Visual walkthrough — Interference — constructive, destructive conditions

1.6.17 · D2 · Physics › Oscillations & Waves › Interference — constructive, destructive conditions

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Step 1 — Ek akeli wave ek hilti-dulti number hai

KYA. Space mein ek point choose karo — use $P$ kaho. Wahan se guzarne wali ek wave us jagah ke medium ko (hawa, paani, ek string) time ke saath upar-neeche hilati hai. Us upar-ya-neeche ki matra ko hum $y$ likhte hain.

Sine kyun. Sabse simple, sabse smooth repeating motion — spring pe lage mass ki motion, jhule ki motion — jab tum displacement ko time ke against plot karo to ek sine curve banti hai. Isliye hum ek wave ko is tarah model karte hain:

$y_1=a\sin (\omega t)$

• $y_1$ — displacement: point $P$ abhi kitna upar ($+$) ya neeche ($-$) hai.
• $a$ — amplitude: sabse bada displacement jo yeh kabhi pahunchta hai. Crest ki unchai.
• $t$ — time, aage badhta hua.
• $\omega$ — angular frequency: wave kitni tez cycle karta hai. Bada $\omega$ = tezi se hilna. Yeh clock-time $t$ ko ek "angle" $\omega t$ mein badalta hai jo har cycle mein ek poora chakkar ($2\pi$) leta hai.

PICTURE. Lal curve time ke saath $y_1$ hai. Dhyan do yeh zero se shuru hoti hai, $+a$ tak uthti hai, $-a$ tak girti hai, aur repeat karti hai.

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Step 2 — Doosri wave, thodi peeche

KYA. Ab usi tarah ki ek doosri wave $P$ tak pahunchti hai, lekin yeh apne source se thodi der baad nikli (ya zyada door se aayi). Wahi shape, wahi amplitude, wahi speed — bas time axis ke saath shift hai.

$\varphi$ kyun. Hum us shift ko ek angle $\varphi$, yaani phase difference ke roop mein measure karte hain. "Phase" matlab hai wave apne cycle mein kahan hai. Agar do waves apne cycle ki usi jagah hain to $\varphi =0$; ek poora cycle peeche ho to $\varphi =2\pi$.

$y_2=a\sin (\omega t+\varphi )$

• $\varphi$ — phase difference: wave 2 apne khud ke cycle mein wave 1 se kitne radians aage hai. Sine ke andar $\varphi$ jodne se curve picture mein baayein (pehle) khisar jaata hai.

PICTURE. Laal $y_1$ hai, teal $y_2$ hai. Dono ki unchai aur rhythm ek jaisi hai; teal bas $\varphi$ jitna side mein khisak gayi hai.

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Step 3 — Superposition: bas dono heights jodo

KYA. Har ek pal, $P$ par medium sirf ek jagah ho sakta hai. Isliye uska actual displacement woh sum hai jo akele har wave demand karti:

$y=y_1+y_2=a\sin (\omega t)+a\sin (\omega t+\varphi )$

Jodna kyun, multiply ya average kyun nahi. Yeh Principle of Superposition hai — chhoti waves ke liye ek physical fact. Ek jhule ko ek saath do baar dhakka do, dhakke literally add ho jaate hain. Isse zyada kuch fancy nahi ho raha.

PICTURE. Plum curve point-by-point sum hai. Jahan laal aur teal dono upar hain, plum bada hai. Jahan ek upar aur doosra neeche hai, woh aadha cancel ho jaate hain aur plum chota hota hai.

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Step 4 — Ek jaisi frequency ke do sines ek SINE mein jud jaate hain

KYA. Hum ek trig identity use karte hain sum ko ek single sine mein compress karne ke liye:

$\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}$

$A=\omega t$ aur $B=\omega t+\varphi$ rakhte hain:

$y=\underset{A_{\text{res}}}{\underbrace{2a\cos \frac{\varphi }{2}}}\sin \left(\omega t+\frac{\varphi }{2}\right)$

Har term:

• $\frac{A+B}{2}=\omega t+\frac{\varphi }{2}$ — nayi sine abhi bhi frequency $\omega$ pe cycle karti hai; bas phase gap ke aadhe se shift hui hai. Wahi rhythm.
• $\frac{A-B}{2}=-\frac{\varphi }{2}$, aur $\cos (-x)=\cos x$, isliye front factor $2a\cos \frac{\varphi }{2}$ hai.
• Woh poora front factor time mein constant hai (usme koi $t$ nahi) — yeh nayi amplitude $A_{\text{res}}$ hai.

Yeh kyun matter karta hai. Ek jaisi frequency ke do sines ka sum abhi bhi usi frequency ki sine hai — bas uski unchai aur starting point badalta hai. Isliye "combined wave kitni badi hai" ki saari physics sirf ek number $A_{\text{res}}=2a\cos (\varphi /2)$ mein hai.

PICTURE. Step 3 ka plum sum phir se draw kiya gaya hai, aur ek dashed envelope uski sachi amplitude $2a\cos (\varphi /2)$ mark karti hai — ek saaf sine.

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Step 5 — Amplitude ko phasor triangle se samjho

KYA. Ek aisi picture hai jo $2a\cos (\varphi /2)$ ko obvious bana deti hai. Har wave ko length $a$ ka ek arrow (ek "phasor") ki tarah draw karo. Wave 1 ek direction mein point karti hai; wave 2 us se angle $\varphi$ pe rotate hai. Resultant woh arrow hai jo unhe tip-to-tail rakhne se milta hai.

Arrows kyun. Ek sine wave ka amplitude-aur-phase bilkul waise behave karta hai jaise ek 2-D arrow ki length-aur-direction. Waves jodna = arrows jodna. Yeh algebra ko geometry mein badal deta hai.

Geometry. Do barabar length $a$ ke arrows jinka beech mein angle $\varphi$ hai, ek rhombus banaate hain. Resultant uska lamba diagonal hai. Woh diagonal angle ko bisect karta hai, rhombus ko do right triangles mein tod deta hai, jinmein se har ek mein resultant ki aadhi length angle $\varphi /2$ ke adjacent side hai:

$\frac{A_{\text{res}}/2}{a}=\cos \frac{\varphi }{2}⟹A_{\text{res}}=2a\cos \frac{\varphi }{2}$

• $\cos \frac{\varphi }{2}$ — us chhote right triangle mein "adjacent over hypotenuse"; yeh measure karta hai ki do arrows kitna ek hi direction mein point kar rahe hain.

PICTURE. Laal arrow $a$, teal arrow $a$ angle $\varphi$ par, plum resultant diagonal ke saath, aur $\varphi /2$ right triangle shaded.

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Step 6 — Amplitude se intensity tak (square kyun karte hain)

KYA. Brightness (light) ya loudness (sound) intensity $I$ hai, aur intensity amplitude ke square ke proportional hoti hai:

$I\propto A_{\text{res}}^2={\left(2a\cos \frac{\varphi }{2}\right)}^2=4a^2{\cos }^2\frac{\varphi }{2}$

Akeli ek wave intensity $I_0\propto a^2$ carry kare. Tab:

$\boxed{I=4I_0{\cos }^2\frac{\varphi }{2}}$

Square kyun. Ek wave ki energy amplitude squared ke saath badhti hai — wave ki unchai double karo to energy chaar guna ho jaati hai (dekho Energy in Waves). Isliye pehle amplitude finish karo, phir square karo. Intensities pehle kabhi mat jodo.

• ${\cos }^2(\varphi /2)$ — hamesha $0$ aur $1$ ke beech: yeh intensity ko poori tarah andhera se $4\times$ bright ke beech dial karta hai.

PICTURE. $I/I_0=4{\cos }^2(\varphi /2)$ ka $\varphi$ ke against plot: $4$ ke peaks, $0$ ki valleys, aur $2$ par dashed average line.

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Step 7 — Har case: bright, dark, aur beech ka neutral

KYA. Ab $\varphi$ ko sab values pe le jao aur boxed law padho.

$\varphi$	$\cos (\varphi /2)$	$I$	matlab
$0,2\pi ,4\pi \dots$	$\pm 1$	$4I_0$	constructive — sabse bright
$\pi /2$	$\frac{1}{\sqrt{2}}$	$2I_0$	neutral — average ke barabar
$\pi ,3\pi \dots$	$0$	$0$	destructive — dark
$2\pi /3$	$1/2$	$I_0$	dim, ek source jitna

Yeh repeat kyun karta hai. ${\cos }^2$ ka $\varphi /2$ mein period $2\pi$ hai, yaani bright aur dark ka pattern bas har $2\pi$ phase pe repeat hota hai — jo path ke zariye matlab har ek wavelength par.

Master link (path ↔ phase). Ek poore wavelength $\lambda$ ka path difference $\Delta x$ ek poora cycle $2\pi$ phase ke barabar hai:

$\varphi =\frac{2\pi }{\lambda }\Delta x$

$\varphi$ ke bright/dark values ko is se wapas feed karo to path conditions milti hain (dekho Coherence and Path Difference):

$\text{Constructive: }\Delta x=n\lambda \text{Destructive: }\Delta x=\left(n+\frac{1}{2}\right)\lambda$

PICTURE. Intensity curve phir se, lekin teen regions flag kiye hue: WOW peaks, HUSH zeros, aur $\varphi =\pi /2$ par neutral crossing.

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Step 8 — Degenerate case: unequal amplitudes poori tarah cancel nahi ho sakti

KYA. "Barabar $a$" ki assumption hatao. Lengths $a_1$ aur $a_2$ ke do arrows angle $\varphi$ par law of cosines se combine hote hain:

$A_{\text{res}}=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \varphi }$

• $2a_1{a}_2\cos \varphi$ — "interference" cross-term; yeh positive (add) ya negative (subtract) swing karta hai jab $\varphi$ ghoomta hai.

Zero kyun nahi aa sakta. $\varphi =\pi$ par (poori tarah opposed) yeh $A_{\text{res}}=|a_1-a_2|$ deta hai. Sirf jab $a_1=a_2$ ho tabhi yeh $0$ banta hai. Unequal waves ek leftover chhodti hain — destructive interference sirf barabar amplitudes ke liye total hota hai.

PICTURE. Arrows $3a$ aur $4a$: $\varphi =0$ par yeh stack hokar $7a$ bante hain; $\varphi =\pi$ par oppose hokar leftover $1a$ reh jaata hai.

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Ek picture mein poora summary

Sab kuch upar se compress karke: length $a$ ke do arrows angle $\varphi$ par → resultant $2a\cos (\varphi /2)$ → intensity $4I_0{\cos }^2(\varphi /2)$ → bright jab $\varphi =2\pi n$, dark jab $\varphi =(2n+1)\pi$.

Recall Feynman: poora walkthrough ek kahani ki tarah sunao

Ek talab mein ek point socho jahan do ripples aate hain. Har ripple, akele, us jagah ko upar-neeche bob karta — ek smooth sine wiggle (Step 1). Doosra ripple usi size ka hai lekin ek beat der se aata hai; woh lateness ek angle $\varphi$ hai (Step 2). Paani sirf ek jagah ho sakta hai, isliye uski motion bas do wiggles add hai (Step 3). Do barabar wiggles jodo aur — trig ka jaadu — tumhe ek wiggle milta hai usi rhythm ka lekin nayi unchai, $2a\cos (\varphi /2)$ (Step 4). Woh unchai bina algebra ke dikhai deti hai: har wave ko length $a$ ka arrow draw karo, unhe angle $\varphi$ par rakho, aur diagonal answer hai — ek right triangle tumhe $\cos (\varphi /2)$ de deta hai (Step 5). Loudness/brightness unchai squared hai, isliye $I=4I_0{\cos }^2(\varphi /2)$ (Step 6). $\varphi$ knob ghoomao: in step → chaar guna bright; opposite → poori khamoshi; quarter-turn → bilkul plain average (Step 7). Aur agar do waves barabar nahi hain, to khamoshi perfect nahi hoti — hamesha leftover $|a_1-a_2|$ rehta hai (Step 8). Yeh interference hai, shuru se ant tak.

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Connections

• Interference — constructive, destructive conditions — parent result jo humne yahan derive kiya.
• Principle of Superposition — Step 3 ka "bas jodo" rule.
• Coherence and Path Difference — Step 7 ka path ↔ phase master link.
• Energy in Waves — Step 6 ka "intensity amplitude squared hai".
• Young's Double Slit Experiment — jahan yeh $\varphi$ conditions fringes paint karti hain.
• Standing Waves — opposite direction mein chalti waves ka interference.
• Beats — kya hota hai jab do frequencies alag hon.