Worked examples — Diffraction — single slit intensity pattern derivation

2.5.14 · D3 · Physics › Optics › Diffraction — single slit intensity pattern derivation

Yeh page single-slit intensity derivation ke liye problem gym hai. Parent note ne master formula bilkul scratch se banaya tha:

Koi bhi number karne se pehle, ek adat: yahan har quantity ek length ya ek angle hai, isliye units har line mein survive karni chahiye. Agar kabhi ek length ko length se divide karo aur "metres" mile, toh galti hui hai.

---

Scenario matrix

Diffraction problems alag-alag lagte hain lekin yeh sab ek grid ke cells hain. Neeche har worked example ko us cell ke saath tag kiya gaya hai jo woh fill karta hai, taaki end tak tune har corner hit kar liya ho.

• Cell A — minimum ka angle dhundho. $\lambda ,a$ diye hain, $a\sin \theta =m\lambda$ solve karo. → Ex 1
• Cell B — screen par minimum ki position dhundho. Screen distance $L$ add karo, $y\approx L\theta$ use karo. → Ex 2
• Cell C — kisi diye gaye angle par intensity. $\beta$ ko $(\sin \beta /\beta {)}^2$ mein plug karo. → Ex 3
• Cell D — secondary maximum. Exactly beech mein nahi: $\tan \beta =\beta$ solve karo. → Ex 4
• Cell E — degenerate / limiting inputs. $\theta =0$; $a=\lambda$; $a<\lambda$ (koi dark fringe nahi); $a\to \infty$ (diffraction khatam). → Ex 5
• Cell F — inverse problem. Pattern diya hai, $a$ ya $\lambda$ dhundho. → Ex 6
• Cell G — real-world word problem. Laser ek baal se guzre, wall par projection. → Ex 7
• Cell H — exam twist: do colours / combined constraint. Do wavelengths ke minima kab coincide karte hain? → Ex 8

Shuru karne se pehle do edge cases par flag lagana zaroori hai, kyunki yeh log ko faansate hain:

---

Cell A — Minimum ka angle

1. Dark-fringe condition likho. Yeh step kyun? Minima — maxima nahi — sharp landmarks hote hain; parent note ne pair-cancellation se dikhaya tha ki $a\sin \theta =m\lambda$. $\sin {\theta }_m=\frac{m\lambda }{a}=\frac{m(650\times 10^{-9})}{0.12\times 10^{-3}}=m\times 5.417\times 10^{-3}.$

2. $m=1$. Kyun: yeh central blob ka inner edge hai. $\sin {\theta }_1=5.417\times 10^{-3}\Rightarrow {\theta }_1=\arcsin (0.005417)=0.3103^{\circ }.$

3. $m=2$. Kyun: agla dark ring baahir ki taraf. $\sin {\theta }_2=1.0833\times 10^{-2}\Rightarrow {\theta }_2=0.6207^{\circ }.$

Forecast answer: haan, almost bilkul double yahan — kyunki $\theta$ bahut chhota hai, $\sin \theta \approx \theta$, isliye $\sin \theta$ double karne par $\theta$ bhi double hota hai. Yeh coincidence bade angles par toot jaata hai.

Recall Verify

Sanity: $a\gg \lambda$ (120 μm vs 0.65 μm), isliye angles bahut chhote hone chahiye — hain bhi (sub-degree). ✓ Ratio check ::: ${\theta }_2/{\theta }_1=0.6207/0.3103=2.00$, small-angle doubling se match karta hai.

---

Cell B — Screen par position

1. Angle ko screen position mein convert karo, aur small-angle chain ko justify karo. Yeh step kyun? Screen flat aur door hai, isliye $\theta$ angle par ek ray $y=L\tan \theta$ height par exactly hit karti hai. Yahan ${\theta }_1=0.31^{\circ }=5.42\times 10^{-3}$ rad — bahut chhota number. Itne chhote angles ke liye radians mein Taylor expansions dete hain $\tan \theta =\theta +\frac{1}{3}{\theta }^3+\cdots \approx \theta$ aur $\sin \theta =\theta -\frac{1}{6}{\theta }^3+\cdots \approx \theta$, dono $10^5$ mein ek hisse se bhi zyada accurate. Isliye $\tan \theta \approx \sin \theta \approx \theta$ aur hum jo convenient ho woh use kar sakte hain — yahan $\sin {\theta }_1$ Ex 1 se pehle se pata hai. Figure dekho: red ray slit se ${\theta }_1$ angle par nikalti hai aur screen par $y_1$ par lagti hai. $y_1\approx L{\theta }_1\approx L\sin {\theta }_1=1.8\times 5.417\times 10^{-3}=9.75\times 10^{-3}\text{m}=9.75\text{mm}.$

2. Full central width. Kyun: central bright band $m=-1$ dark fringe se $m=+1$ dark fringe tak jaata hai — symmetric, isliye $y_1$ ka double. $W=2y_1=19.5\text{mm}.$

3. Outer fringes ki spacing se compare karo. Kyun: outer dark fringes $\Delta y=L\lambda /a=9.75\text{mm}$ se spaced hain. Toh central band ($19.5$ mm) kisi bhi do neighbouring outer fringes ke beech gap se double wide hai.

Forecast answer: wider — central maximum har diffraction pattern ka signature "mota beech" hota hai.

Recall Verify

Units ::: $[\text{m}]\times [\text{dimensionless}]=[\text{m}]$ ✓ — $\sin \theta$ (aur $\theta$ radians mein) koi units nahi rakhta. $W=2\times 9.75=19.5$ mm, aur outer spacing $=L\lambda /a=9.75$ mm, ratio exactly $2$. ✓

---

Cell C — Arbitrary angle par intensity

1. $\beta$ dhundho. Yeh step kyun? Intensity sirf $\beta$ par depend karti hai, seedha $\theta$ par nahi. $\beta =\frac{\pi a\sin \theta }{\lambda }=\pi \cdot \frac{3}{4}=\frac{3\pi }{4}.$

2. Sinc-squared mein plug karo. Kyun: yahi master formula HAI. $\frac{I}{I_0}={\left(\frac{\sin (3\pi /4)}{3\pi /4}\right)}^2={\left(\frac{0.7071}{2.356}\right)}^2=(0.3001{)}^2=\mathrm{0.0901.}$

3. Interpret karo. Kyun: $\approx 9\%$ — pehle se bahut dim, $\beta =\pi$ par zero ki taraf tezi se ja raha hai.

Forecast answer: below half. Sinc-squared curve tezi se girta hai, isliye "angle mein teen-chauthaayi raaste par" pehle hi ~9% tak gir chuka hai, ~25% nahi.

Recall Verify

$\sin (3\pi /4)=\frac{\sqrt{2}}{2}=0.7071$, $2.356$ se divide karo, square karo → $0.0901$. ✓ Parent ke Ex 3 se cross-check: $a\sin \theta =\lambda /2$ par ($\beta =\pi /2$), $I/I_0=4/{\pi }^2=0.405$. Hamara point aage hai, isliye dimmer hona chahiye — $0.09<0.405$ ✓.

---

Cell D — Secondary maximum ("not halfway" case)

1. Differentiate karo — slope zero kahan hai dekho. Yeh step kyun? $f(\beta )={\left(\frac{\sin \beta }{\beta }\right)}^2$ ka peak wahan hai jahan $f^{\prime }(\beta )=0$. $g(\beta )=\frac{\sin \beta }{\beta }$ likho; phir $f=g^2$ aur $f^{\prime }=2g{g}^{\prime }$, isliye hume $g=0$ chahiye (yeh dark fringes hain, woh nahi chahiye) YA $g^{\prime }=0$. $g=\sin \beta /\beta$ par quotient rule use karke: $g^{\prime }(\beta )=\frac{\beta \cos \beta -\sin \beta }{{\beta }^2}.$ Numerator zero karo: $\beta \cos \beta -\sin \beta =0$. $\cos \beta$ se divide karo: $\beta -\tan \beta =0\Rightarrow \boxed{\tan \beta =\beta }.$ Geometrically (figure dekho) yeh wahan hai jahan straight line $y=\beta$, $y=\tan \beta$ ki branch ko cross karti hai. Kyunki $\tan \beta$ $\pi$ se tezi se $1.5\pi$ par apne asymptote ki taraf shoot karta hai, yeh gently-rising line $y=\beta$ se $1.5\pi$ pahunchne se pehle milta hai — isliye peak centre ki taraf pull hoti hai.

2. Numerically solve karo. Kyun: $\tan \beta =\beta$ ka koi clean closed form nahi hai. ${\beta }_1\approx 1.4303\pi =4.4934\text{rad}.$

3. Wahan intensity. Kyun: ${\beta }_1$ ko envelope mein plug karo. $\frac{I}{I_0}={\left(\frac{\sin (4.4934)}{4.4934}\right)}^2=0.04718\approx 4.7\%.$

Forecast answer: $1.5\pi$ par NAHI — yeh $1.430\pi$ par hai, andar ki taraf shift hua. Side lobe central brightness ka 5% se kam carry karta hai, isliye diffraction blobs mein ek bright core hota hai aur faint ears.

Recall Verify

${\beta }_1=4.4934$: check karo $\tan (4.4934)=4.4934$ (self-consistent root). Intensity $(\sin (4.4934)/4.4934{)}^2=0.0472$. ✓ Parent ke stated $\approx 4.5\%$ se match karta hai (parent ne $\beta$ ko $1.5\pi$ tak round kiya tha; exact root se $4.7\%$ milta hai).

---

Cell E — Degenerate & limiting inputs

(a) Dead centre, $\theta =0$. Kyun: sab maan lete hain sabse bright hai lekin prove karna chahiye. $\beta =0\Rightarrow \frac{\sin \beta }{\beta }\to 1\Rightarrow I=I_0.$ Ratio $\frac{\sin \beta }{\beta }\to 1$ as $\beta \to 0$ small-angle limit hai ($\sin x\approx x$). Sare wavelets in phase → peak. Dark fringe nahi (yeh $m=0$ wala trap parent note se hai).

(b) $a=\lambda$. Kyun: pehle minimum ko $\sin \theta =m\lambda /a=1$ chahiye. $\sin {\theta }_1=\frac{\lambda }{a}=1\Rightarrow {\theta }_1=90^{\circ }.$ Pehla dark fringe "forward" ke bilkul edge tak push ho gaya. Sirf ek minimum ($m=1$) physically reachable hai; $m=2$ ko $\sin \theta =2>1$ chahiye → impossible.

(c) $a=0.5\lambda$. Kyun: existence check karo. $\sin {\theta }_1=\frac{\lambda }{a}=\frac{\lambda }{0.5\lambda }=2>1.$ Koi solution nahi. Zero dark fringes hain — central maximum puri forward hemisphere mein fail jaata hai. Slit itni tight hai ki barely ek beam mein diffract karti hai; yeh almost ek single point source ki tarah behave karta hai jo har jagah radiate karta hai.

(d) $a\to \infty$ (bahut wide slit). Kyun: opposite extreme — kya hota hai jab slit $\lambda$ ke comparison mein bahut bada ho jaata hai? Pehla minimum hai $\sin {\theta }_1=\frac{\lambda }{a}\underset{a\to \infty }{\to }0\Rightarrow {\theta }_1\to 0.$ Central bright band zero angular width tak collapse ho jaata hai — saare dark fringes $\theta =0$ ki taraf bheed jaate hain, isliye pattern ek single infinitely-narrow spike seedha aage tak shrink ho jaata hai. Physically: diffraction khatam ho jaata hai aur light seedha ek plain beam ki tarah travel karti hai (ray-optics / geometric-optics limit). Wavelength se bahut bada hole — ek khidki, ek darwaza — sharp shadow dalta hai aur koi visible fringes nahi hote. Isliye hum kabhi bhi roz ke openings se diffraction nahi dekhte: unke liye $a\gg \lambda$ hai aur $\lambda /a$ unmeasurably chhota hai.

Forecast answer: $a\to \infty$ par light seedha through march karti hai — diffraction disappear ho jaata hai aur geometry (sharp shadows) le leti hai.

Recall Verify

(a) ${\lim }_{\beta \to 0}\sin \beta /\beta =1$ ✓. (b) $\lambda /a=1\Rightarrow \theta =90^{\circ }$ ✓. (c) $\lambda /a=2>1$, koi real $\arcsin$ nahi ✓. (d) ${\lim }_{a\to \infty }\lambda /a=0\Rightarrow {\theta }_1\to 0$ ✓.

---

Cell F — Inverse problem

1. Pehle half-width. Yeh step kyun? Hamare formulas ek fringe tak ki distance use karte hain; $W$ full width hai. $y_1=\frac{W}{2}=13\text{mm}=0.013\text{m}.$

2. Slit se relate karo. Kyun: $y_1=L\lambda /a$ (small-angle first minimum). $a$ ke liye rearrange karo. $a=\frac{L\lambda }{y_1}=\frac{2.5\times 546\times 10^{-9}}{0.013}.$

3. Compute karo. Kyun: ab sirf arithmetic hai. $a=\frac{1.365\times 10^{-6}}{0.013}=1.05\times 10^{-4}\text{m}=0.105\text{mm}=105\mu \text{m}.$

Forecast answer: ek millimetre ka roughly tenth — sau microns, ek typical lab slit.

Recall Verify

Units ::: $[\text{m}][\text{m}]/[\text{m}]=[\text{m}]$ ✓. Back-substitute ::: $y_1=L\lambda /a=2.5\times 546\times 10^{-9}/1.05\times 10^{-4}=0.013$ m ✓.

---

Cell G — Real-world word problem

1. Fringe spacing = neighbouring minima ke beech distance. Yeh step kyun? Adjacent dark fringes ($m$ aur $m+1$) small-angle regime mein $\Delta y=L\lambda /a$ se separated hote hain. $\Delta y=\frac{L\lambda }{a}.$

2. $a$ ke liye solve karo. Kyun: hair width wahi hai jo hum chahte hain. $a=\frac{L\lambda }{\Delta y}=\frac{3.0\times 632.8\times 10^{-9}}{0.021}=9.04\times 10^{-5}\text{m}=90\mu \text{m}.$

Forecast answer + reality check: ~90 μm. Real hair 50–100 μm hota hai — bilkul sahi. Yeh genuinely ek laser aur ruler se hair measure karne ka tarika hai.

Recall Verify

$a=3.0\times 632.8\times 10^{-9}/0.021=9.04\times 10^{-5}$ m $=90.4\mu$m ✓, human-hair range ke andar.

---

Cell H — Exam twist: do colours saath mein

1. Coincidence condition. Yeh step kyun? Blue ka dark fringe order $m_1$ par aur orange ka order $m_2$ par ek angle share karte hain jab $a\sin \theta =m_1{\lambda }_1=m_2{\lambda }_2.$

2. Ratio reduce karo. Kyun: $m_1{\lambda }_1=m_2{\lambda }_2$ satisfy karne wale sabse chhote integers $\frac{m_1}{m_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{640}{480}=\frac{4}{3}$ se aate hain. Isliye sabse chhota pair hai $m_1=4$ (blue), $m_2=3$ (orange).

3. Angle compute karo. Kyun: koi bhi colour plug karo — dono construction se agree karte hain. $\sin \theta =\frac{m_1{\lambda }_1}{a}=\frac{4\times 480\times 10^{-9}}{80\times 10^{-6}}=\mathrm{0.0240.}$ $\theta =\arcsin (0.0240)=1.3752^{\circ }.$

4. Orange se cross-check karo. Kyun: jab tak dono colours match na karein, kisi par trust mat karo. $\frac{m_2{\lambda }_2}{a}=\frac{3\times 640\times 10^{-9}}{80\times 10^{-6}}=\mathrm{0.0240.}✓$

Forecast answer: ratio $4:3$ hai, isliye blue ka 4th dark fringe orange ke 3rd se coincide karta hai, $\theta \approx 1.38^{\circ }$ par.

Recall Verify

$4\times 480=1920=3\times 640$ ✓ (dono $=1920$ nm·order). $\sin \theta =1920\times 10^{-9}/80\times 10^{-6}=0.024$, $\theta =1.3752^{\circ }$ ✓.

---

Connections

• Har "bright vs dark" rule ko Young's Double Slit Experiment se contrast karo — $a\sin \theta =m\lambda$ ka matlab opposite hai.
• Pair-cancellation logic Huygens Principle aur Phasor Addition of Waves par depend karti hai.
• Far-screen "$y=L\tan \theta$" geometry Fraunhofer limit assume karti hai.
• Many-slit generalisation → Diffraction Grating; do blobs resolve karna → Rayleigh Criterion and Resolution.
• "Narrow slit ⇒ wide spread" ka trend Heisenberg Uncertainty Principle ki wave root hai.