Thin film interference — reflected and transmitted: Visual walkthrough

2.5.12 · D2 · Physics › Optics › Thin film interference — reflected and transmitted

Pehli line se pehle, teen words plain language mein:

Neeche sab kuch do rays ke baare mein hai jo ek se janmi hain, aur kya wo waapis in step aati hain.

Step 1 — Ek ray do mein split hoti hai

KYA. Air mein ek single ray of light ek transparent film ke top se takraati hai. Uska kuch hissa top surface se seedha reflect ho jaata hai (ise Ray 1, blue kaho). Baaki film mein enter karti hai, neeche jaati hai, bottom se reflect hoti hai, aur waapis upar aati hai (ise Ray 2, orange kaho).

KYUN. Ye do rays ek hi wave se shuru hui hain, isliye ye coherent hain — ek doosre ke relative ek fixed rhythm mein locked hain. Yahi ek reason hai ki ek stable bright/dark pattern exist kar sakta hai. Do unrelated flashlights kabhi aisa nahi karti.

PICTURE. Split ko follow karo. Ray 2 clearly zyada travel karti hai — wo film ke through ek round trip leta hai jo Ray 1 kabhi nahi leta. Wo extra distance hi sab kuch ka beej hai.

Step 2 — Film ke andar slant path measure karo

KYA. Ray 2 angle $θ_{r}$ par neeche jaati hai aur waapis aati hai. Har leg ek slanted line hai jiska length $t / cos θ_{r}$ hai, toh dono legs milke $2 t / cos θ_{r}$ hain.

KYUN ye length? Neeche wali leg jo right triangle banati hai use dekho: film thickness $t$ wo side hai jo normal ke saath hai (angle $θ_{r}$ ki "adjacent" side), aur slant leg hypotenuse hai.

Do legs (neeche + upar) ⇒ total slant $= \frac{2 t}{cos θ _{r}}$ .

PICTURE. Neeche ke do shaded triangles mirror images hain. Vertical side hamesha $t$ hai; diagonal wo leg hai jo hum ne abhi measure ki.

Step 3 — Fairly compare karo: ek common wavefront drop karo

KYA. Ek wavefront wo line hai jo rays ke perpendicular hai jahan har wave ka same phase hota hai — ek fair race ke liye "starting line." Hum wavefront ko wahan se drop karte hain jahan Ray 2 exit karti hai aur mark karte hain jahan ye Ray 1 ko cross karti hai. Sirf us line tak ka path "extra" count hota hai.

KYUN. Phase is baare mein hai ki do waves ko kitne wavelengths alag karti hain. Agar hum Ray 2 ka poora slant measure karein lekin Ray 1 ko free chalte dein, toh hum runners ko alag starting lines se time kar rahe hain — meaningless. Wavefront shared finish line hai.

PICTURE. Green segment wo piece hai Ray 1 ke path ka (aur Ray 2 ke slant ka) jo subtract off hoti hai. Jo bachta hai wo round trip ka normal par projection hai.

Recall Ek-line algebra jo ise cosine banati hai

Sawaal: kyun $\frac{2 t}{c o s θ _{r}} - 2 t sin θ_{r} tan θ_{r}$ equal hota hai $2 t cos θ_{r}$ ke? ::: Common denominator par rakh do: $\frac{2 t - 2 t s i n ^{2} θ _{r}}{c o s θ _{r}} = \frac{2 t ( 1 - s i n ^{2} θ _{r} )}{c o s θ _{r}} = \frac{2 t c o s ^{2} θ _{r}}{c o s θ _{r}} = 2 t cos θ_{r}$ , using $1 - sin^{2} θ_{r} = cos^{2} θ_{r}$ .

Step 4 — Distance ko "optical distance" mein convert karo

KYA. Geometric extra path ko $n$ se multiply karo: $Δ_{opt} = 2 n t cos θ_{r} .$

KYUN ye tool — kyun $n$ se multiply karo, add ya square nahi? Film ke andar light slow ho jaati hai, isliye uska wavelength shrink ho ke $λ_{film} = λ / n$ ho jaata hai. Interference ke liye jo matter karta hai wo hai kitne wavelengths path mein fit hote hain, aur shorter wavelengths ka matlab hai unme se zyada fit honge. Wavelengths count karte hain: $\frac{geometric path}{λ _{film}} = \frac{2 t c o s θ _{r}}{λ / n} = \frac{n \cdot 2 t c o s θ _{r}}{λ} .$ Factor $n$ naturally appear hota hai — ye ek real distance ko "as-if-in-vacuum" distance mein convert karta hai taaki hum doosri side par ordinary air wavelength $λ$ use karte rahe.

PICTURE. Same geometric path, lekin film mein crammed wave faster wiggle karti hai (chhota $λ$ ), isliye wo har millimetre mein zyada phase accumulate karti hai. $n$ isko count karta hai.

Step 5 — Reflection par half-wave hiccup

KYA. Reflection khud wave ko ulta palat sakti hai, half a wavelength ( $λ /2$ , phase of $π$ ) add karte hue — lekin sirf kabhi kabhi.

KYUN / rule. Denser medium se bounce karna (low $n$ se high $n$ jaana, "rarer → denser") wave ko flip karta hai. Rarer medium se bounce karna (high $n$ → low $n$ ) ise flip nahi karta. (Yeh hai Phase change on reflection.)

PICTURE. Ray 1 air→film (rarer→denser) hit karta hai ⇒ flip, drawn as ek peak trough mein turn karti hai. Ray 2 ka bottom bounce depend karta hai ki film ke neeche kya hai — picture dono possibilities dikhati hai.

Step 6 — Case A: air–film–air (soap bubble), ONE flip

KYA. Upar air, beech mein film, neeche air (ek free soap bubble).

Top bounce: air→film = rarer→denser ⇒ flip.
Bottom bounce: film→air = denser→rarer ⇒ no flip.
Net = ek flip ⇒ $+ \frac{λ}{2}$ rakho.

KYUN yeh matter karta hai. Ek flip usual bright/dark rules ko swap karta hai, isliye "do flips" ke liye memorised formula yahan exactly opposite deta hai.

PICTURE. Flip tags har surface par; sirf top wala fire karta hai.

Step 7 — Case B: anti-reflection coating, TWO flips

KYA. Air ( $1.0$ ) → coating ( $n_{c} = 1.38$ ) → glass ( $1.52$ ). Ab dono bounces rarer→denser hain ( $1 < 1.38$ aur $1.38 < 1.52$ ), isliye do flips.

KYUN. Do flips = poora $λ$ shift = koi relative shift nahi bilkul. $λ /2$ term gayab ho jaata hai, aur standard rules waapis aate hain — soap bubble ke relative flip back ho jaate hain. Yahi poora idea hai Anti-reflection coatings ke peeche.

PICTURE. Dono surfaces ek flip dikhate hain; milke cancel ho jaate hain.

Step 8 — Degenerate case: vanishing film ( $t \to 0$ )

KYA. Soap bubble ko itna thin hone do ki wo pop hone wala ho: $t \to 0$ , toh $2 n t cos θ_{r} \to 0$ .

KYUN yeh sabse gehri test hai. Zero geometric path ke saath, naive thinking kehti hai "koi path difference nahi ⇒ sab in step ⇒ bright." Lekin one-flip soap case mein total hai $Δ = 0 + \frac{λ}{2} = \frac{λ}{2}$ — exactly out of step ⇒ dark. Isliye ek dying bubble ka top ek saath har colour ke liye black ho jaata hai: akela flip reflection ko khatam kar deta hai.

PICTURE. Ray 2 ka round trip kuch nahi reh jaata, phir bhi single flip dono waves ko half a step apart rakhta hai.

Step 9 — Worked numbers taaki yeh pakka ho jaaye

Ek-picture summary

Upar sab kuch, ek canvas par: split, slant, wavefront cut, $n$ -stretch, flip count, aur final boxed rule.

$Δ = 2 n t cos θ_{r} + (odd flips ? \frac{λ}{2} : 0)$

Recall Poore walkthrough ki Feynman retelling

Light ki ek beam ek soap wall se takraati hai aur do runners mein split ho jaati hai. Runner 1 front se bounce karta hai. Runner 2 andar ghusta hai, back wall tak sprint karta hai, bounce karta hai, aur waapis chadhta hai — toh Runner 2 ne extra daura. Us extra ko fairly measure karne ke liye hum ek finish line (wavefront) dono runners ke across square draw karte hain; extra distance jo actually count hoti hai woh $2 t cos θ_{r}$ nikalta hai, lambi diagonal nahi, kyunki dive ka kuch hissa finish line se pehle hota hai. Film ke andar runner chhote chhote kadam leta hai ( $λ / n$ ), isliye hum distance ko $n$ se multiply karte hain kadam sahi count karne ke liye — woh hai $2 n t cos θ_{r}$ . Phir twist: harder cheez se bounce karna (denser medium) runner ko half-stride hiccup deta hai. Hiccups count karo. Ek (soap bubble) ⇒ half a stride add karo aur bright/dark rules swap ho jaate hain. Do (coating on glass) ⇒ hiccups cancel aur plain rules waapis. Aur agar wall paper-thin ho, extra distance zero hai lekin akela hiccup bana rehta hai — toh runners exactly out of step milte hain aur spot black ho jaata hai. Woh single hiccup hi poora game hai.

Recall Quick self-test

Extra path cosine kyun hai, secant kyun nahi? ::: Kyunki do rays ko ek shared wavefront par compare kiya jaata hai; pre-wavefront slice subtract karne se $2 t / cos θ_{r}$ $2 t cos θ_{r}$ ban jaata hai. $n$ se multiply kyun karte hain? ::: Film ke andar wavelength shrink ho ke $λ / n$ ho jaati hai, isliye zyada wavelengths fit hote hain; $n$ geometric path ko optical path mein convert karta hai taaki hum air $λ$ rakh sakein. Ek soap film jo pop hone wali hai reflection mein black kyun dikhti hai? ::: $t \to 0$ geometric path $0$ deta hai, lekin single reflection flip $λ /2$ chhod deti hai ⇒ perfect cancellation.

Prerequisites revisited: Interference of light · Young's double slit · Refractive index · Snell's law · Phase change on reflection · Conservation of energy.