Thin film interference — reflected and transmitted: Worked examples

2.5.12 · D3 · Physics › Optics › Thin film interference — reflected and transmitted

Kuch bhi count karne se pehle, hume teen simple definitions chahiye.

Definition Refracted angle

$θ_{r}$ Jab light film mein enter karti hai toh woh bend hoti hai. ==Refracted angle $θ_{r}$ == woh angle hai jo ray film ke andar normal ke saath banati hai — woh angle nahi jo aap bahar measure karte hain. Yeh Snell's law se aata hai. Isse index-by-index poora likho: $n_{i} sin θ_{i} = n_{t} sin θ_{r},$ jahan $n_{i}$ us medium ka index hai jis se light aa rahi hai aur $n_{t}$ us medium ka index jisme woh ja rahi hai. Air se film mein enter hone wali light ke liye, $n_{i} = n_{air} = 1$ aur $n_{t} = n_{film}$ , toh yeh $sin θ_{i} = n_{film} sin θ_{r}$ mein collapse ho jaata hai — us shortcut mein $n$ aslaan mein ratio $n_{film} / n_{air}$ hai. Ray ko normal ki taraf tilt hote socho jab woh dense film mein slow hoti hai. Hum har jagah $θ_{r}$ use karte hain (na ki $θ_{i}$ ) kyunki extra path film ke andar hai. Dekhein bhi Refractive index.

Recall Examples se pehle vocabulary check karo

Teen quick reveals — karo inhe, phir aage padho. Yeh kya decide karta hai ki hum standard ya swapped reflection condition use karte hain? ::: Flip count $F$ ki parity (odd/even). Right-hand side par kaun sa wavelength jaata hai, film ka ya vacuum ka? ::: Vacuum/air wavelength $λ$ — film ka index $n$ pehle hi $2 n t cos θ_{r}$ ke andar hai. $n_{i} sin θ_{i} = n_{t} sin θ_{r}$ mein, air→film ke liye kaun sa index kaun sa hai? ::: $n_{i} = 1$ (air, jis se aa raha hai), $n_{t} = n_{film}$ (jisme ja raha hai). Yahan $t$ ka kya matlab hai? ::: Normal ke saath measure ki gayi film thickness.

The scenario matrix

Har film problem in cells mein se ek mein hoti hai. Neeche ke worked examples mein har ek apna cell naam karta hai.

Cell	Kya vary karta hai	Flip count $F$	Extra feature
A	Air–film–air (soap)	1 (odd)	normal incidence
B	Air–coat–glass (AR coating)	2 (even)	hum chahte hain dark reflection
C	Air–oil–water (index in between)	1 (odd)	angled view, $cos θ_{r}$
D	Degenerate $t \to 0$	any	limiting behaviour
E	Transmitted pattern	complement	energy conservation check
F	Non-integer result	—	nearest condition classify karo
G	Angled light + Snell	1 (odd)	pehle $θ_{r}$ find karna padega
H	Exam twist: woh $λ$ find karo jo missing hai	1 (odd)	colour ke liye solve karo
I	Total internal reflection edge case	—	$θ_{i}$ critical angle se aage

A — Air–film–air, normal incidence (soap bubble)

Figure 1 (Cell A geometry). Incoming orange ray split hoti hai ray 1 mein (top se reflect hoti hai, magenta, flip carry karta hai) aur ray 2 mein (andar jaati hai, bottom se bounce karti hai bina flip ke, magenta, extra path $2 t$ travel karke wapas aati hai). Top se nikalti do magenta rays wohi hain jinhe aapki aankh combine karti hai. Steps se pehle labelled flips study karo.

Figure — Thin film interference — reflected and transmitted

Caption: air–soap–air, normal incidence — upar ek flip, neeche koi nahi, isliye $F = 1$ .

Worked example Cell A: sabse patli bright-red soap film

Soap $n = 1.33$ , light seedhi girti hai ( $cos θ_{r} = 1$ ), red $λ = 650$ nm. Sabse patli film kaun si hai jo bright red reflect kare?

Forecast: guess karo — kya "thinnest bright" $mλ$ use karega ya $(m + \frac{1}{2}) λ$ ? Apna guess note karo.

Flips count karo. Top: air→soap, $1.00 \to 1.33$ , rarer→denser → flip. Bottom: soap→air, $1.33 \to 1.00$ , denser→rarer → no flip. Toh $F = 1$ (odd). (Yeh Figure 1 mein flip labelling hai.) Yeh step kyun? $F$ decide karta hai ki hum kaun si master line use karte hain — isse hamesha pehle karo.
Bright line choose karo. $F$ odd → bright hai $2 n t = (m + \frac{1}{2}) λ$ . Yeh step kyun? Odd flip constructive ko half-integer condition par swap karta hai.
Smallest $t$ ⇒ $m = 0$ . $t = \frac{λ}{4 n} = \frac{650}{4 ( 1.33 )} = 122.2 nm$ Yeh step kyun? $m = 0$ sabse chhota non-negative integer hai, isliye yeh sabse patli film deta hai.

Verify: $t = 122.2$ nm wapas rakho: $2 (1.33) (122.2) = 325 = \frac{1}{2} (650)$ ✓ — exactly half wavelength, $m = 0$ bright se match karta hai. Units: nm in, nm dono sides par. ✓

B — Air–coat–glass (AR coating), do flips

Worked example Cell B: MgF₂ anti-reflection coating

Coating $n_{c} = 1.38$ air ( $1.00$ ) aur glass ( $1.52$ ) ke beech mein. Sabse patli coating find karo jo $λ = 550$ nm ke liye reflection cancel kare (taaki lens zyada transmit kare). Dekhein Anti-reflection coatings.

Forecast: do flips — kya conditions swap hongi ya standard rahengi?

Flips count karo. Top: air→coat, $1.00 \to 1.38$ → flip. Bottom: coat→glass, $1.38 \to 1.52$ → flip. $F = 2$ (even). Yeh step kyun? Dono interfaces rarer→denser hain kyunki $1.00 < 1.38 < 1.52$ poora ek taraf climb karta hai.
Dark line choose karo. $F$ even → dark reflection hai $2 n_{c} t = (m + \frac{1}{2}) λ$ . Yeh step kyun? Even flips standard conditions rakhte hain; standard dark half-integer wali hai.
Thinnest ⇒ $m = 0$ . $t = \frac{λ}{4 n _{c}} = \frac{550}{4 ( 1.38 )} = 99.6 nm$ Yeh step kyun? Cell A jaisi hi reasoning: sabse chhota integer, sabse patli layer.

Verify: $2 (1.38) (99.6) = 274.9 \approx \frac{1}{2} (550) = 275$ ✓ — half a wavelength, toh do reflected waves cancel ho jaati hain. Note karo ki yeh Cell A ke formula jaisa lagta hai lekin bilkul alag flip count ke through aata hai — coincidence yeh hai kyunki dono " $λ /4$ " cases hain.

C — Air–oil–water, angled view (index beech mein hai)

Worked example Cell C: ek angle par oil on water

Oil $n = 1.40$ on water ( $n = 1.33$ ), $λ = 600$ nm, internal angle se $cos θ_{r} = 0.80$ milta hai, thickness $t = 500$ nm. Reflection mein bright hai ya dark?

Forecast: oil paani se denser hai — kya bottom reflection flip karta hai?

Flips count karo. Top: air→oil, $1.00 \to 1.40$ → flip. Bottom: oil→water, $1.40 \to 1.33$ , denser→rarer → no flip. $F = 1$ (odd). Yeh step kyun? Paani oil se less dense hona doosra flip khatam kar deta hai — yeh ek common trap hai.
Path compute karo. $Δ = 2 n t cos θ_{r} = 2 (1.40) (500) (0.80) = 1120 nm .$ Yeh step kyun? Yeh raw optical path hai jise hum $λ$ se compare karte hain.
Wavelengths mein express karo. $Δ/ λ = 1120/600 = 1.867$ . Yeh step kyun? Interference sirf yeh care karta hai ki kitne wavelengths fit hote hain.
Classify karo. $F$ odd → bright at $(m + \frac{1}{2}) = 1.5, 2.5, \dots$ ; dark at $m = 1, 2, \dots$ . $1.867$ nearest hai $2$ se (distance $0.133$ ) versus $1.5$ se (distance $0.367$ ). Toh yeh dark condition ke sabse kareeb hai. Yeh step kyun? Real thicknesses rarely exact maxima hit karti hain; hum nearest band report karte hain.

Verify: $∣1.867 - 2∣ = 0.133 < ∣1.867 - 1.5∣ = 0.367$ ✓ — dark jeet ta hai. Units: $Δ/ λ$ mein saare nm cancel ho jaate hain. ✓

D — Degenerate limit $t \to 0$

Worked example Cell D: kyun ek bubble about to pop black ho jaata hai

Air–soap–air, $F = 1$ . Jab draining soap bubble ka top $t \to 0$ ki taraf thin hota hai, reflection kya karta hai?

Forecast: zero path — zaroor bright hoga (waves in step)? Dekho dhyan se.

Path term ka limit lo. $Δ = 2 n t cos θ_{r} \to 0$ jab $t \to 0$ . Yeh step kyun? Geometric extra path vanish ho jaata hai jab cross karne ke liye koi film nahi hoti.
Lekin flip apply karo. $F = 1$ odd hai, isliye bright ke liye $Δ = (m + \frac{1}{2}) λ$ chahiye aur dark ke liye $Δ = mλ$ . $Δ = 0 = 0 \cdot λ$ ke saath hum $m = 0$ dark hit karte hain. Yeh step kyun? Half-wave flip ek $λ /2$ mismatch supply karta hai even zero path ke saath — do waves exactly out of step hain.
Conclusion. Reflection $\to$ har colour ke liye ek saath dark (black). Yeh step kyun? $Δ = 0$ condition wavelength-independent hai, isliye har colour cancel hota hai → black, coloured nahi.

Verify: $t = 0$ par, $2 n t = 0$ aur $m = 0$ dark extinction predict karta hai; akela flip $λ /2$ hai — perfect destructive interference. Yeh dying bubble par famous black top hai. ✓

E — Transmitted pattern (complement + energy)

Worked example Cell E: wohi Cell C oil film neeche se kaisi dikhti hai

Same oil film jaise Cell C ( $Δ/ λ = 1.867$ , $F = 1$ , nearest $m = 2$ ). Kya transmitted light wahan bright hai ya dark?

Forecast: agar reflection dark hai, toh transmission...?

Transmitted condition use karo. Master transmission formula se, transmitted bright hai $Δ = mλ$ (plain condition, is film ke swapped reflection rule ke opposite). Yeh step kyun? Jo transmitted beam interfere karta hai woh bottom surface par do internal reflections se bana hota hai; woh same type ki reflection hain, isliye unka net flip even hai — transmission ke liye koi $λ /2$ introduce nahi hota. Reflection mein yahan ek flip tha (odd), transmission mein koi nahi (even); Phase change on reflection guarantee karta hai ki yeh dono exactly $λ /2$ se differ karte hain, isliye do patterns opposite hain.
Evaluate karo. $Δ/ λ = 1.867$ nearest to $m = 2$ → transmission mein bright. Yeh step kyun? Same number, lekin ab integer condition bright wali hai.
Energy se cross-check karo. Reflection dark tha ⇒ almost koi light bounce nahi hui ⇒ woh light transmitted ⇒ transmission bright. Yeh step kyun? Conservation of energy dono ko dark hone se forbid karta hai; patterns negatives hain.

Verify: reflection nearest $m = 2$ dark, transmission nearest $m = 2$ bright — complementary ✓. Numerically $1.867$ identically classify hota hai ( $m = 2$ ) lekin bright/dark label flip ho jaata hai. ✓

F — Non-integer result, carefully classify karo

Worked example Cell F: kya yeh film maximally bright hai, maximally dark hai, ya beech mein hai?

Air–soap–air ( $n = 1.33$ , $F = 1$ ), normal incidence, $λ = 500$ nm, $t = 300$ nm. Reflection classify karo.

Forecast: bright, dark, ya mid-grey guess karo.

Path. $Δ = 2 (1.33) (300) = 798$ nm.
Wavelengths mein. $798/500 = 1.596$ . Yeh step kyun? Fractional part decide karta hai ki hum ek band ke kitne kareeb hain.
Classify karo ( $F$ odd). Bright targets $1.5, 2.5$ ; dark targets $1.0, 2.0$ . $d_{b} = ∣1.596 - 1.5∣ = 0.096$ ; $d_{d} = ∣1.596 - 2.0∣ = 0.404$ . Koi bhi $\leq 0.05$ nahi hai, isliye upar ke rule se yeh partial intensity leaning bright hai (near the $m = 1$ maximum). Yeh step kyun? Hum explicit cutoff rule apply karte hain taaki "near" ek vague guess na rahe.

Verify: $d_{b} = 0.096 < d_{d} = 0.404$ ✓ — leans bright; aur $0.096 > 0.05$ isliye yeh partially bright hai, perfect maximum nahi, exactly jaisa rule classify karta hai. ✓

G — Angled light: pehle $θ_{r}$ Snell se find karo

Figure 2 (Cell G geometry). Orange incident ray $θ_{i} = 30°$ par film ke andar smaller refracted angle $θ_{r}$ par bend ho jaati hai (magenta). Dotted line normal hai; vertical double arrow thickness $t$ mark karta hai. Dhyan do ki path mein sirf normal par projection enter hoti hai, isliye $cos θ_{r}$ appear karta hai (na ki $sec θ_{r}$ ).

Caption: angled incidence — Snell se $θ_{r}$ find karo, phir $t = λ / (4 n cos θ_{r})$ .

Worked example Cell G: aapko

bahar ka angle diya gaya hai Air–soap–air, $n = 1.33$ , $λ = 550$ nm. Light incidence angle $θ_{i} = 30°$ par aati hai. $cos θ_{r}$ find karo, phir sabse patli bright-reflecting thickness. Snell's law aur Refractive index use karta hai.

Forecast: kya angled path film ko normal-incidence answer se thicker ya thinner banata hai?

Internal angle ke liye Snell's law. $n_{i} = 1$ (air) aur $n_{t} = 1.33$ (film) ke saath: $n_{i} sin θ_{i} = n_{t} sin θ_{r} \Rightarrow sin θ_{r} = \frac{sin 30°}{1.33} = \frac{0.5}{1.33} = 0.3759$ . Yeh step kyun? Extra path film ke andar traverse hota hai (Figure 2 mein magenta ray), isliye hume $θ_{r}$ chahiye, $θ_{i}$ nahi.
Refracted angle ki cosine. $cos θ_{r} = 1 - 0.375 9^{2} = 1 - 0.1413 = 0.9268$ . Yeh step kyun? Path formula $cos θ_{r}$ use karta hai; isse Pythagorean identity se nikalo, calculator angle ki zaroorat nahi.
Thinnest bright ( $F = 1$ , $m = 0$ ). $2 n t cos θ_{r} = \frac{1}{2} λ \Rightarrow$ $t = \frac{λ}{4 n c o s θ _{r}} = \frac{550}{4 ( 1.33 ) ( 0.9268 )} = 111.5 nm .$ Yeh step kyun? Odd- $F$ bright condition ko $t$ ke liye rearrange kiya.

Verify: normal incidence par ( $cos θ_{r} = 1$ ) hume $103.4$ nm milta; tilt karne par thicker film chahiye ( $111.5$ nm) kyunki $cos θ_{r} < 1$ path ko shrink karta hai, isliye zyada thickness isse restore karti hai ✓. Check: $2 (1.33) (111.5) (0.9268) = 274.9 \approx \frac{1}{2} (550)$ ✓.

H — Exam twist: reflection se kaun sa colour missing hai?

Worked example Cell H: colour reverse-engineer karo

Air–soap–air, $n = 1.33$ , $t = 400$ nm, normal incidence. Reflection mein kaun sa visible wavelength (400–700 nm) sabse dark (destroyed) hai?

Forecast: odd $F$ ke liye dark matlab $2 n t = mλ$ . Kitne $m$ visible range mein aate hain?

Dark condition ( $F = 1$ odd). $2 n t = mλ \Rightarrow λ = \frac{2 n t}{m} = \frac{2 ( 1.33 ) ( 400 )}{m} = \frac{1064}{m}$ nm. Yeh step kyun? Dark equation ko $t$ ki jagah $λ$ ke liye solve karo — twist hai "colour find karo."
Visible answers ke liye $m$ scan karo. $m = 1 : 1064$ nm (infrared, out). $m = 2 : 532$ nm (green, in). $m = 3 : 354.7$ nm (UV, out). Yeh step kyun? Sirf integer $m$ jo $400 \leq λ \leq 700$ deta hai woh real missing colour count karta hai.
Answer. $λ = 532$ nm (green) missing hai → film reflection mein apna complementary magenta-ish dikhta hai. Yeh step kyun? White light se green remove karna perceived colour shift karta hai.

Verify: $1064/2 = 532$ nm, 400–700 ke andar ✓; $m = 1$ (1064) aur $m = 3$ (354.7) dono visible se bahar hain ✓, isliye green unique missing colour hai.

I — Edge case: total internal reflection path khatam kar deta hai

Angled-incidence problems mein ek hidden trap hai: agar light kisi denser medium se rarer mein bahut steep angle par jaane ki koshish karti hai, toh woh refract hi nahi hoti — woh poori reflect ho jaati hai. Yeh total internal reflection (TIR) hai, aur yeh bottom surface kya karta hai woh change ho jaata hai.

Worked example Cell I: oil-on-water bahut steeply dekha gaya

Oil $n = 1.40$ on water $n = 1.33$ . Oil ke andar, neeche jaane wali ray oil→water bottom surface se hit hoti hai. Kis internal angle par yeh paani mein transmit karna band kar deti hai, aur phir interference ka kya hoga?

Forecast: bottom ray oil ( $1.40$ ) → water ( $1.33$ ) hai, denser→rarer — isliye TIR possible hai. Guess karo: critical angle se upar, kya transmission mein abhi bhi two-beam interference pattern hoga?

Oil→water interface par critical angle. $n_{1} = 1.40$ (oil), $n_{2} = 1.33$ (water): $sin θ_{c} = \frac{1.33}{1.40} = 0.95 \Rightarrow θ_{c} = arcsin (0.95) = 71.8°.$ Yeh step kyun? Yeh internal angle hai jisse aage bottom surface sab kuch reflect karta hai.
$θ_{c}$ ke upar kya hota hai. Internal angle $θ_{r} > 71.8°$ ke liye, bottom reflection total ho jaata hai — koi bhi light paani mein leak nahi hoti, isliye interfere karne ke liye koi transmitted beam nahi hoti. Cell E ka transmitted pattern simply vanish ho jaata hai. Yeh step kyun? Conservation of energy: agar $0%$ transmit hota hai, toh $100%$ reflect hota hai; "transmission negative" ke paas negate karne ke liye kuch nahi hota.
Reflected pattern survive karta hai — lekin ek twist ke saath. Reflected ray abhi bhi wahan hai (ab full strength par), isliye ek reflection pattern exist karta hai, lekin TIR apna khud ka phase shift add karta hai (simple $λ /2$ nahi), isliye naive flip count ab clean bright/dark bands nahi deta. Exams ke liye practical takeaway: transmitted-light formula trust karne se pehle $θ_{r}$ ko $θ_{c}$ ke against check karo. Yeh step kyun? Yeh exactly flag karta hai jab Cells A–H ki machinery apply hona band ho jaati hai.

Verify: $sin θ_{c} = 1.33/1.40 = 0.95 \leq 1$ ✓ (ek real angle exist karta hai kyunki oil paani se denser hai); $θ_{c} = 71.8°$ , aur is se neeche $θ_{r}$ ke liye ordinary formulas hold karte hain, upar transmission zero hai ✓.

Recall Whole-page self-test

Saare nine cells, ek ek line. Cell A thinnest bright, air–soap–air, $λ = 650$ ? ::: $t = λ /4 n = 122.2$ nm. Cell B AR coating thinnest dark, $λ = 550$ , $n_{c} = 1.38$ ? ::: $99.6$ nm. Cell C oil film $Δ/ λ = 1.867$ reflection? ::: nearest $m = 2$ → dark. Cell D bubble $t \to 0$ reflection? ::: saare colours ke liye black (dark). Cell E wohi oil transmission mein? ::: bright (reflection ka complement). Cell F soap $t = 300$ , $λ = 500$ : $Δ/ λ = 1.596$ , classify karo? ::: partial, leans bright (near $m = 1$ max, exact nahi). Cell G angled soap $θ_{i} = 30°$ thinnest bright? ::: $111.5$ nm. Cell H missing colour, $t = 400$ nm? ::: green, $532$ nm. Cell I oil→water critical angle? ::: $θ_{c} = arcsin (1.33/1.40) = 71.8°$ ; uske upar transmission vanish ho jaata hai.

Related builds: Interference of light · Young's double slit · Newton's rings · Phase change on reflection.