Newton's rings — derivation: Visual walkthrough — Hinglish deep dive

2.5.13 · D2 · Physics › Optics › Newton's rings — derivation

Step 1 — Stage set karo: actually kya kya touch kar raha hai?

KYA. Ek plano-convex lens lo (upar se flat, neeche se halki curved — convex) aur use, curved side neeche karke, ek flat glass plate par rakh do. Woh exactly ek point par touch karte hain. Baaki har jagah curved glass aur flat glass ke beech mein ek kaagaz-jaisi patli air ki gap hai.

KYUN. Interference ke liye do surfaces itni close chahiye ki dono se bounce karne wali light "in step" rahe (yeh closeness wala idea hai coherence). Yeh trapped air layer hi hamari thin film hai. Setup mein aur kuch abhi matter nahi karta.

PICTURE.

Cyan curve lens ka neeche wala hissa hai.
White line flat plate hai.
Amber dot single contact point hai — iske through jaane wali vertical line ko axis kaho.
$r$ = axis se horizontal distance (ise hum baar baar use karenge).
$t$ = distance $r$ par air gap ki height. Axis par, $t = 0$ .

Step 2 — Do rays paida hoti hain: A aur B

KYA. Monochromatic light (ek hi colour → ek wavelength $λ$ ) seedha neeche daalo. Lens ke bottom face par beam do mein split ho jaati hai:

Ray A us glass→air surface se seedha wapas reflect hoti hai.
Ray B aage jaati hai, air gap cross karti hai, flat plate ke top (air→glass) se reflect hoti hai, aur wapas upar aati hai.

KYUN. Yeh dono rays ek hi beam se shuru huin, isliye perfectly in step (coherent) hain. Jab yeh tumhari aankhon mein milti hain to ya reinforce karein ya cancel — yahi decision poora pattern hai.

PICTURE.

Ray A (white) — chhota bounce, gap ke top se.
Ray B (amber) — lamba chakkar, gap se neeche aur phir upar.
Shaded band thickness $t$ wali air film hai.

Step 3 — Ray B ne jo extra distance travel ki, use measure karo

KYA. Light ke almost seedha neeche girane par (near-normal incidence), ray B film se neeche jaati hai aur wapas upar aati hai. Woh thickness $t$ ko do baar cross karti hai. To ray A ke upar ray B ka extra geometric travel hai:

$neeche aur upar 2 \times film ka index μ \times film thickness t = 2 μ t$

Har symbol kyun.

2 — kyunki B gap mein round trip karta hai, one-way nahi. Ise bhool jaao to har answer aadha ho jaata hai (ek classic galti).
$μ$ (mu) — gap mein jo bhi hai uska refractive index. Light denser medium mein "distance dheerey count karti hai", isliye hum physical thickness ko $μ$ se weight karte hain. Plain air ke liye, $μ \approx 1$ , to $2 μ t \to 2 t$ .
$t$ — Step 1 wali thickness.

PICTURE.

Amber path thoda tilt draw kiya hai taaki down-leg aur up-leg dikh sakein; har leg $t$ lambi hai, total $2 t$ (times $μ$ ).

Step 4 — Chhupa hua half-wavelength (Stokes' flip)

KYA. Ek aur effect hai jiska distance se koi lena dena nahi. Jab light kisi surface se reflect hoti hai jahan agla medium denser ho (higher index), to wave palat jaati hai — jaise usne aadha wavelength khoya ho. Yeh ek Stokes phase change hai $π$ ka.

Ray A: glass → air reflection. Air kam dense hai → koi flip nahi.
Ray B: plate par air → glass reflection. Glass zyada dense hai → flip, worth $+ \frac{λ}{2}$ .

Dono rays mein se exactly ek flip hoti hai. To total optical path difference hai:

$Δ = distance part (Step 3) 2 μ t + sirf B par Stokes flip \frac{λ}{2}$

Centre par kyun sabse zyada matter karta hai. Contact point par $t = 0$ , distance part zero ho jaata hai lekin $\frac{λ}{2}$ rehta hai. Yeh bacha hua half-wave hi centre ko dark banata hai.

PICTURE.

Left wave: koi flip nahi se reflect hoti hai (crest crest rehti hai).
Right wave: denser glass se reflect ho kar inverted wapas aati hai — amber crest trough ban gayi. Yeh inversion hi extra $λ /2$ hai.

Step 5 — "Path difference" ko bright / dark rules mein badlo

KYA. Do waves reinforce (bright) hoti hain jab B, A se poore wavelengths ki sankhya peeche ho, aur cancel (dark) hoti hain jab lag ek whole-plus-a-half ho.

$Bright: Δ = mλ Dark: Δ = (m + \frac{1}{2}) λ$

Ab $Δ = 2 μ t + \frac{λ}{2}$ substitute karo. Left side ka $\frac{λ}{2}$ dono conditions ko swap kar deta hai:

Kyun swap hi punchline hai. Dark condition mein $t = 0$ daalo: $0 = 0 \cdot λ$ ✓. To exact centre dark rule satisfy karta hai — observed dark spot explain ho gaya, purely Stokes flip se.

PICTURE.

Do ripples ek doosre ke upar draw ki hain: left mein, crest-on-trough = flat line = dark; right mein, crest-on-crest = tall wave = bright.

Step 6 — Geometry link: $t$ kyun $r^{2}$ ki tarah badhta hai

KYA. Lens surface radius $R$ wale sphere ka hissa hai. Sphere ko plate par rakh do; contact point se horizontal distance $r$ par, surface height $t$ tak uth jaati hai. Sphere par Pythagoras deta hai:

$r^{2} = t (2 R - t)$

Kyunki gap microns mein hai aur $R$ metres mein, $t ≪ R$ , isliye $2 R - t \approx 2 R$ :

$t = \frac{r ^{2}}{2 R}$

Har symbol kyun.

$R$ — jis sphere se lens grind ki gayi uska radius; bada hota hai ( $\sim$ ek metre).
Poori form $r^{2} = t (2 R - t)$ intersecting-chord theorem se aati hai: $2 R$ lambi vertical chord $t$ aur $(2 R - t)$ mein split hoti hai, aur iske across half-chord $r$ hai.
$2 R$ ke saath $t$ ko drop karna key approximation hai — valid hai kyunki $t / R \sim 1 0^{- 6}$ .

Answer ki important shape: $t$ , $r$ ke proportional nahi, $r^{2}$ ke proportional hai. Double radius → chaar guna gap.

PICTURE.

Sphere ke andar right triangle: horizontal leg $r$ , aur chhota sa sagitta $t$ surface tak upar jaata hua. Dekho curve centre ke paas kitna flat hai — isliye $t$ pehle bahut chhota rehta hai aur phir speed up karta hai.

Step 7 — Geometry ko ring rule mein daalo → payoff formulas

KYA. $t = \frac{r ^{2}}{2 R}$ ko dark condition $2 μ t = mλ$ mein substitute karo:

$2 μ \cdot \frac{r _{m}^{2}}{2 R} = mλ ⟹ \frac{μ r _{m}^{2}}{R} = mλ ⟹ r_{m}^{2} = \frac{mλ R}{μ}$

Air ke liye ( $μ = 1$ ):

Shape kyun aisi hai. $r_{m} \propto m$ . Ring 1 se 2 jaane par radius $1 \to 2$ ki tarah badhti hai; 100 se 101 tak barely hilti hai. To rings bahar ki taraf simat jaati hain — ruler-se-equal spacing kabhi nahi.

PICTURE.

Dark centre ke saath concentric dark rings; tick marks dikhate hain ki gaps $m$ ke badhne ke saath simat rahi hain — proof aankhon se ki woh crowd karti hain.

Step 8 — Centre aur offset-centre problem (degenerate cases)

KYA — centre, $m = 0$ . $m = 0$ ke saath dark rule deta hai $r_{0} = 0$ : bahut centre ek dark spot hai, Step 4 se match karta hai. Yeh zero radius ka degenerate ring hai.

KYA — radii nahi, diameters measure karna. Practice mein contact point squashed aur fuzzy hota hai, isliye tum "centre" se $r$ trust nahi kar sakte. Iski jagah ek ring ka poora diameter $D$ measure karo, aur do rings $n$ aur $m$ ka difference lo:

$D_{n}^{2} - D_{m}^{2} = 4 nλ R - 4 mλ R = 4 (n - m) λ R$

$λ = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) R}$

Subtract kyun karte hain. Estimated centre ka koi bhi offset har $D^{2}$ mein same constant add karta hai, aur Stokes $λ /2$ bhi har ring mein same tarah baithta hai. Dono subtract karne se dono nuisances ek saath khatam — ek clean straight line milti hai.

PICTURE.

Ek ring jisme true centre guessed centre se offset hai; do half-widths alag hain, lekin poora diameter (edge to edge) offset se immune hai.

Ek-picture summary

Ek diagram, poori chain: rays A & B mein split → extra path $2 μ t$ → plus Stokes $λ /2$ → condition $2 μ t = mλ$ → geometry $t = r^{2} /2 R$ → radius $r_{m} = mλ R$ → crowding rings.

Recall Feynman retelling — plain words mein wapas bolo

Ek round lens ek flat sheet of glass par baitha hai; beech mein ek patli si air hai jo zero-thick hai jahan woh touch karte hain aur bahar jaane par mothi hoti jaati hai. Ek pure colour ki light neeche daalo. Kuch air ke top se bounce ho jaati hai; kuch ghus ke plate se bounce karti hai aur wapas aati hai — yeh doosra beam extra down-and-up travel karta hai, jo hai $2 t$ (gap mein jo bhi hai uske index $μ$ se multiply karke). Upar se, jo beam denser plate se bounced hai woh palat jaati hai, jo aadha wavelength add karne jaisa hai. To jahan gap zero hai wahan bhi, do beams aadhe wavelength out hain — woh cancel karte hain, aur bilkul middle dark hai. Bahar jaao: gap distance ke square ki tarah badhta hai ( $t = r^{2} /2 R$ , sphere ki geometry se), to jab bhi extra path ek aur poora wavelength clock karta hai tumhe ek dark ring milti hai, aur beech mein ek bright ring. Kyunki $r$ ring number ki square root jaisa jaata hai, bahari rings ek saath pile up ho jaati hain. Wavelength cleanly measure karne ke liye tum fuzzy centre par trust nahi karte — rings par diameters measure karo aur do subtract karo, jo centre error aur half-wavelength dono cancel kar deta hai, bacha rehta hai $λ = (D_{n}^{2} - D_{m}^{2}) / [4 (n - m) R]$ .

Recall Quick self-test

Dark centre kyun? ::: $t = 0$ par sirf Stokes $λ /2$ ka difference hai → destructive interference. $2 μ t$ mein factor 2 kyun? ::: Ray B film ko neeche aur upar cross karti hai. $t \propto r^{2}$ kyun? ::: Sphere geometry: $r^{2} \approx 2 R t$ . Bahar crowd kyun karte hain? ::: $r_{m} = mλ R \propto m$ , isliye spacing ghatti hai. Diameters subtract kyun? ::: Unknown centre offset aur $λ /2$ constant dono cancel ho jaate hain.