ko ek picture se jodein.

Neeche sab kuch ek kahani hai ki yeh do fixed numbers, $n_o$ aur $n_e$, kaise blend hote hain jab angle $\theta$, $0$ se $90^{\circ }$ tak swing karta hai.

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Step 1 — Do "stiffnesses" ko do lengths ki tarah draw karo

KYA. Hum optic axis ko vertically ($z$) aur sideways directions ko horizontally ($x,y$) rakhte hain. $z$ ke along crystal $n_e$ index ke saath respond karta hai; sideways $n_o$ ke saath.

KYU. Dielectric tensor and anisotropic media humein batata hai ki crystal ke electrons springs pe baithe hain: ek taraf stiff, doosri taraf soft. Stiffness $\to$ permittivity $\to$ index. Toh koi bhi geometry karne se pehle, physics ne humein exactly do numbers de diye hain, ek per direction. Hume bas ek aisi picture chahiye jo dono store kare.

PICTURE. Centre se do arrows: ek chhota/lamba wala axis ke upar ($n_e$ length ka) aur ek sideways ($n_o$ length ka). Do lengths, abhi kuch aur nahi.

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Step 2 — Un do lengths ko ek ellipse mein grow karo

KYA. Do lengths ko centre ke around sweep karo. Jahan direction horizontal hai, radius $n_o$ hai; jahan vertical hai, radius $n_e$ hai; beech mein, yeh smoothly interpolate karta hai. Result ek ellipse hai (3D mein, ek ellipsoid) — index ellipsoid.

KYU. Hum sirf do directions nahi chahte; hum ek aisa rule chahte hain jo har possible wiggle direction ke liye ek index de. Ellipse sabse simple curve hai jo sideways $n_o$ aur vertically $n_e$ ko hit karta hai aur beech mein blend karta hai. Yeh akela curve ab ek saath sab directions ka jawab store karta hai.

PICTURE. Neeche ka ellipse: horizontal semi-axis $=n_o$, vertical semi-axis $=n_e$. Curve pe har point ek possible wiggle direction hai; centre se uski doori us wiggle ke liye index hai.

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Step 3 — Light ko point karo, aur ellipsoid ko slice karo

KYA. Ek travel direction $\hat{k}$ choose karo jo optic axis se angle $\theta$ banata ho. Ellipsoid ko us flat plane se kato jo centre se guzre aur $\hat{k}$ ke perpendicular ho. Kaat ek ellipse hai.

KYU. Light wave ka wiggle ($\vec{E}$, aur uska partner $\vec{D}$) travel direction ke perpendicular hona chahiye — yahi "transverse wave" ka matlab hai (dekho Polarization of light). Toh jo bhi wiggle directions wave allowed hai use karne ke liye, woh sab usi perpendicular plane mein hain. Ellipsoid ko us plane se slice karne se exactly woh allowed directions collect ho jaati hain.

PICTURE. Travel arrow $\hat{k}$ vertical se $\theta$ tilt karta hai. Grey disk perpendicular slice hai; uski rim allowed wiggle directions ka chhota ellipse hai.

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Step 4 — Do semi-axes = do rays

KYA. Kisi bhi ellipse mein ek sabse lamba radius aur ek sabse chhota radius hota hai — uske do semi-axes, right angles pe. Hamare sliced ellipse pe yeh do lengths do allowed indices hain: ek o-ray hai, doosra e-ray hai.

KYU. Nature sirf usi wave ko travel karne deti hai jiska wiggle is slice ke ek semi-axis ke along ho — woh "stable" polarizations hain jo apni shape chalte waqt maintain karti hain. Toh ek akela beam do independent waves mein forced ho jaata hai, har ek ki apni index hoti hai. Yahi woh splitting hai jiske naam pe birefringence rakha gaya hai.

PICTURE. Slice pe: ek semi-axis hamesha horizontal rehta hai (uski length exactly $n_o$ hai — yeh o-ray hai); perpendicular semi-axis tilted hai aur uski koi beech wali length $n_e(\theta )$ hai — yeh e-ray hai.

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Step 5 — E-ray ke liye right triangle set up karo

KYA. E-ray ki wiggle direction $\vec{D}$ vertical ($xz$) plane mein hoti hai aur sideways direction se angle $\theta$ banati hai — equivalent mein, uska perpendicular axis se $\theta$ banata hai. Is unit direction ko ek piece along $n_o$-direction aur ek piece along $n_e$-direction mein resolve karo.

KYU. Humein e-ray ke index ke liye ek number chahiye. Ellipse ko pata hai kaise jawab dena hai agar hum usse apne do natural axes mein split direction feed karein. Toh hum e-ray direction ko do axes pe project karte hain — woh projection sirf ek right triangle pe trigonometry hai.

PICTURE. Ek unit arrow angle $\theta$ pe; do legs pane ke liye perpendiculars daalo: horizontal ($n_o$) axis ke along $\cos \theta$ aur vertical ($n_e$) axis ke along $\sin \theta$.

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Step 6 — Pieces ko inverse squares ki tarah combine karo

KYA. Un do legs ko ellipse mein feed karo. Semi-axes $n_o$ aur $n_e$ wale ek ellipse pe ek point, jiska direction $\cos \theta$ aur $\sin \theta$ components rakhe, distance $n_e(\theta )$ pe baitha hai jo ellipse rule obey karta hai: $\frac{1}{n_e(\theta {)}^2}=\frac{{\cos }^2\theta }{n_o^2}+\frac{{\sin }^2\theta }{n_e^2}.$

KYU. Step 2 ki ellipse equation ne kaha tha "$(\text{sideways}{)}^2/n_o^2+(\text{axis}{)}^2/n_e^2=1$". Direction $(\cos \theta ,\sin \theta )$ mein distance $r=n_e(\theta )$ pe ek point ka sideways part $r\cos \theta$ aur axis part $r\sin \theta$ hota hai. Inhe plug karo, aur $r^2$ divide ho jaata hai left pe sirf $1/r^2$ chhod ke — exactly woh boxed formula. Inverse squares magic nahi hain; woh ellipse ke apne $1/n^2$ weights hain.

PICTURE. Final ellipse jisme e-ray radius drawn hai aur uski length $n_e(\theta )$ label ki gayi hai, $n_e$ aur $n_o$ ke beech mein baitha hua.

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Step 7 — Edge cases (koi corner kabhi mat chhodo)

KYA. Do extreme angles test karo aur confirm karo ki picture sahi behave karti hai.

KYU. Woh formula jise tum apne endpoints pe check nahi kar sakte, woh ek aisa formula hai jis par tum trust nahi karte. Yeh do limits $n_o$, $n_e$ aur optic axis ki definitions reproduce karni chahiye.

PICTURE. Left panel: $\theta =0$, slice horizontal hai, dono semi-axes $n_o$ ke barabar hain — ek circle, koi splitting nahi. Right panel: $\theta =90^{\circ }$, slice vertical hai, e-ray semi-axis full height $n_e$ tak pahunchti hai — maximum splitting.

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Ek-picture summary

KYA. Ek figure jo poori chain carry karta hai: ellipse (semi-axes $n_o$, $n_e$), ek tilted $\hat{k}$, $\theta$ triangle, aur e-ray radius $n_e(\theta )$ dono ke beech mein land karta hua. Ise left se right padhna hi derivation hai.

Recall Feynman: plain words mein poori walk

Ek squashed balloon ki picture banao — side-to-side round (radius $n_o$), upar-neeche thoda flatter (radius $n_e$). Woh balloon hi crystal ki memory hai ki "har wiggle direction ke liye light kitni slow hai." Ab usme se kisi slant angle $\theta$ pe ek beam shoot karo. Beam sirf usi flat plate mein wiggle kar sakta hai jo seedha uske path ke across baitha ho, toh us plate se balloon ko slice karo. Kaat ek chhoti oval ban jaati hai jisme ek lamba side aur ek chhota side hota hai — woh do sides hi sirf do allowed ways hain jisme light wiggle kar sakti hai. Ek side hamesha flat rehti hai aur fixed length $n_o$ maintain karti hai: yeh shant ordinary ray hai. Doosri side beam ke saath tip karti hai, aur uski length $n_o$ aur $n_e$ ka ek blend hai jo sirf angle se set hoti hai — yeh restless extraordinary ray hai. Tilted direction ko ek "sideways" part ($\cos \theta$) aur ek "up-the-axis" part ($\sin \theta$) mein split karo, balloon ko har part ko apne $1/n^2$ se weight karne do, unhe add karo, aur $\frac{1}{n_e(\theta {)}^2}=\frac{{\cos }^2\theta }{n_o^2}+\frac{{\sin }^2\theta }{n_e^2}$ nikal aata hai. Beam ko seedha axis ke upar point karo aur blend $n_o$ wapas deta hai — do rays fuse ho jaati hain. Ise dead sideways point karo aur tumhe full $n_e$ milta hai — maximum split. Birefringence jo kuch bhi karta hai woh sab us ek squashed balloon mein rehta hai.

Recall Quick self-test

O-ray index sabhi $\theta$ ke liye constant kyun hoti hai? ::: Uska wiggle semi-axis hamesha $xy$-plane mein flat rehta hai, kabhi optic axis ki taraf tilt nahi karta, toh hamesha $n_o$ ke barabar hota hai. Formula mein ${\cos }^2\theta$ kya weight karta hai? ::: E-ray index mein sideways ($n_o$) contribution. $\theta =0$ pe $n_e(\theta )$ kya hai? ::: Exactly $n_o$ — optic-axis case jisme koi splitting nahi. $n_e(\theta )$ hamesha kin do values ke beech trapped rehta hai? ::: $n_o$ aur $n_e$ ke beech.