Exercises — Interference — constructive, destructive conditions

Kya karein: $\Delta x$ ko $\lambda$ se compare karo. Kyun: parent condition kehti hai ki poore number of wavelengths ka matlab hai waves crest-on-crest align hoti hain. $\frac{\Delta x}{\lambda }=\frac{3\lambda }{\lambda }=3(\text{a whole number, order }n=3)$ Whole number $\Rightarrow$ constructive. (Mnemonic: Whole = Wow.)

Kya/Kyun: in phase matlab $\varphi =0$, toh ise $I=4I_0{\cos }^2(\varphi /2)$ mein feed karo. $I=4I_0{\cos }^2(0)=4I_0\cdot 1=4I_0$ Yeh sabse bright possible point hai — ek single wave se chaar guna, do guna nahi. "Extra" energy dark points se borrow hoti hai, jaisa Energy in Waves insist karta hai.

Kya karein: $\Delta x$ ko $\lambda$ se compare karo aur fractional part padho. Kyun: path mein half extra wavelength ka matlab hai ek wave exactly half cycle peeche hai — crest wahan pahunchta hai jahan doosri wave ka trough hota hai, toh woh cancel ho jaate hain (yeh parent note ki $\Delta x=(n+\frac{1}{2})\lambda$ destructive condition hai). $\frac{\Delta x}{\lambda }=2.5=\left(2+\frac{1}{2}\right)$ Half-integer (order $n=2$, plus ek half) $\Rightarrow$ destructive. (Half = Hush.)

Pehle units: $90^{\circ }=\pi /2\text{ rad}$. Dono form theek hain, lekin calculator ka mode match karna chahiye — yahan hum degrees mein kaam karte hain, toh DEG set karo (ya convert karke RAD use karo). Phir phase ko aadha karo aur cosine ko square karo. $I=4I_0{\cos }^2\left(\frac{90^{\circ }}{2}\right)=4I_0{\cos }^{2}45^{\circ }=4I_0{\left(\frac{1}{\sqrt{2}}\right)}^2=4I_0\cdot \frac{1}{2}=2I_0$ Bilkul average intensity — $90^{\circ }$ ($=\pi /2$) "neutral" phase hai, na brightening na darkening.

Kya: master link use karo distance ko phase mein badalne ke liye. Kyun: intensity law $\varphi$ mein bolta hai. Link mein $2\pi$ ka matlab hai $\varphi$ radians mein niklega. $\varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{0.40}\times 0.60=2\pi \times 1.5=3\pi \text{ rad}$ $3\pi =(2\cdot 1+1)\pi$, $\pi$ ka odd multiple $\Rightarrow$ destructive — ek quiet spot. (Periodicity use karke, $3\pi$ reduce hoke $\pi$ ho jaata hai — same result.) Ratio se check karo: $\Delta x/\lambda =1.5$, ek half-integer. ✓

Units: phase degrees mein quoted hai, $\varphi =120^{\circ }$, toh calculator DEG mein rakhte hain aur poora kaam degrees mein karte hain — convert karne ki zaroorat nahi. (Agar radians prefer ho, toh $\varphi =2\pi /3$ use karo aur ${\cos }^2(\pi /3)$ RAD mode mein; same answer milega.) $I=4I_0{\cos }^2\left(\frac{120^{\circ }}{2}\right)=4(5){\cos }^{2}60^{\circ }=20{\left(\frac{1}{2}\right)}^2=20\cdot \frac{1}{4}=5{\text{W m}}^{-2}$ Ek source akele jaisi hi — ek acchi reminder ki overlap karna guaranteed brightening nahi hai.

Kyun general formula — picture se banao. Har wave ko ek phasor ki tarah draw kiya gaya hai: ek chhota arrow jiska length amplitude hai aur direction phase hai. Figure dekho: pehla arrow $a_1$ flat pada hai; doosra arrow $a_2$ phase angle $\varphi$ se ghuma hua hai aur pehle par tip-to-tail rakha gaya hai. Resultant $A$ (red arrow) woh arrow hai jo bilkul shuruwaat se bilkul ant tak jaata hai. Uski length nikalne ke liye hum do arrows se bane triangle use karte hain. Triangle ke andar ka angle, $a_1$ (extended) aur $a_2$ ke beech, $180^{\circ }-\varphi$ hai, toh law of cosines deta hai $A^2=a_1^2+a_2^2-2a_1{a}_2\cos (180^{\circ }-\varphi )=a_1^2+a_2^2+2a_1{a}_2\cos \varphi ,$ kyunki $\cos (180^{\circ }-\varphi )=-\cos \varphi$. Isliye $A=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \varphi }.$ Ab arrows se extremes padho:

• Max at $\varphi =0$ ($\cos \varphi =1$): arrows ek hi direction mein point karte hain aur simply add ho jaate hain — $A_{\max }=a_1+a_2=8$.
• Min at $\varphi =\pi$ ($\cos \varphi =-1$): arrows opposite directions mein point karte hain aur partially cancel ho jaate hain — $A_{\min }=|a_1-a_2|=2$, zero nahi, kyunki chhota arrow lamba arrow ko erase nahi kar sakta. $\frac{I_{\max }}{I_{\min }}={\left(\frac{A_{\max }}{A_{\min }}\right)}^2={\left(\frac{8}{2}\right)}^2=16.$

Kya: $I=\frac{1}{4}{I}_{\max }=\frac{1}{4}(4I_0)=I_0$. Law ko equal set karo aur $\varphi$ ke liye solve karo. $4I_0{\cos }^2\frac{\varphi }{2}=I_0\Rightarrow {\cos }^2\frac{\varphi }{2}=\frac{1}{4}\Rightarrow \cos \frac{\varphi }{2}=\frac{1}{2}$ $\frac{\varphi }{2}=60^{\circ }\Rightarrow \varphi =120^{\circ }=\frac{2\pi }{3}\text{ rad}.$ Kyun positive root liya: sabse chhota positive $\varphi$ positive cosine branch se aata hai. (Har doosra solution is se poore $2\pi$ se alag hoga, jo periodicity se same intensity deta hai.)

Destructive points $\Delta x=(n+\frac{1}{2})\lambda$ par hote hain, jahan order $n$ ek dark point se doosre tak ek step upar jaata hai. Consecutive waale bilkul ek $\lambda$ se alag hote hain: $\Delta x_{n+1}-\Delta x_n=\left[(n+1)+\frac{1}{2}\right]\lambda -\left[n+\frac{1}{2}\right]\lambda =\lambda =600\text{ nm}.$ Successive dark points ke beech spacing ek wavelength hoti hai — yeh fact Young's Double Slit Experiment fringe spacing mein convert karta hai.

Kya: listener tak har speaker ki distance nikalo, $\Delta x$ ke liye subtract karo. Kyun: condition sirf path length ke difference par depend karti hai. $S_1$ se distance (seedha saamne): $r_1=L=4.0\text{ m}$. $S_2$ se distance: yeh ek right triangle banaata hai legs $L=4.0$ aur $d=3.0$ ke saath: $r_2=\sqrt{L^2+d^2}=\sqrt{4.0^2+3.0^2}=\sqrt{16+9}=\sqrt{25}=5.0\text{ m}.$ $\Delta x=r_2-r_1=5.0-4.0=1.0\text{ m}=1\lambda .$ Ek poori wavelength (order $n=1$) $\Rightarrow$ constructive — ek loud spot ($I=4I_0$).

Geometry unchanged hai, toh $\Delta x=1.0\text{ m}$ abhi bhi same hai. Ab $\frac{\Delta x}{\lambda }=\frac{1.0}{2.0}=0.5=\left(0+\frac{1}{2}\right),$ ek half-integer (order $n=0$) $\Rightarrow$ destructive. Formally, $\varphi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{2.0}(1.0)=\pi \text{ rad},I=4I_0{\cos }^2\frac{\pi }{2}=0.$ Same room, same speakers — wavelength double karne se loud silent ho gaya. Pattern $\lambda$ se chipka hua hai, sirf geometry se nahi.

Kya: $\varphi$ ke ek full cycle mein $I=4I_0{\cos }^2(\varphi /2)$ ka average karo. Kyun: slow uniform drift ka matlab hai har phase equally likely hai, toh hum mean value lete hain — aur periodicity se $2\pi$ ka ek poora turn har possible behaviour capture karta hai. ${\cos }^2\theta$ ka mean ek full cycle mein $\frac{1}{2}$ hota hai, toh $⟨I⟩=4I_0\cdot \frac{1}{2}=2I_0.$ Yeh bilkul $I_0+I_0$ hai — incoherent result: intensities simply add ho jaati hain aur koi pattern nahi bachta. Coherent maximum $4I_0$ se compare karke, drifting case at best aadha bright hai, aur koi fringes nahi dikhate. Isliye hi Coherence and Path Difference ek steady $\varphi$ demand karta hai.

Kya/Kyun: parent note se equal-amplitude resultant $A=2a\cos (\varphi /2)$ use karo har fixed phase par.

• $\varphi =\pi$ par: $A=2a\cos (\pi /2)=2a\cdot 0=0$. Point har time zero displacement par rehta hai — ek node.
• $\varphi =0$ par (ya $2\pi$ — periodicity se same): $A=2a\cos 0=2a$, sabse bada possible swing — ek antinode.

Kyunki do waves ek doosre ki taraf travel karti hain, permanently-still nodes aur maximally-swinging antinodes ka yeh pattern space mein fixed hai aur string ke saath move nahi karta. Woh frozen interference pattern ek standing wave hai.

Kyun amplitudes: intensity $\propto$ amplitude${}^2$, toh $\frac{a_1}{a_2}=\sqrt{\frac{I_1}{I_2}}=\sqrt{4}=2$. Lo $a_2=1$, $a_1=2$. $A_{\max }=a_1+a_2=3,A_{\min }=|a_1-a_2|=1.$ $\frac{I_{\max }}{I_{\min }}={\left(\frac{A_{\max }}{A_{\min }}\right)}^2={\left(\frac{3}{1}\right)}^2=9.$ Toh $I_{\max }:I_{\min }=9:1$. (General formula: ${\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2$.)

Kyunki frequencies alag hain, phase difference $\varphi$ constant nahi hai — yeh time ke saath steadily grow karta hai, constructive ($\varphi =2\pi n$, loud) aur destructive ($\varphi =(2n+1)\pi$, silent) se clock tick karte karte sweep karta hai. Result hai loudness jo badhti aur ghatti hai: Beats. Spatial interference ko equal frequencies chahiye; temporal beats unequal frequencies se aate hain.