Thin film interference — reflected and transmitted: Question bank

2.5.12 · D5 · Physics › Optics › Thin film interference — reflected and transmitted

Drills se pehle, har ek symbol aur term pin down kar lete hain jo neeche milenge, taaki kuch unexplained na lage.

Definition Is page par use hone waale symbols

$t$ ::: film thickness — transparent layer kitni gehri hai, seedha neeche (surfaces ke perpendicular) measure karte hain.
$n$ ::: film ka Refractive index — light andar kitni slow hoti hai; wavelength is factor se andar shrink hoti hai.
$n_{c}$ ::: ek coating layer ka refractive index specifically (anti-reflection item mein use hota hai); ye wahi role play karta hai jo $n$ karta hai, lekin generic film ki jagah coating ko name karta hai.
$λ$ ::: light ki wavelength, hamesha vacuum/air mein measure ki jaati hai (woh value jo aap "green light" ke liye quote karte, jaise $550$ nm).
$θ_{i}$ ::: angle of incidence, incoming ray film ke bahar surface normal ke saath jo angle banati hai — woh angle jo aap upar wali air mein measure karte hain.
$θ_{r}$ ::: refracted angle, ray film ke andar surface normal ke saath jo angle banati hai; denser film ke liye $θ_{i}$ se chhota hota hai. Dono Snell's law se linked hain: $sin θ_{i} = n sin θ_{r}$ .
$m$ ::: interference order, ek whole number $0, 1, 2, \dots$ jo count karta hai ki path difference mein kitne full wavelengths fit hote hain.
$Δ$ ::: do interfering rays ke beech total path difference, optical (wavelength-counting) terms mein measure kiya gaya. Ye geometric path $2 n t cos θ_{r}$ ko kisi bhi half-wavelength flip ke saath bundle karta hai: $Δ = 2 n t cos θ_{r} \pm \frac{λ}{2}$ jab ek flip ho.

Neeche ke do figures woh pictures hain jinhe baaki page baar-baar point karta hai — ek path geometry ke liye, ek flip rule ke liye. Drills se pehle inhe study karo.

Figure — Thin film interference — reflected and transmitted

Ab teen levers ki checklist jo har thin-film question pull karta hai:

Recall Teen levers jo har thin-film question pull karta hai

Path ::: Lower ray ka geometric extra path $2 t cos θ_{r}$ hai (pehla figure), optical mein $2 n t cos θ_{r}$ ho jaata hai. Flip ::: == $π$ ka phase change== ( $= λ /2$ ) tabhi hota hai jab denser medium (low $n \to$ high $n$ ) se reflect karo — doosra figure dekho. Count ::: Flips ki odd number (1) ⇒ conditions swap hoti hain; even (0 ya 2) ⇒ standard conditions.

True or false — justify

Recall Ek soap film jo pop hone wali ho reflected light mein black dikhti hai.

T/F? True. Jab $t \to 0$ geometric path khatam ho jaata hai, lekin single air→film flip phir bhi $λ /2$ contribute karta hai. Do rays exactly out of step ⇒ perfect destructive interference ⇒ dark.

Recall

$2 n t cos θ_{r}$ mein $n$ ka matlab ye hai ki right side par in-film wavelength $λ / n$ bhi use karni chahiye. T/F? False. Factor $n$ already geometric path ko optical path mein convert kar deta hai. Right side par $λ$ vacuum/air wavelength hi rehti hai — $λ / n$ use karna medium ko double-count kar deta hai.

Recall Paani par oil film, reflection mein dekhi jaaye, ke liye "bright when

$2 n t cos θ_{r} = mλ$ " sahi constructive rule hai. T/F? False. Air→oil flip karta hai, oil→water (denser→rarer) nahi karta: ek net flip. Isliye conditions swap ho jaati hain aur bright ban jaata hai $2 n t cos θ_{r} = (m + \frac{1}{2}) λ$ .

Recall Jahan film reflection mein bright ho, wahan transmission mein bhi bright hogi.

T/F? False. Reflection aur transmission complementary hain (photographic negatives). Jo energy reflect nahi hoti woh transmit hoti hai, isliye reflection-bright band, transmission-dark hoti hai. Dekho Conservation of energy.

Recall Viewing angle badhane se effective path difference

$2 n t cos θ_{r}$ hamesha badh jaata hai. T/F? False. Jab angle badhta hai, $θ_{r}$ badhta hai aur $cos θ_{r}$ ghat'ta hai, isliye path difference shrink karta hai. Isliye jab aap soap bubble tilt karte ho toh film colors shift ho jaate hain.

Recall Formula mein external angle of incidence

$θ_{i}$ use karna chahiye. T/F? False. Extra path film ke andar travel hoti hai, isliye relevant angle refracted wala $θ_{r}$ hai, jo $θ_{i}$ se Snell's law se linked hai: $sin θ_{i} = n sin θ_{r}$ .

Recall

Zero net flips (0 ya 2) wali film aur $t \to 0$ reflection mein bright dikhti hai. T/F? True. Koi flip term nahi, isliye $Δ = 2 n t cos θ_{r} \to 0$ , do rays in step hain ⇒ constructive ⇒ bright. Ye soap film ka exact opposite hai, purely flip count ki wajah se.

Recall Thin-film interference ke liye

Young's double slit ki tarah do alag light sources chahiye. T/F? False. Dono interfering rays usi incident ray se aati hain jo surfaces par split hoti hai (division of amplitude), isliye ye automatically coherent hain. Young's double slit ek wavefront ko do slits se split karta hai (division of wavefront).

Recall Thick (

$t \sim 1$ mm) glass plate par white light vivid interference colors dikhati hai. T/F? False. Do reasons stack up hoti hain. Pehli, white light ki coherence length bahut chhoti hoti hai (sirf microns): front-surface aur back-surface reflections us se kaafi zyada separated hoti hain, isliye woh fixed phase relationship maintain nahi kar sakti aur simply interfere nahi karti. Doosri, agar kar bhi sakti, toh itne zyada wavelengths bright/dark conditions satisfy karti hain ek saath ki saare colors overlap ho jaate aur white ho jaate. Vivid color ke liye $t$ coherence length ke andar honi chahiye, $λ$ ke comparable.

Spot the error

Recall "Dono surfaces reflect karti hain, isliye dono

$λ /2$ flip deti hain; $λ$ add karo." — galti kahan hai? Flip tabhi hota hai jab denser medium mein jaao (doosra figure). Bottom reflection aksar denser→rarer hoti hai (koi flip nahi). Aapko har interface ka index ordering alag se check karna hoga, ye assume nahi karna ki dono flip karti hain.

Recall "Extra path poori slant length

$2 t / cos θ_{r}$ hai." — fix karo. Do emerging rays ko ek common wavefront (rays ke perpendicular) ke along compare karna hota hai, jaise pehle figure mein hai. Film normal par slant travel ko project karne se woh $2 t cos θ_{r}$ ho jaata hai; secant survive nahi karta.

Recall "Anti-reflection ke liye, hum reflection bright chahte hain, isliye

$2 n_{c} t = mλ$ ." — dono errors pakdo. Ek AR coating reflection ko cancel karna chahti hai (destructive), isliye woh minimize kar rahi hai, maximize nahi. Aur do flips ke saath phases $π + π = 2 π$ ho jaate hain; kyunki $2 π$ turn wave ko uske starting phase par waapas laata hai, isliye ye no shift ke equivalent hai, jo destructive reflection deta hai $2 n_{c} t cos θ_{r} = (m + \frac{1}{2}) λ$ par (jo normal incidence par $2 n_{c} t = (m + \frac{1}{2}) λ$ hai, jahan $cos θ_{r} = 1$ ).

Recall "Bahut thin film reflection mein hamesha dark dikhti hai." — ye kab fail hota hai?

Sirf odd flip count ke liye true hai (jaise air–soap–air). Air aur glass ke beech coating ke liye (do flips), vanishingly thin film bright dikhti hai. Ye blanket claim flip parity ko ignore karta hai.

Recall "Kyunki light film ke andar hai,

$c / n$ use karo aur alag $λ$ milti hai, phir path ko $n$ se bhi multiply karo." — kya galat hai? Ye medium correction do baar apply kar deta hai. Ya toh in-film wavelength $λ / n$ aur geometric path $2 t cos θ_{r}$ ke saath kaam karo, ya vacuum $λ$ aur optical path $2 n t cos θ_{r}$ ke saath — kabhi dono ek saath nahi.

Recall "

Newton's rings ka center bright hota hai kyunki wahan path difference zero hai." — correct karo. Contact point par geometric path $\to 0$ hai, isliye $Δ \to λ /2$ single flip se: air-gap mein ek net flip hota hai (glass→air→glass, flip sirf doosre, air→glass, interface par hoti hai). Isliye center dark hai, bright nahi — wahi $t \to 0$ logic jaisi soap film mein.

Why questions

Recall Derivation mein

$cos θ_{r}$ kyun use hota hai, $sec θ_{r}$ kyun nahi? Kyunki do exiting rays ko unke perpendicular ek common wavefront ke along compare karna hota hai (pehla figure). Jab lower ray film se zig-zag karti hai toh woh sideways bhi $2 t sin θ_{r}$ drift karti hai; us sideways offset ki wajah se upper ray ka wavefront ek aisa point reach karta hai jo uski slant par $2 t sin θ_{r} \cdot tan θ_{r}$ aage hai, aur hume woh already-counted lateral piece subtract karna hota hai. Algebra $\frac{2 t}{c o s θ _{r}} - 2 t sin θ_{r} tan θ_{r} = 2 t cos θ_{r}$ phir secant ko cosine mein collapse kar deta hai.

Recall Reflection aur transmission patterns complementary kyun hone chahiye?

Conservation of energy se, har photon ya toh reflect hota hai ya transmit. Agar ek wavelength reflection mein destructively interfere karti hai, toh woh missing energy transmitted beam mein appear karni chahiye, aur vice versa.

Recall Soap film ek color ki jagah

spectrum of colors kyun dikhati hai? Alag-alag wavelengths bright condition $2 n t cos θ_{r} = (m + \frac{1}{2}) λ$ alag-alag thicknesses par satisfy karti hain. Jab film thickness uski surface par vary hoti hai, har region alag color ko brightly reflect karta hai.

Recall Hum

Refractive index kyun invoke karte hain, sirf distances measure karne ki jagah? Phase wavelengths mein count hota hai, aur wavelength film ke andar $n$ se shrink hoti hai. Do waves jo geometric paths mein equal hain lekin alag media mein hain, phir bhi out of step ho sakti hain, isliye optical path ( $n \times$ distance) hi interference ko govern karta hai.

Recall Flip specifically

denser medium se reflect karne par kyun aata hai? Ye electromagnetic wave ki boundary conditions se aata hai (dekho Phase change on reflection): higher-index medium se reflect karna reflected wave ko invert karne par majboor karta hai, exactly jaise ek rope par wave ek fixed (heavier) end se reflect hote waqt karti hai.

Recall

Young's double slit yaad rakhna yahan useful cousin kyun hai? Dono do coherent beams ki Interference of light par rely karte hain, aur dono "path difference = whole $λ$ ⇒ bright" use karte hain. Thin films bas reflection-flip twist add karti hain jo double-slit interference mein nahi hoti.

Edge cases

Recall Exactly

$t = 0$ (perfect contact) par reflected condition ka kya hota hai? Geometric path zero hai, isliye sirf flip term bachti hai. Odd flips ⇒ $Δ = λ /2$ ⇒ dark; even flips ⇒ $Δ = 0$ ⇒ bright. Film ki indices ki ordering outcome decide karti hai.

Recall Kya hoga agar film index uske upar wale medium ke barabar ho (

$n_{f i l m} = n_{ab o v e}$ )? Top surface par koi index step nahi, isliye wahan koi reflection (aur koi flip) nahi — "film" upar se optically invisible hai aur koi two-ray interference produce nahi karti.

Recall Grazing incidence (

$θ_{r} \to 9 0^{\circ}$ ) par, path difference kya approach karta hai? $cos θ_{r} \to 0$ , isliye geometric part $2 n t cos θ_{r} \to 0$ aur $Δ$ sirf flip term ki taraf collapse ho jaata hai. Sirf flip matter karta hai, isliye reflected fringe behaviour $t \to 0$ (flip-dominated) case ke paas pahunch jaata hai.

Recall Oblique incidence par, kya simple "denser medium se reflection par flip" rule dono polarizations ke liye bhi hold karta hai?

Sirf approximately. Reflection phase actually polarization par depend karti hai: electric field incidence ke plane mein wali component (p) aur uske perpendicular wali (s) oblique angles par alag-alag phase shifts pick up karti hain, aur Brewster's angle ke paas p-reflection dusri taraf flip bhi kar sakta hai. Near-normal incidence par s/p difference vanish ho jaata hai aur simple $π$ -flip rule exact hota hai — isliye introductory treatments ise assume karte hain.

Recall Do identical glass plates ke beech air wedge ke liye, kya thin edge reflection mein bright hai ya dark?

Dark. Glass→air (denser→rarer) koi flip nahi deta; air→glass (rarer→denser) ek flip deta hai: ek net flip. Thin edge par $t \to 0$ , ek flip ⇒ $Δ \to λ /2$ ⇒ destructive ⇒ dark line.

Recall Kya hoga agar ek film do colors se illuminate ho aur ek bright ho jabki doosri dark ho same spot par?

Bilkul possible hai: bright/dark condition $λ$ par depend karti hai, isliye fixed thickness $2 n t cos θ_{r}$ par ye $(m + \frac{1}{2}) λ_{1}$ (bright) aur $m^{'} λ_{2}$ (dark) simultaneously equal ho sakta hai. Isliye films tinted dikhti hain.

Recall Agar aap monochromatic light mein ek soap film ke liye

$t$ ko zero se slowly badhate ho, toh reflected brightness ka kya sequence dikhta hai? Dark se shuru (sirf flip), $t = λ /4 n$ par pehle maximum mein bright ho jaata hai, $t = λ /2 n$ par dark ho jaata hai, $t = 3 λ /4 n$ par phir bright ho jaata hai, thickness mein har $λ /2 n$ increase par cycle karta hai.