Young's double slit — fringe width derivation: Visual walkthrough

2.5.11 · D2 · Physics › Optics › Young's double slit — fringe width derivation

Step 0 — Woh words jo hum use karenge (taaki koi symbol anjaan na lage)

Kisi bhi maths se pehle, characters se milte hain. Yeh sab lengths hain jo tum ruler se measure kar sakte ho.

Do ideas hum doosre notes se lete hain:

Do slits se aane wali light coherent hai — dekho Coherence and coherent sources — matlab dono waves ek doosre ke saath fixed rhythm maintain karti hain, isliye unka milna hamesha ek jaisi kahani hoti hai.
Waves add up ya cancel hoti hain is baat par depend karke ki woh kaise align hoti hain — dekho Interference of light aur Path difference and phase difference.

Step 1 — Stage: do slits, ek screen

KYA. Hum do slits $S_{1}$ (neecha) aur $S_{2}$ (upar) rakhte hain, $d$ gap ke saath. Seedha saamne, $D$ door, screen khadi hai. Slits ke midpoint ke bilkul saamne wale point ko hum $O$ kehte hain — "centre".

KYU. Interference ek geometry ka problem hai: do waves do starting points se ek hi finish point ki taraf daudti hain. Yeh measure karne ke liye ki kaun zyada door travel karta hai, pehle hume sab kuch pin down karna hoga.

PICTURE. Neeche, do slits baayein taraf hain, screen daayein taraf. $O$ bilkul beech mein hai. Point $P$ woh jagah hai jahan hum dekhna chahte hain, $O$ se $y$ upar.

Step 2 — Do right triangles do distances dete hain

KYA. Do light-paths $S_{1} P$ aur $S_{2} P$ draw karo. Har path ek right triangle ki slanted side (hypotenuse) hai jiska flat side $D$ hai aur vertical side "P us slit se kitna upar hai" hai.

KYU. Hume har path ki length chahiye. Right triangle ek aisa tool hai jahan do sides jaante ho toh teesri exactly mil jaati hai — yahi Pythagoras karta hai: $(hypotenuse)^{2} = (base)^{2} + (height)^{2}$ . Hum ise isliye use karte hain kyunki dono paths naturally right triangles hain (horizontal $D$ , vertical offset).

PICTURE. Do shaded triangles dekho. Dono ke bases $D$ hain. Unki heights alag hain: $P$ , $S_{2}$ se zyada upar hai? Nahi — $P$ , $S_{1}$ (neecha slit) se $S_{2}$ se zyada upar hai. Vertical legs dekho.

Step 3 — Chalak subtraction: pehle square karo, phir difference lo

KYA. Do ugly square roots subtract karne ki jagah, pehle unke squares subtract karte hain.

KYU. Square roots ko directly subtract karna mushkil hai; squares aasan hain. Messy $D^{2}$ dono mein hai, isliye subtract karne par cancel ho jayega — kuch clean bachega.

PICTURE. Animation-style panel mein $S_{2} P^{2}$ aur $S_{1} P^{2}$ areas ki tarah dikhaye gaye hain; shared $D^{2}$ block cancel ho jata hai, aur sirf height-difference bachta hai.

Step 4 — Factor karke path difference $Δ$ reveal karo

KYA. Hum left side par algebra identity $a^{2} - b^{2} = (a - b) (a + b)$ use karte hain.

KYU. Kyunki $(a - b)$ exactly $Δ$ hai — woh cheez jo hume chahiye. Factoring use surgically extract karta hai.

PICTURE. Bar dikhata hai $S_{2} P^{2} - S_{1} P^{2}$ do factors mein split: difference $Δ$ (jo hum dhundh rahe hain) times sum $S_{2} P + S_{1} P$ (jise hum agle step mein approximate karenge).

Step 5 — Far-screen approximation

KYA. Hum kehte hain: kyunki screen door hai ( $D$ bahut bada) aur slit-gap aur look-height tiny hain, dono paths ki length almost exactly $D$ hai. Toh unka sum $\approx 2 D$ .

KYU. Ek metre lambi path zyada nahi badlti jab tum uska endpoint ek millimetre se nudge karo. Yeh small-angle regime hai — dekho Small angle approximation. Hum ise sirf sum mein use karte hain (jahan chhoti si error harmless hai) aur kabhi nahi delicate difference $Δ$ mein (jahan har cheez matter karti hai).

PICTURE. Dono slanted paths horizontal $D$ par flatten ho jaati hain; panel dikhata hai ki jab $D ≫ d, y$ , dono lengths visually ek hi $D$ par collapse ho jaati hain.

Step 6 — Path difference se bright fringe positions nikalo

KYA. Bright fringe wahan hoti hai jahan do waves in step pahunchti hain — yaani jahan extra path $Δ$ wavelengths ki ek whole number ho.

KYU. Agar ek wave exactly ek (ya do, ya teen...) poori ripples peeche hai, uska crest phir bhi doosre ke crest pe aata hai — woh reinforce karte hain. Yahi bright condition hai $Δ = nλ$ .

PICTURE. Screen bright bands dikhata hai; $n$ -th band wahan hoti hai jahan $Δ = nλ$ . Har band ke neeche, do aane wali waves crest-on-crest draw ki hain.

Step 7 — Fringe width: neighbours subtract karo, dekho $n$ gayab ho jata hai

KYA. Fringe width $β$ band $n$ aur band $n + 1$ ke beech ka gap hai.

KYU. Kyunki bands $y_{n} = \frac{nλ D}{d}$ par hain, evenly spaced, neighbours ke beech ki distance ko hum fringe width kehte hain. Subtract karne se $n$ khatam ho jata hai.

PICTURE. Screen par ek ruler rakha: consecutive bright bands equally spaced hain; arrow $β$ mark karta hai, har jagah same length.

Step 8 — Degenerate & edge cases (reader ko kabhi surprise nahi hona chahiye)

KYA. Hum un corners check karte hain jo sundar formula chupa sakta hai.

KYU. Ek derivation jis par trust kar sako use apni extremes mein survive karna chahiye: dead centre, below-centre, huge slit-gap, aur small-angle failure.

PICTURE. Char mini-panels, har corner case ke liye ek.

Ek-picture summary

Yahan poora safar ek canvas par hai: do right triangles → squares subtract karo → $Δ$ factor out karo → far-screen approximation → bright condition → neighbours subtract karo → $β = \frac{λ D}{d}$ .

Recall Feynman retelling — walkthrough simple words mein

Do chhote lamps ek baal ki duri par hain aur perfect time mein blink karte hain. Door wali wall par tum ek spot dekhte ho jo thoda off-centre hai. Upar wale lamp ki light ko neche wale lamp ki light se thoda zyada chalna padta hai — woh extra walk $Δ$ hai. Use find karne ke liye, main do right triangles draw karta hoon (har lamp se spot tak) aur, do square roots se ladne ki jagah, main unke squares subtract karta hoon — bada shared side $D^{2}$ politely cancel ho jaata hai, sirf $2 y d$ bachta hai. Factoring se woh " $Δ$ times (do lambi paths ka sum)" ban jaata hai. Kyunki wall door hai, dono paths basically $D$ hain, toh sum $2 D$ hai aur $Δ = \frac{y d}{D}$ . Ab, ek bright stripe tab hoti hai jab extra walk poori ripples ki ek whole number ho, $nλ$ , jo har stripe ko $y_{n} = \frac{nλ D}{d}$ par pin karta hai. Stripes barabar steps mein chalti hain, toh unke beech ka gap — fringe width — bas $\frac{λ D}{d}$ hai: ripple-size times wall-distance, divided by lamp-gap. Aur bilkul centre mein, jahan dono walks equal hain, koi bhi colour ho wahan bright hoga.

Active Recall

Recall Hum paths ko directly subtract karne ki jagah unke

squares kyun subtract karte hain? Kyunki squares square roots ko khatam kar dete hain aur shared $D^{2}$ cancel ho jaata hai, clean $2 y d$ bachta hai — do roots ko directly subtract karna intractable hai.

Recall Small-angle approximation exactly kahan use hoti hai, aur kahan NAHI hoti?

Sirf sum $S_{1} P + S_{2} P \approx 2 D$ mein use hoti hai; difference $Δ$ mein kabhi nahi, jahan full precision rakhi jaati hai.

Recall Fringes equally spaced kyun aati hain?

Kyunki $y_{n} = \frac{nλ D}{d}$ $n$ mein linear hai, isliye $y_{n + 1} - y_{n}$ constant $\frac{λ D}{d}$ hai — $n$ cancel ho jaata hai.

D^{2}

aur

\frac{d ^{2}}{4}

cancel karne ke baad surviving path-difference kya hai?

2 y d

Master path difference formula?

Δ = \frac{y d}{D}

Final fringe width?

β = \frac{λ D}{d}

Centre hamesha bright kyun hota hai?

O

par dono paths equal hain, isliye

Δ = 0 = 0 \cdot λ

— har wavelength ke liye constructive.

Connections

Young's double slit — fringe width derivation — woh parent jise yeh page illustrate karta hai.
Interference of light — crest-on-crest kyun bright karta hai.
Coherence and coherent sources — ek source se do slits kyun kaam karti hain.
Path difference and phase difference — $Δ$ ka matlab.
Small angle approximation — $S_{1} P + S_{2} P \approx 2 D$ ko license deta hai.
Refractive index — setup ko immerse karne se $λ$ aur $β$ kaise shrink hote hain.
Diffraction grating — is two-slit story ka many-slit cousin.