Exercises — Diffraction — single slit intensity pattern derivation

Pehla dark fringe → sabse chhota non-zero $m$, toh $m=1$. Use karo $a\sin \theta =m\lambda$: $\sin \theta =\frac{m\lambda }{a}=\frac{1\cdot 500\times 10^{-9}}{0.20\times 10^{-3}}=2.5\times 10^{-3}.$ Kyunki yeh bahut chhota hai, $\theta \approx \sin \theta$ radians mein: $\theta \approx 2.5\times 10^{-3}\text{rad}=0.1432^{\circ }.$ Kaisa dikhta hai: bahut chhota opening angle — pehla dark line almost straight ahead hai, kyunki $a$, $\lambda$ se bahut bada hai.

$a\sin \theta =2\lambda$, $a\sin \theta =m\lambda$ se match karta hai jab $m=2$ ho, jo ek pura number hai aur $\ne 0$. $\Rightarrow \text{DARK (second minimum).}$ $\beta$ se check karo: $\beta =\frac{\pi a\sin \theta }{\lambda }=\frac{\pi \cdot 2\lambda }{\lambda }=2\pi$, aur $\sin (2\pi )=0$, toh $I=0$. Dark confirm hua.

Screen par angle $\theta$ par ek point height $y=L\tan \theta$ par hota hai (exact geometry). Reference figure mein yahi woh jagah hai jahan slit se red ray screen par strike karta hai. Kyunki yahan diffraction angles bahut chhote hain ($\theta \sim 10^{-3}$ rad), hum small-angle approximation use karte hain: $\tan \theta \approx \sin \theta \approx \theta (\theta \text{ radians mein, }\theta \ll 1),$ toh $y\approx L\theta$. Yeh woh ek approximation hai jo hume screen geometry ($y=L\tan \theta$) ko diffraction formula ($a\sin {\theta }_1=\lambda$) ke saath consistently combine karne deta hai. Half-width $y_1$ pehle minimum tak pahunchti hai jahan ${\theta }_1\approx \lambda /a$: $y_1=L{\theta }_1=L\frac{\lambda }{a}=1.5\cdot \frac{500\times 10^{-9}}{0.20\times 10^{-3}}=3.75\times 10^{-3}\text{m}.$ Full width = dono sides: $W=2y_1=7.5\times 10^{-3}\text{m}=7.5\text{mm}.$

Hum wahi small-angle approximation reuse karte hain jo L2.1 mein thi: chhote angles ke liye $\tan \theta \approx \sin \theta \approx \theta$, toh screen height aur diffraction angle dono same $\theta$ share karte hain. Phir ${\theta }_1\approx \lambda /a$ aur $y_1=L{\theta }_1$ combine hoke dete hain: $y_1=\frac{L\lambda }{a}\Rightarrow a=\frac{L\lambda }{y_1}=\frac{2.0\cdot 650\times 10^{-9}}{6.5\times 10^{-3}}.$ $a=2.0\times 10^{-4}\text{m}=0.20\text{mm}.$ Approximation valid hai yeh check karo: ${\theta }_1\approx \lambda /a=650\times 10^{-9}/0.20\times 10^{-3}=3.25\times 10^{-3}$ rad — actually $\ll 1$ hai, toh $\tan \approx \sin \approx \theta$ $0.001\%$ se bhi behtar hold karta hai.

Path difference ko $\beta$ mein convert karo: $\beta =\frac{\pi a\sin \theta }{\lambda }=\frac{\pi \cdot \frac{2}{3}\lambda }{\lambda }=\frac{2\pi }{3}.$ Ab $I/I_0=(\sin \beta /\beta {)}^2$ mein plug karo: $\frac{I}{I_0}={\left(\frac{\sin (2\pi /3)}{2\pi /3}\right)}^2={\left(\frac{\sqrt{3}/2}{2\pi /3}\right)}^2={\left(\frac{3\sqrt{3}}{4\pi }\right)}^2\approx \mathrm{0.1710.}$ Toh central peak ka lagbhag $17\%$ — hum falling shoulder par hain, pehle dark fringe $a\sin \theta =\lambda$ ($\beta =\pi$) ki taraf ja rahe hain.

Lo $\beta =1.4303\pi \approx 4.4934$ rad. $\frac{I}{I_0}={\left(\frac{\sin (4.4934)}{4.4934}\right)}^2\approx {\left(\frac{-0.9775}{4.4934}\right)}^2\approx \mathrm{0.0472.}$ Yeh lagbhag $4.7\%$ hai — side bands centre se dramatically zyada faint hote hain. Envelope $1/{\beta }^2$ hi inhe crush karta hai. Kaisa dikhta hai: figure mein, sabse uncha side bump barely axis clear karta hai, towering central peak ke comparison mein.

Ek double-slit bright fringe disappear ho jaata hai jab woh exactly ek single-slit minimum par land karta hai.

• Double-slit bright: $d\sin \theta =p\lambda$.
• Single-slit dark: $a\sin \theta =m\lambda$.

Dono conditions divide karo (same $\theta$): $\frac{d}{a}=\frac{p}{m}\Rightarrow \frac{p}{m}=\frac{0.50}{0.10}=5.$ Toh $p=5m$: orders $p=\pm 5,\pm 10,\dots$ missing hain (pehla missing order $p=5$ hai, $m=1$ se kill hota hai). Kyun: us angle par dono slits in phase hain (double-slit kehta hai bright), lekin har individual slit ka output zero hai (single-slit envelope dark hai), toh bright $\times$ zero $=$ zero.

Wavelength $\lambda$ ka ek photon total momentum $p=h/\lambda$ carry karta hai. Half-angle ${\theta }_1$ ke central cone mein emerge hote hue, uska transverse (sideways) momentum roughly span karta hai: $\Delta p_y\approx p{\theta }_1=\frac{h}{\lambda }\cdot \frac{\lambda }{a}=\frac{h}{a}.$ Position spread $\Delta y\sim a$ se multiply karo: $\Delta y\Delta p_y\sim a\cdot \frac{h}{a}=h.$ Yeh kya kehta hai aur kya nahi kehta. $a$ cancel ho jaata hai — narrower slit → wider angular spread — aur product Planck's constant ke scale par aata hai, yaani $\Delta y\Delta p_y\sim h\sim 2\pi \hslash$. Yeh sirf ek order-of-magnitude argument hai: hamara $\Delta y\sim a$ aur $\Delta p_y\sim h/a$ rough "sizes" hain, precise statistical standard deviations nahi. Uncertainty principle ka exact, rigorous statement hai: $\Delta y\Delta p_y\ge \frac{\hslash }{2},$ aur hamara estimate ($\sim h=2\pi \hslash$) us floor ke comfortably upar baithta hai — jaisa hona chahiye. Takeaway yeh scaling hai ($\Delta p_y\propto 1/a$), yeh claim nahi ki product exactly $h$ ke barabar hai.

Hum kya karte hain: small $\beta$ ke liye $\sin \beta$ ko Taylor-expand karo: $\sin \beta =\beta -\frac{{\beta }^3}{6}+\frac{{\beta }^5}{120}-\cdots$ $\beta$ se divide karo: $\frac{\sin \beta }{\beta }=1-\frac{{\beta }^2}{6}+\cdots \stackrel{\beta \to 0}{\to }1.$ Square karne se limit $1$ rehta hai, toh $I(0)=I_0\cdot 1=I_0$. Kyun / kaisa dikhta hai (phasor picture). Parent derivation se yaad karo ki slit ko $N$ thin strips ka ek bada set imagine kiya jaata hai, har ek ek tiny wavelet of amplitude $A_0/N$ radiate karta hai, jahan $A_0$ woh total amplitude hai jo tumhe milti agar saare $N$ wavelets perfectly in phase hote. Jab $\theta \to 0$ path difference across slit vanish ho jaata hai, toh har wavelet ek saath chal raha hota hai. In $N$ equal-length arrows ko tip-to-tail add karna, bina kisi phase turn ke, ek full length $A_0$ ki straight line deta hai — kuch bhi cancellation se nahi khota. Woh maximal, undiminished sum ($I_0=A_0^2$) hi towering central peak hai.

Pehle minimum ko chahiye $\sin {\theta }_1=\lambda /a$.

• $a=\lambda$: $\sin {\theta }_1=1\Rightarrow {\theta }_1=90^{\circ }$. Pehla dark fringe visible half-plane ke bilkul edge tak push ho jaata hai. Central bright band ab puri forward screen fill karta hai.
• $a<\lambda$: $\sin {\theta }_1=\lambda /a>1$, jiska koi real solution nahi hai. Koi dark fringe bilkul nahi — intensity sirf sides mein smoothly fall off hoti hai ek single broad hump ke saath. Slit itni narrow hai ki woh almost ek point source (ek akela Huygens wavelet) ki tarah behave karta hai, bina kisi cancellation ke saare forward directions mein radiate karta hai.

Lesson: "fringe" picture ke liye $a>\lambda$ chahiye. Jab $a\le \lambda$ ho jaaye toh pattern degenerate hokar ek single spread-out glow ban jaata hai — "narrow slit ⇒ wide pattern" ki extreme.

L1.1 first minimum ::: $\theta \approx 0.1432^{\circ }$ L1.2 $a\sin \theta =2\lambda$ ::: dark (2nd minimum) L2.1 full central width ::: $7.5\text{mm}$ L2.2 slit width ::: $0.20\text{mm}$ L3.1 $I/I_0$ at $a\sin \theta =\frac{2}{3}\lambda$ ::: $\approx 0.171$ L3.2 first secondary peak ::: $\approx 0.047$ (4.7%) L4.1 first missing double-slit order ::: $p=5$ (family $5m$) L4.2 $\Delta y\Delta p_y$ ::: $\sim \hslash$ (order of magnitude; exact bound $\ge \hslash /2$) L5.1 $\lim (\sin \beta /\beta {)}^2$ ::: $1$, toh $I(0)=I_0$ L5.2 $a\le \lambda$ ::: koi dark fringes nahi (single broad hump)