Huygens' principle — wavefront propagation: Visual walkthrough

2.5.10 · D2 · Physics › Optics › Huygens' principle — wavefront propagation

Step 1 — Asal mein kya move kar raha hai? Wavefront draw karo

KYA. Kisi bhi bending se pehle, hume woh cheez chahiye jo bend hoti hai. Wavefront ek line hai (2D mein) jo wave ke un sab points ko join karti hai jo "ek hi waqt mein ek hi kaam kar rahe hain" — sab saath mein crest ho rahe hain. Ek seedhi rank of soldiers ko imagine karo jo step mein march kar rahe hain: woh rank hi wavefront hai.

KYO. Hum is equal-phase line ko track karte hain, individual light "particles" ko nahi, kyunki Huygens ki poori trick yahi hai: is line ka har point ek chota naya source hai. Agar hume pata hai abhi line kahan hai, toh hum build kar sakte hain ki aage woh kahan hogi.

PICTURE. Blue line $A B$ hamari wavefront hai. Pale-yellow arrow, jo iske perpendicular draw hua hai, woh ray hai — woh direction jisme poori rank march karti hai. Yaad rakho: ray aur wavefront hamesha right angles par hote hain (kabhi parallel nahi).

Step 2 — Isko march karne do: distance = speed × time

KYA. Wave ko ek stopwatch do. Time $t$ mein rank ek distance aage badhti hai. Agar wave medium 1 mein speed $v_{1}$ (metres per second) se move karti hai, toh $t$ seconds mein woh cover karti hai

$d = medium 1 mein speed v_{1} \times gaya hua time t$

KYO. Speed ka matlab hi hai "distance per unit time", toh distance bas speed times time hai. Yahi ek physics input hai — iske baad sab geometry hai.

PICTURE. Do blue lines: purani wavefront aur, $v_{1} t$ aage, nayi wavefront. Woh parallel rehti hain kyunki har point ne same $v_{1} t$ march kiya. Yahi parent note ka "plane stays plane" hai.

Step 3 — Wavefront ko ek boundary par tilt karo

KYA. Ab marching rank ko ek slanted line $X Y$ se takrao — medium 1 (upar, speed $v_{1}$ ) aur medium 2 (neeche, slower, speed $v_{2}$ ) ke beech ki surface. Kyunki wavefront slanted hai, ek end — isko $A$ kaho — surface par pehle pahunchta hai, jabki doosra end abhi medium 1 mein upar hai, ek point par jo hum $B$ kehte hain.

KYO. Agar wavefront surface se bilkul seedha takraati, toh dono end saath cross karte aur kuch interesting nahi hota. Tilt hi woh cheez hai jo ek end ko rukwaata hai jabki doosra cross karta hai — aur rukna hi light ko bend karwata hai.

PICTURE. $A$ boundary par baitha hai. $B$ abhi uske upar hang kar raha hai. Incoming wavefront aur surface ke beech ka angle (equivalently, incoming ray aur normal ke beech ka angle, $X Y$ par dashed perpendicular) angle of incidence $i$ hai.

Step 4 — Race: end $B$ medium 1 mein finish karta hai jabki end $A$ medium 2 mein start karta hai

KYA. Ek hi time interval $t$ mein do cheezein hoti hain:

End $B$ , abhi medium 1 mein, surface par ek point $C$ tak neeche daurta hai. Woh travel karta hai $B C = v_{1} t (v_{1} = fast top speed) .$
End $A$ , already cross kar chuka, apna Huygens wavelet medium 2 mein aage bhejta hai. Usi $t$ mein woh sirf itna phailta hai $A D = v_{2} t (v_{2} = slow bottom speed, so A D < B C) .$

KYO. Yahi poori baat ki jaan hai. $A$ aur $B$ ko ek hi clock time $t$ milta hai, lekin $A$ slow medium mein hai aur $B$ fast mein. Same time, alag speeds ⟹ alag distances. Yahi mismatch bend hai.

PICTURE. $B C$ ek lamba blue segment hai (fast). $A D$ radius $v_{2} t$ ka ek chota pink arc hai — $A$ se secondary wavelet (Huygens: har point ek source hai). Kyunki $v_{2} < v_{1}$ , arc $B C$ se chhota hai.

Step 5 — Nayi wavefront draw karo: common tangent

KYA. Time $t$ tak, end $B$ abhi $C$ par pahuncha hai (toh uska wavelet wahan zero radius ka hai), jabki $A$ ka wavelet radius $A D = v_{2} t$ tak badh gaya hai. $A$ aur $C$ ke beech ka har point kisi beech ke moment mein cross hua, toh uske wavelet ka radius beech-beech ka hai. Medium 2 mein nayi wavefront woh seedhi line hai $C$ se jo in sab wavelets ki tangent hai — yeh $A$ ke circle ko exactly $D$ par touch karti hai.

KYO. Huygens' rule: agla wavefront sab secondary wavelets ka common tangent (envelope) hota hai. Tangent se door, wavelets out of step hain aur cancel karte hain; tangent par woh reinforce karte hain.

PICTURE. Refracted wavefront $C D$ (blue) $D$ par pink wavelet ke saath lean karti hai, $D$ par right angle ke saath ( $A D ⊥ C D$ , kyunki tangent, touch point par radius ke perpendicular hoti hai).

Recall

$A D$ kyun $C D$ ke perpendicular hai? Ek tangent line circle ko exactly ek point par touch karti hai, aur us point par radius hamesha tangent se $9 0^{\circ}$ par milta hai. Yahan $C D$ , $A$ ke wavelet par $D$ par tangent hai, isliye $A D ⊥ C D$ . ::: Yahi right angle hai jo triangle $A C D$ ko Step 6 mein right triangle banata hai. Refracted wavefront surface par kis point se pin hoti hai? ::: Point $C$ se, jahan end $B$ abhi pahuncha hai.

KYA. Surface segment $A C$ ke saath milne wale do triangles dekho:

$△ A B C$ : right angle $B$ par, hypotenuse $A C$ , aur side $B C = v_{1} t$ , $A$ par angle ke opposite hai, jo $i$ ke barabar hai.
$△ A D C$ : right angle $D$ par, hypotenuse $A C$ (wahi same segment), aur side $A D = v_{2} t$ , $C$ par angle ke opposite hai, jo angle of refraction $r$ ke barabar hai.

Ab sine ki definition use karo — "opposite over hypotenuse":

$sin i = \frac{opposite}{hyp} = \frac{B C}{A C} = \frac{v _{1} t}{A C}, sin r = \frac{A D}{A C} = \frac{v _{2} t}{A C} .$

KYO sine, aur kyun yeh do angles? Sine ek angle ko sides ke ratio mein convert karta hai, jo exactly woh language hai jo triangle bolti hai. Hum $i$ aur $r$ choose karte hain kyunki dono angles ek hi hypotenuse $A C$ par ride karte hain — toh jab hum divide karte hain, $A C$ cancel ho jaata hai aur messy geometry gayab ho jaati hai.

PICTURE. Dono triangles shared base $A C$ par draw hain: yellow highlights do "opposite" sides $B C$ aur $A D$ ; angle $i$ , $A$ par rehta hai, angle $r$ , $C$ par.

Step 7 — Divide karo, aur Snell's law nikal aata hai

KYA. Do sine expressions ko divide karo. Common time $t$ aur common hypotenuse $A C$ dono cancel ho jaate hain:

$\frac{s i n i}{s i n r} = \frac{B C / A C}{A D / A C} = \frac{B C}{A D} = \frac{v _{1} t}{v _{2} t} = \frac{v _{1}}{v _{2}} .$

KYO. Har woh cheez jo depend karti thi ki wave kitni door ya kitne time mein travel ki ( $t$ , $A C$ ) gayab ho gayi. Jo bacha woh sirf do speeds ka ratio hai — do media ki fixed property. Ek constant ratio exactly wahi hai jo ek "law" hota hai.

$n = \frac{c}{v}$ use karte hue (refractive index = vacuum mein light speed ÷ medium mein speed):

PICTURE. Final annotated diagram: incident ray, bent refracted ray, aur boxed result — dekho Snell's Law and Refractive Index aur Phase and Path Difference jahan yeh aage feed hota hai.

Step 8 — Edge cases (reader ko kabhi stranded mat chodo)

KYA & KYO & PICTURE, teeno ek saath, un situations ke liye jo clean derivation mein gloss over ho gayi thi:

Seedha incidence ( $i = 0$ ). Wavefront surface ke parallel, dono ends saath cross karte hain, koi tilt develop nahi hota. Tab $sin i = 0 \Rightarrow sin r = 0 \Rightarrow r = 0$ . Light seedhi nikal jaati hai, bina bend ke — lekin andar slower zaroor hoti hai.
Same medium ( $v_{1} = v_{2}$ ). Ratio $1$ hai, toh $sin i = sin r$ , $i = r$ : koi bending nahi. Koi boundary hi nahi thi jis par bend karna worth it ho.
Slow → fast ( $v_{2} > v_{1}$ ). Ab $A D > B C$ : wavelet end $B$ se aage nikal jaata hai. Toh $sin r > sin i$ , $r > i$ : ray normal se door bend karti hai.
Critical angle & total internal reflection. Slow→fast jaate hue, $i$ badhate jao. Required $sin r = \frac{v _{2}}{v _{1}} sin i$ $1$ ki taraf badhta hai. Critical angle $i_{c}$ par jahan $sin i_{c} = \frac{v _{1}}{v _{2}}$ , hume milta hai $r = 9 0^{\circ}$ — refracted wavefront surface ke saath-saath skim karta hai. $i_{c}$ se aage, $sin r > 1$ impossible hai: medium 2 mein koi tangent wavefront draw nahi ho sakta, toh saari light reflect ho jaati hai. (Dekho Laws of Reflection.)

Ek-picture summary

Sab ek slate par: tilted incident wavefront $A B$ , race ( $B C = v_{1} t$ lamba, $A D = v_{2} t$ chhota), do right triangles jo hypotenuse $A C$ share karte hain, aur bent wavefront $C D$ — neeche boxed law ke saath.

Recall Feynman: poora walkthrough simple words mein

Doston ki ek line step mein march karti hai (wavefront). Woh ek keechi (muddy) field tak pahunchti hai jo slant par kata hai. Mud ke sabse paas ka dost pehle andar ghusta hai aur slow ho jaata hai; mud mein uske neighbours bhi slow ho jaate hain, lekin firm ground par wale dost sprint karte rehte hain. Usi kuch seconds mein, sprinters lamba stretch cover karte hain jabki mud-walkers thoda hi shuffle karte hain. Toh poori line pivot karti hai — ab woh mud mein aur teedhi taraf aim kar rahi hai. Yahi pivot bend hai. "Sprinter distance = $v_{1} t$ " aur "mud distance = $v_{2} t$ " ek slanted baseline par likho; do right triangles jo woh banate hain baseline share karte hain, toh unke sines ko divide karne se baseline aur time cancel ho jaate hain, sirf $v_{1} / v_{2}$ bachta hai. Slow karne ki jagah speed up karo (mud → firm ground) aur line doosri taraf pivot karti hai. Itna slow karo ki koi forward line ban hi na sake, aur poori rank bounce back karti hai — total internal reflection.

Connections

Huygens' principle — wavefront propagation (parent)
Snell's Law and Refractive Index
Laws of Reflection
Phase and Path Difference
Wave Optics — Interference
Young's Double Slit Experiment
Diffraction