Diffraction — single slit intensity pattern derivation

2.5.14 · Physics › Optics

Intuition Core idea kya hai

Jab wavelength $λ$ ki light, width $a$ ki ek slit se guzarti hai, to slit ek single source nahi hoti — woh actually infinitely many choti Huygens wavelets ka collection hoti hai jo poori width mein pheli hoti hain. Har wavelet ek door screen ke kisi point tak travel karti hai, lekin kyunki woh slit ke across alag-alag positions se start karti hain, woh alag-alag path lengths ke saath pahunchti hain, isliye alag-alag phases ke saath. Screen ki brightness in saari wavelets ke interfere karne ka result hai. Jahan woh add hoti hain → bright. Jahan woh ek closed loop mein cancel hoti hain → dark.

EK SINGLE SLIT PATTERN BANATI HI KYUN HAI? Kyunki finite width ki slit actually kaafi saare sources ki tarah hai. Two-slit interference "do point sources" ka special case hai; single-slit diffraction hai "ek continuous strip of sources."

Hum kya derive kar rahe hain

Step 1 — Slit ke across phase set up karna (first principles se)

Slit ko $N$ patli strips mein divide karo, har ek ki width $Δ y = a / N$ hai. Position $y$ ko top edge se measure karo ( $0$ se $a$ tak).

Position $y$ par ek strip ka path difference, top strip ke relative mein, angle $θ$ ki taraf jaate hue: $Δ (y) = y sin θ .$

Toh $y$ par strip ka phase hai $ϕ (y) = \frac{2 π}{λ} y sin θ .$

Top ( $y = 0$ ) aur bottom ( $y = a$ ) edges ke beech total phase difference ek important quantity hai: $Φ = \frac{2 π}{λ} a sin θ = 2 β, β \equiv \frac{π a s i n θ}{λ} .$

Step 2 — Wavelets ko add karna (phasor / integral method)

Har strip ek tiny field contribute karti hai. $N$ strips ke saath, ek strip ki amplitude $A_{0} / N$ hai, jahan $A_{0}$ woh total amplitude hai agar sab in phase hote ( $θ = 0$ ). Complex phasors use karke:

$E = \frac{A _{0}}{N} \sum_{n = 0}^{N - 1} e^{i ϕ_{n}}, ϕ_{n} = \frac{2 π}{λ} sin θ \cdot \frac{na}{N} .$

$N \to \infty$ le lo (continuous slit). Sum integral ban jaata hai: $E = \frac{A _{0}}{a} \int_{0}^{a} exp (i \frac{2 π}{λ} y sin θ) d y .$

Maano $k^{'} = \frac{2 π}{λ} sin θ$ . Tab $E = \frac{A _{0}}{a} \int_{0}^{a} e^{i k^{'} y} d y = \frac{A _{0}}{a} \cdot \frac{e ^{i k^{'} a} - 1}{i k ^{'}} .$

Simplify kaise karein: half-angle factor out karo (standard "phasor chord" trick): $e^{i k^{'} a} - 1 = e^{i k^{'} a /2} (e^{i k^{'} a /2} - e^{- i k^{'} a /2}) = e^{i k^{'} a /2} \cdot 2 i sin (\frac{k ^{'} a}{2}) .$

Ab $\frac{k ^{'} a}{2} = \frac{π a sin θ}{λ} = β$ . Substitute karo: $E = \frac{A _{0}}{a} \cdot \frac{e ^{i β} 2 i s i n β}{i k ^{'}} = A_{0} e^{i β} \frac{s i n β}{β}$ jahan humne $k^{'} a /2 = β \Rightarrow k^{'} = 2 β / a$ use kiya, toh $\frac{1}{a} \cdot \frac{2}{k ^{'}} = \frac{1}{a} \cdot \frac{a}{β} = \frac{1}{β}$ .

Step 3 — Intensity

Intensity = (amplitude)². Phase factor $e^{i β}$ ki magnitude 1 hai, isliye woh disappear ho jaata hai: $I (θ) = I_{0} (\frac{sin β}{β})^{2}, β = \frac{π a sin θ}{λ}, I_{0} = A_{0}^{2} .$

Figure — Diffraction — single slit intensity pattern derivation

Step 4 — Formula se physics padhna

Minima (dark fringes): $I = 0$ jab $sin β = 0$ lekin $β \neq = 0$ , yaani $β = mπ$ for $m = \pm 1, \pm 2, \dots$ $\frac{π a s i n θ}{λ} = mπ \Rightarrow a sin θ = mλ, m = \pm 1, \pm 2, \dots$

Secondary maxima: approximately jahan $tan β = β$ , giving $β \approx 1.43 π, 2.46 π, \dots$ Pehle secondary peak ki intensity: $\frac{I}{I _{0}} = (\frac{s i n ( 1.43 π )}{1.43 π})^{2} \approx \frac{1}{( 1.5 π ) ^{2}} \approx 0.045 = 4.5%.$

Central maximum ki angular width: $m = \pm 1$ minima ke beech, $sin θ = \pm \frac{λ}{a} \Rightarrow half-width θ_{1} \approx \frac{λ}{a} (small θ) .$

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

#flashcards/physics

Single-slit intensity formula kya hai?

I = I_{0} (\frac{sin β}{β})^{2}

with

β = \frac{π a sin θ}{λ}

β

physically kya represent karta hai?

Slit ke top aur bottom edges ke beech total phase difference ka aadha.

Dark fringes (minima) ke liye condition?

a sin θ = mλ

, with

m = \pm 1, \pm 2, \dots

(NOT

m = 0

a sin θ = mλ

dark kyun hai, bright kyun nahi?

Wavelets slit ke across pairs mein ban jaate hain aur exactly cancel ho jaate hain (pairs mein

λ /2

apart).

θ = 0

par kya hota hai?

β \to 0

sin β / β \to 1

, intensity

= I_{0}

: central maximum (sabse bright).

Central maximum ki angular half-width?

θ_{1} \approx λ / a

(small angle).

Slit width

a

kam karne ka effect?

Pattern wider spread ho jaata hai (

θ_{1} \propto 1/ a

); zyada diffraction.

Pehle secondary maximum ki relative intensity?

Approximately

4.5%

I_{0}

(

β \approx 1.43 π

par).

Secondary maxima ke liye exact condition?

tan β = β

(not

β = (m + \frac{1}{2}) π

a sin θ = λ /2

par intensity?

(2/ π)^{2} \approx 0.405 I_{0}

Slit problems mein

a

aur

d

mein difference?

a

=slit width → diffraction envelope;

d

=slit separation → interference fringes.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek wide doorway hai aur tiny torch-logon ki bheed us par shoulder to shoulder khadi hai, saari ek saath door wall ki taraf light flash kar rahi hain. Agar tum seedha saamne khade ho, har torch ki light saath pahunchti hai aur woh super bright hoti hai. Lekin agar tum sideways chalo, doorway ke ek edge ki light doosre edge ki light se thodi zyada door travel karti hai. Jab woh extra distance exactly ek full "wave step" hoti hai, left side ki torches right side ki torches ke exactly opposite line up ho jaati hain — left kehti hai "upar," right kehti hai "neeche," aur woh cancel ho kar zero ho jaati hain: ek dark stripe. Doorway ko narrow karo aur wall par bright patch wider ho jaata hai, kyunki narrow gaps waves ko aur zyada fan out karti hain. Yahi phailna diffraction hai.

Connections

Young's Double Slit Experiment — interference; single-slit uska continuous-source generalization hai.
Huygens Principle — woh wavelets provide karta hai jinhein humne sum kiya.
Phasor Addition of Waves — Step 2 mein use kiya gaya chord-of-an-arc method.
Diffraction Grating — kaafi saari slits; yahan envelope, grating ke sharp peaks ko multiply karta hai.
Rayleigh Criterion and Resolution — $θ \approx 1.22 λ / D$ use karta hai, wahi $λ / a$ spreading idea.
Fraunhofer vs Fresnel Diffraction — humara far-screen (parallel ray) assumption Fraunhofer hai.
Heisenberg Uncertainty Principle — narrow slit ⇒ wide angular spread uska optical analogue hai.