Interference — constructive, destructive conditions: Worked examples

1.6.17 · D3 · Physics › Oscillations & Waves › Interference — constructive, destructive conditions

Yeh parent note ka companion hai. Wahan humne do master laws derive kiye the. Yahan hum har tarah ke questions drill karenge jo wo laws produce kar sakte hain — $n$ ke har sign ke saath, zero aur degenerate cases, limits, ek word problem, aur ek exam twist. Parent page par jo kuch hai usse koi contradiction nahi; hum bas wider aur slower jaate hain.

Shuru karne se pehle, ek reminder plain words mein taaki koi symbol unearned na rahe:

Recall Do tools jo hum har jagah reuse karte hain

$ϕ$ (phase difference) = "kitne radians out of step hain dono waves." Ek full step-around ek circle ka $2 π$ radians hota hai.
$Δ x$ (path difference) = "ek wave ne doosre se kitna zyada travel kiya," metres mein measure kiya.
$λ$ (wavelength) = "ek full wave ki length," metres mein.
$I_{0}$ = ek wave akele ki intensity (brightness/loudness). Intensity kitni energy wave deliver karta hai, aur yeh amplitude ke square ke saath scale karta hai.
Dono worlds ke beech ka bridge: $ϕ = \frac{2 π}{λ} Δ x$ .
Do equal sources: $I = 4 I_{0} cos^{2} \frac{ϕ}{2}$ . Unequal sources $a_{1}, a_{2}$ : $A = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos ϕ$ .

Scenario matrix

Interference ke baare mein har exam question in cells mein se kisi ek mein aata hai. Neeche har example apni cell ke saath tagged hai.

#	Cell (case class)	Isme kya special hai	Example
A	Phase given, ordinary angle	$ϕ$ ko $4 I_{0} cos^{2} (ϕ /2)$ mein plug karo	Ex 1
B	Path given, integer ratio	$Δ x / λ$ whole hai → constructive	Ex 2
C	Path given, half-integer ratio	$Δ x / λ$ half hai → destructive	Ex 2
D	The zero case ( $Δ x = 0$ , $ϕ = 0$ )	central maximum, sabse bright, sign trap	Ex 3
E	Negative order $n$	source doosri side par; $n$ ka sign	Ex 4
F	In-between value (na max na min)	answer $4 I_{0}$ ka fraction hoga	Ex 5
G	Degenerate: unequal amplitudes	fully cancel nahi ho sakta; law of cosines	Ex 6
H	Degenerate limit: $a_{2} \to 0$	ek wave khatam → no fringes	Ex 6b
I	Real-world word problem	speakers / thin film, $Δ x$ extract karo	Ex 7
J	Exam twist: intensity ratio → amplitude ratio	pehle $I \propto a^{2}$ invert karo	Ex 8

Example 1 — Cell A: phase given, ordinary angle

Example 2 — Cells B & C: path given, integer vs half-integer

Worked example Do speakers sound emit karte hain

$λ = 0.40$ m ka. Listener P ka path difference $Δ x = 1.20$ m hai; listener Q ka $Δ x = 1.00$ m hai. Dono ke liye loud ya quiet? Forecast: Har $Δ x$ ko $λ$ se divide karo. Whole → loud, half → quiet. Compute karne se pehle P aur Q ke liye guess karo.

Step 1 — P: $\frac{Δ x}{λ} = \frac{1.20}{0.40} = 3.0$ . Yeh step kyun? Yeh count karna ki extra path mein kitne whole wavelengths fit hote hain, batata hai ki waves re-line-up hoti hain ya nahi. Whole number matlab exactly re-line-up. $3.0 = 3 λ$ → integer → constructive → P loud hai ( $n = 3$ ).

Step 2 — Q: $\frac{Δ x}{λ} = \frac{1.00}{0.40} = 2.5 = (2 + \frac{1}{2})$ . Yeh step kyun? Half-extra wavelength ek wave ke crest ko doosre ke trough ke against flip kar deta hai. Half-integer → destructive → Q quiet hai ( $n = 2$ ).

Verify: Phase mein convert karke double-check karte hain. $ϕ_{P} = \frac{2 π}{0.40} (1.20) = 6 π = 2 π \cdot 3$ → $2 π$ ka multiple → constructive ✓. $ϕ_{Q} = \frac{2 π}{0.40} (1.00) = 5 π = (2 \cdot 2 + 1) π$ → $π$ ka odd multiple → destructive ✓. Dono routes agree hain — Coherence and Path Difference dekho ki kyun steady $Δ x$ ise stable banata hai.

Example 3 — Cell D: zero case (classic trap)

Example 4 — Cell E: negative order $n$ (doosri side)

Worked example Ek fringe pattern mein (dekho

Young's Double Slit Experiment) ek bright fringe wahan hai jahan $Δ x = - 2 λ$ hai. Kya yeh valid hai? Konsa order? Kya yeh $+ 2 λ$ jaisa bright hai? Forecast: Kya minus sign brightness change karta hai? Guess karo.

Step 1 — Sign ko geometrically samjho.

Negative

Δ x

ka matlab sirf yeh hai ki doosra slit farther wala hai — point centre ke opposite side par hai. Figure dekho: do coral rays ne swap kar liya hai ki kaun zyada lamba hai. Yeh step kyun?

Δ x

mein sign sirf isliye hai ki hume pata chale kaun sa wave lead kar raha hai; brightness kisi label ki parwah nahi kar sakti.

Step 2 — Condition apply karo. $Δ x = nλ$ with $n = - 2$ . Kyunki $- 2$ integer hai, yeh constructive hai — ek valid bright fringe, order $n = - 2$ . Yeh step kyun? Constructive condition $n = 0, \pm 1, \pm 2, \dots$ allow karta hai — negatives bhi included hain.

Step 3 — Brightness. $ϕ = \frac{2 π}{λ} (- 2 λ) = - 4 π$ ; $cos^{2} (- 4 π /2) = cos^{2} (- 2 π) = 1$ , toh $I = 4 I_{0}$ . Yeh step kyun? $cos$ ek even function hai: $cos (- θ) = cos θ$ . Sign wash out ho jaata hai.

Verify: $cos^{2} (- 2 π) = cos^{2} (2 π) = 1$ ✓. $+ 2 λ$ fringe jaisi hi brightness — pattern centre ke baare mein symmetric hai, exactly jaisa double-slit fringes ke figures mein dikhta hai.

Example 5 — Cell F: ek in-between value

Example 6 — Cell G: unequal amplitudes (degenerate — fully cancel nahi ho sakta)

Worked example Sources ki amplitude

$a_{1} = 5 a$ aur $a_{2} = 2 a$ hai. Maximum aur minimum resultant amplitude nikalo, aur max/min intensity relative to $I_{0}$ (ek single unit- $a$ wave ka). Forecast: Unequal waves ke saath, kya destructive interference exactly zero tak pahunch sakta hai? Pehle yes/no guess karo.

Step 1 — General resultant (law of cosines).

A = (5 a)^{2} + (2 a)^{2} + 2 (5 a) (2 a) cos ϕ

Yeh step kyun? Jab amplitudes differ karti hain toh hum

sin A + sin B

use nahi kar sakte; instead hum dono waves ko phasors ke roop mein add karte hain —

5 a

aur

2 a

length ke chhote arrows jinke beech

ϕ

angle hai (figure dekho). Sum ki length law of cosines hai.

Step 2 — Maximum ( $ϕ = 0$ , arrows aligned). $cos 0 = 1$ : $A_{m a x} = 25 a^{2} + 4 a^{2} + 20 a^{2} = 49 a^{2} = 7 a .$ Yeh step kyun? Aligned arrows apni lengths simply add kar dete hain: $5 a + 2 a = 7 a$ .

Step 3 — Minimum ( $ϕ = π$ , arrows opposed). $cos π = - 1$ : $A_{m i n} = 25 a^{2} + 4 a^{2} - 20 a^{2} = 9 a^{2} = 3 a = ∣5 a - 2 a ∣.$ Yeh step kyun? Opposed arrows subtract karte hain: bacha hua difference hota hai, jab lengths differ karein toh kabhi zero nahi.

Step 4 — Intensities. Ek unit- $a$ wave ka $I_{0} \propto a^{2}$ ke saath: $I_{m a x} \propto (7 a)^{2} = 49 a^{2} = 49 I_{0}, I_{m i n} \propto (3 a)^{2} = 9 a^{2} = 9 I_{0} .$

Verify: $A_{m i n} = 3 a \neq = 0$ ✓ — forecast "fully cancel nahi ho sakta" sahi hai: full cancellation ke liye equal amplitudes chahiye. Sanity: $A_{m a x} = 7 a = 5 a + 2 a$ aur $A_{m i n} = 3 a = ∣5 a - 2 a ∣$ , triangle side ke do extremes ✓.

Example 6b — Cell H: limit $a_{2} \to 0$

Worked example Upar wale setup mein,

$a_{2} \to 0$ hone do (doosra source switch off ho gaya). Fringes ka kya hoga? Forecast: Koi second wave nahi — kya phir bhi bright/dark alternation ho sakta hai? Guess karo.

Step 1 — Limit lo. $a_{2} = 0$ ke saath: $A = a_{1}^{2} + 0 + 0 = a_{1}$ har $ϕ$ ke liye. Yeh step kyun? $cos ϕ$ term $a_{2}$ se multiply hota hai; $a_{2}$ kill karo aur phase dependence bilkul khatam ho jaati hai.

Step 2 — Interpret karo. $A$ ab $ϕ$ par depend nahi karta, toh $I = a_{1}^{2}$ har jagah — uniform, na maxima, na minima. Yeh step kyun? Interference hai hi $A$ ki $ϕ$ ke saath variation. Ek wave remove karo aur interfere karne ke liye kuch nahi bachta; pattern flat ho jaata hai (dekho Energy in Waves — ek source ke saath energy sirf evenly spread hoti hai).

Verify: $A_{m a x} = A_{m i n} = a_{1}$ toh fringe contrast $(A_{m a x} - A_{m i n}) = 0$ ✓ — koi visible fringes nahi, exactly degenerate limit.

Example 7 — Cell I: real-world word problem (thin film / speakers)

Worked example Ek listener equidistant khada hai... phir chalta hai jab tak sound pehli baar silent na ho jaaye. Do speakers (

$λ = 0.68$ m) ka ab path difference $Δ x$ hai. Pehla positive $Δ x$ kya hai jo pehla silence deta hai, aur wahan $ϕ$ kya hai? Forecast: "Pehla silence" — destructive condition ka kaun sa order $n$ hai? $n = 0$ ya $n = 1$ ? Guess karo.

Step 1 — Sahi condition chunno. Silence = destructive: $Δ x = (n + \frac{1}{2}) λ$ . Yeh step kyun? Silence cancellation hai, jo destructive (half-integer) family hai, integer wali nahi.

Step 2 — Sabse chhota positive $Δ x$ . Sabse chhota non-negative choice $n = 0$ hai: $Δ x = (0 + \frac{1}{2}) (0.68) = 0.34 m .$ Yeh step kyun? Centre se bahar chalne par, pehla dead spot $n = 0$ destructive wala hota hai, half a wavelength ka extra path.

Step 3 — Wahan phase. $ϕ = \frac{2 π}{0.68} (0.34) = π$ . Yeh step kyun? Half a wavelength ka path exactly half a cycle hai = $π$ radians = perfect opposition.

Verify: $cos^{2} (π /2) = 0$ → $I = 0$ ✓ genuine silence. Units: $0.34$ m ek physical distance hai, aur yeh exactly $λ /2$ hai jaise dead spot hona chahiye ✓.

Example 8 — Cell J: exam twist (pehle intensity ratio → amplitude ratio)

Worked example Do coherent sources ki

intensities ratio mein hain $I_{1} : I_{2} = 9 : 1$ . Interference pattern ka $I_{m a x} : I_{m i n}$ ratio nikalo. Forecast: Tempting hai $9 + 1 = 10$ aur $9 - 1 = 8$ likhna. Lekin intensities aise add nahi hoti. Guess karo ki sach mein answer $10 : 8$ se bada hai ya chhota.

Step 1 — Intensity ratio ko amplitude ratio mein convert karo. Kyunki $I \propto a^{2}$ , $a \propto I$ : $\frac{a _{1}}{a _{2}} = \frac{I _{1}}{I _{2}} = 9 = 3 \Rightarrow a_{1} : a_{2} = 3 : 1.$ Yeh step kyun? Interference amplitudes add karta hai, intensities nahi. Combine karne se pehle hume amplitude-land mein aana hoga — is poore topic ka recurring lesson.

Step 2 — Extreme amplitudes. $A_{m a x} = a_{1} + a_{2} = 3 + 1 = 4, A_{m i n} = ∣ a_{1} - a_{2} ∣ = 3 - 1 = 2.$ Yeh step kyun? Aligned phasors add hote hain ( $ϕ = 0$ ); opposed phasors subtract hote hain ( $ϕ = π$ ) — Ex 6 jaisa hi.

Step 3 — Wapas intensities mein (square karo). $\frac{I _{m a x}}{I _{m i n}} = \frac{A _{m a x}^{2}}{A _{m i n}^{2}} = \frac{4 ^{2}}{2 ^{2}} = \frac{16}{4} = 4.$ Yeh step kyun? Ab hum amplitude-land chhodne ke liye square karte hain, kyunki question brightness pooch raha hai.

Verify: $I_{m a x} : I_{m i n} = 4 : 1$ , nahi $10 : 8$ ✓ — forecast trap se bacha. General formula check: $\frac{( I _{1} + I _{2} ) ^{2}}{( I _{1} - I _{2} ) ^{2}} = \frac{( 3 + 1 ) ^{2}}{( 3 - 1 ) ^{2}} = 4$ ✓.

Recall

Recall Matrix par quick fire

Q: $Δ x = 0$ — bright ya dark? ::: Bright — central maximum ( $n = 0$ ), $I = 4 I_{0}$ . Q: Kya unequal-amplitude waves $I = 0$ tak pahunch sakti hain? ::: Nahi — minimum $(a_{1} - a_{2})^{2} \neq = 0$ hota hai. Q: $I_{1} : I_{2} = 9 : 1$ diya ho, $I_{m a x} : I_{m i n}$ kya hai? ::: $4 : 1$ (pehle $a$ -ratio $3 : 1$ mein convert karo). Q: Sabse chhota destructive path difference? ::: $λ /2$ (the $n = 0$ dead spot). Q: Kya negative order $n$ brightness change karta hai? ::: Nahi — $cos^{2}$ even hai, sign wash out ho jaata hai.

Connections

Parent: conditions — do laws yahan drill ki gayi hain.
Principle of Superposition — kyun amplitudes (intensities nahi) add hote hain.
Young's Double Slit Experiment — jahan $\pm n$ orders (Ex 4) ek screen par rehte hain.
Coherence and Path Difference — kyun steady $Δ x$ Ex 2 ko stable banata hai.
Energy in Waves — $a_{2} \to 0$ limit ka flat pattern (Ex 6b).
Standing Waves aur Beats — sibling superposition patterns.