Diffraction — single slit intensity pattern derivation: Visual walkthrough

2.5.14 · D2 · Physics › Optics › Diffraction — single slit intensity pattern derivation

Step 1 — Ek slit ek source nahi hai, yeh sources ka ek beach hai

KYA HAI. Hum ek opaque screen mein kati hui opening ("slit") jiske width $a$ hai, uspe ek pure colour ki light daaltey hain. Slit par ki light ko ek single cheez ke roop mein sochne ke bajaye, hum opening ko bahut saari patli vertical strips mein kaat dete hain.

KYUN. Ek famous idea — Huygens Principle — kehta hai ki jahan bhi wave pahunchti hai woh jagah ripples ("wavelets") ki ek nayi choti source ban jaati hai jo baahir ki taraf failti hai. Toh real width ki ek opening actually ek continuous row hai choti-choti sources ki jo ek dusre ke saath side by side baithti hain. Yahi puri wajah hai ki ek single slit interaction bana sakti hai: bahut saari sources interfere kar sakti hain.

PICTURE. Figure dekho. Slit (height $a$ ) strips mein kati hui hai. Har red dot ek wavelet source hai. Hum measure karte hain ki ek strip top edge se kitni neeche hai aur usse letter $y$ se darshate hain — toh $y = 0$ top par hai, $y = a$ bottom par.

Figure — Diffraction — single slit intensity pattern derivation

Step 2 — Angle par nikalne wali rays unequal distances tay karti hain

KYA HAI. Hum ek viewing direction choose karte hain: seedhe-aage ki line se ek angle $θ$ tilted. Har strip usi door direction ki taraf ek wavelet bhejti hai. Kyunki strips alag-alag heights par hain, unki rays alag-alag jagahon se shuru hoti hain — toh woh screen tak alag-alag total distances cover karti hain.

KYUN. Screen bahut door hai (yeh "far screen" assumption Fraunhofer condition hai), toh ek point ki taraf jaane wali saari rays effectively parallel hain. Parallel rays ke liye hum ek ray ke liye extra distance ek clean right triangle se measure kar sakte hain — pure geometry.

PICTURE. Top strip se ek depth $y$ par strip se nikalne wali ray par perpendicular daalo. Joh chota segment katta hai woh — extra path — ek right triangle ki choti side hai jiska lamba side $y$ hai aur angle $θ$ hai.

Triangle padhna: angle $θ$ ke opposite wali side, hypotenuse $y$ se divided, yahi $sin θ$ ka matlab hai. Toh

$y par strip ka extra path Δ (y) = strip kitni gehri hai y \times viewing direction ka tilt sin θ .$

Step 3 — Extra distance extra phase ban jaati hai

KYA HAI. Do waves jo alag-alag distances tay karti hain woh "out of step" pahuncho. Hum extra distance $Δ (y)$ ko lag ki ek quantity mein convert karte hain, jise phase kehte hain.

KYUN. Extra path ka ek poora wavelength $λ$ ek wave ko exactly ek full cycle out of step push karta hai. Hum ek full cycle ko $2 π$ radians ke angle ke roop mein measure karte hain (ek circle ke around ek baar). Toh extra path aur phase sirf "kitna out of step" ke liye do units hain, yeh idea se linked hain ki " $λ$ of path = $2 π$ of phase."

PICTURE. Wheel picture: jab ek wave travel karti hai, imagine karo ek dot circle pe ghoom raha hai. Ek wavelength move karna = ek full lap = $2 π$ . Aadha wavelength = aadha lap = $π$ = ek crest wahan land karta hai jahan trough tha.

$ϕ (y) = phase per metre of path \frac{2 π}{λ} \times extra path Δ (y) y sin θ .$

Sabse bada phase gap bilkul top ( $y = 0$ ) aur bilkul bottom ( $y = a$ ) strips ke beech hai:

$Φ = \frac{2 π}{λ} a sin θ = 2 β, β \equiv \frac{π a sin θ}{λ} .$

Step 4 — Waves add karna = choti arrows (phasors) add karna

KYA HAI. Har strip same choti strength ki lekin alag phase ki ek wave bhejti hai. Hum in saari waves ko add karna chahte hain. Hum yeh har wave ko ek choti arrow ("phasor") ki tarah draw karke aur unhe tip-to-tail rakh ke karte hain.

KYUN. Ek wave ki ek size aur ek phase hoti hai — exactly wahi jo ek arrow ke paas hoti hai: ek length aur ek direction. Waves add karna phir bas arrows add karna hai. Yeh phasor trick hai, aur yeh poori cheez ko algebraic ki jagah visual bana deti hai.

PICTURE. Har arrow ki same length hai (har strip equally bright hai) lekin har ek pehle wale se thoda zyada ghuma hua hai (phase depth $y$ ke saath steadily badhti hai). Tip-to-tail rakhe gaye, equal arrows with steadily-turning direction ek arc of a circle mein bend ho jaate hain. Resultant wave woh single arrow hai jo pehle ki start se aakhri ki tip tak jaata hai — us arc ka chord.

Step 5 — Chord ki length $sin β / β$ deti hai

KYA HAI. Hum ek arc ki chord length compute karte hain jo total angle $2 β$ se bend hota hai, arc ki length ko fixed rakhte hue (woh length total amplitude $A_{0}$ hai jo milti agar saare arrows ek direction mein point karte).

KYUN. Chord hi resultant wave ka amplitude hai. Use $β$ mein ek formula ke roop mein uski length nikalna poori derivation hai ek geometric move mein.

PICTURE. Arc ki length $A_{0}$ hai aur $2 β$ se bend hota hai, toh yeh radius $R = A_{0} / (2 β)$ ke circle par rehta hai (arc length = radius × angle). Chord wohi angle $2 β$ centre par subtend karta hai, aur ek arc ka chord $2 R sin (half the bend)$ hai.

$E = = A_{0} / β 2 R half-bend sin β = \frac{A _{0}}{β} sin β = A_{0} \frac{s i n β}{β} .$

Term by term: $R = \frac{A _{0}}{2 β}$ "arc $=$ radius $\times$ angle" se aata hai jahan arc $= A_{0}$ aur angle $= 2 β$ ; chord formula $2 R sin β$ bend ka aadha use karta hai, isliye $β$ (na ki $2 β$ ) $sin$ ke andar aata hai.

Step 6 — Square karo brightness pane ke liye

KYA HAI. Amplitude (arrow length) ko intensity mein convert karo — jo ek aankh ya sensor actually measure karta hai.

KYUN. Brightness amplitude squared ke proportional hoti hai (wave dwara carry ki gayi energy uski size ke square ki tarah badhti hai). Squaring direction-only factor $e^{i β}$ ko bhi khatam kar deta hai, kyunki uski length $1$ hai.

PICTURE. Wahi chord as before, ab uski length squared ke saath — central spike tall appear hota hai kyunki $θ = 0$ par arc bilkul seedha hai aur chord sabse lamba hai.

$I (θ) = I_{0} (\frac{sin β}{β})^{2}, β = \frac{π a sin θ}{λ}, I_{0} = A_{0}^{2}$

$I (θ)$ ::: viewing angle $θ$ par brightness.
$I_{0}$ ::: bilkul centre mein brightness ( $θ = 0$ ).
$(sin β / β)^{2}$ ::: " $sinc$ -squared" shape — chord length squared, $A_{0}$ ki units mein.

Step 7 — Edge case: seedhe aage ( $θ = 0$ )

KYA HAI. Check karo ki formula centre par kya karta hai, jahan $θ = 0$ toh $β = 0$ — aur $sin β / β$ forbidden $0/0$ jaisa lagta hai.

KYUN. Ek formula ko apna degenerate input survive karna chahiye. Physically $θ = 0$ ka matlab hai ki har strip perfect step mein hai, toh hum sabse bright point expect karte hain, na ki ek hole.

PICTURE. Arc ek seedhi line mein unbend ho jaata hai: saare arrows ek hi direction mein point karte hain, chord poore arc ke barabar hai, resultant sabse lamba hai. Jab $β \to 0$ toh ratio $sin β / β \to 1$ .

$lim_{β \to 0} \frac{s i n β}{β} = 1 \Rightarrow I (0) = I_{0} \cdot 1^{2} = I_{0} .$

Step 8 — Edge case: dark fringes ( $β = mπ$ )

KYA HAI. Dhundho kahan brightness exactly zero ho jaati hai. Uske liye $sin β = 0$ chahiye $β \neq = 0$ ke saath, yaani $β = mπ$ for $m = \pm 1, \pm 2, \dots$

KYUN. $sin$ ke zeros wahan hain jahan arc poora ghoomke ek closed loop ban jaata hai — chord kuch bhi nahi reh jaata. Yeh ek continuous slit ka signature "cancel in pairs" hai.

PICTURE. Jab $β = π$ toh total bend $2 β = 2 π$ hai: arrows ek full circle mein curl ho jaate hain, tip tail se milti hai, chord $= 0$ . Slit ko top half aur bottom half mein split karo — top wala har strip ek partner rakhta hai jo slit mein aadha neeche hai aur exactly $λ /2$ out of step hai, toh woh annihilate ho jaate hain.

$β = mπ \Rightarrow \frac{π a s i n θ}{λ} = mπ \Rightarrow a sin θ = mλ, m = \pm 1, \pm 2, \dots$

Step 9 — Edge case: dim secondary peaks

KYA HAI. Do dark fringes ke beech light vanish nahi hoti — yeh ek chote "secondary" peak tak rise karti hai. Yeh wahan hote hain jahan chord locally sabse lamba hota hai, exactly halfway nahi.

KYUN. $I = I_{0} (sin β / β)^{2}$ ke peaks dhundhe jaate hain yeh pooch ke ki curve kahan chadna band karta hai. Aisa karne par (slope ko zero set karne par) condition $tan β = β$ milti hai, jiska pehla solution $β \approx 1.43 π$ hai — shrinking $1/ β$ envelope se centre ki taraf khicha gaya hai, toh nahi tidy $1.5 π$ par.

PICTURE. Poora $sinc^{2}$ curve: ek giant central hump, phir bumps jo fast fall off karte hain. Pehla side bump centre ka sirf lagbhag $4.5%$ tak pahunchta hai.

$\frac{I}{I _{0}}_{β \approx 1.43 π} = (\frac{s i n ( 1.43 π )}{1.43 π})^{2} \approx 0.0472 \approx 4.5%.$

Worked checks (numbers jo tum trust kar sakte ho)

Ek-picture summary

Yeh figure poori logic stack karta hai: slit → path difference $y sin θ$ → phase → arc of phasors → chord $\propto sin β / β$ → square it → $sinc^{2}$ pattern apne centre ke saath, dark fringes $a sin θ = mλ$ par, aur dim side peaks.

Recall Feynman retelling — ise seedhe shabd mein wapas bolo

Ek slit actually ek row of tiny light sources hai (Huygens). Apni nazar ko angle $θ$ se tilt karo: har source ki ray thodi alag distance travel karti hai, extra bit $y sin θ$ hai ek source ke liye jo top se $y$ neeche hai. Extra distance matlab extra lag, aur lag ka ek wavelength ek full turn hai ( $2 π$ ). Toh har source ki wave uske upar wale se thodi zyada ghumi hui point karti hai. Unhe equal arrows ke roop mein draw karo, har ek thoda aur ghuma hua — woh ek arc mein curl ho jaate hain, aur sum us par seedha chord hai. Geometry karo aur chord ki length $A_{0} sin β / β$ hai, jahan $β$ slit ke across half the total twist hai. Ise square karo (brightness $=$ amplitude squared) aur $I_{0} (sin β / β)^{2}$ milta hai. Seedhe aage arc flat hai → longest chord → brightest spot. Jab arc ek full circle mein curl ho jaata hai ( $β = mπ$ , yaani $a sin θ = mλ$ ) toh chord gayab → dark. Loops ke beech arc thoda bend karta hai aur chord short par zero nahi → faint side peaks. Patli slit, angle per zyada curling, wider fan — light fan out ho gayi.

Recall

Ek single slit DARK fringes deta hai ::: $a sin θ = mλ$ par jahan $m = \pm 1, \pm 2, \dots$ ( $m = 0$ central bright peak hai). Variable $β$ physically hai ::: slit ke across half the total phase spread, $β = π a sin θ / λ$ . Resultant amplitude $\propto sin β / β$ kyun hai ::: yeh length $A_{0}$ aur total angle $2 β$ se bend hote ek circular arc ka chord hai. Pehla secondary peak brightness ::: $I_{0}$ ka lagbhag $4.5%$ , $β \approx 1.43 π$ ke paas.