Exercises — Thin film interference — reflected and transmitted

Top interface: air ($n=1.00$) → soap ($n=1.33$). Yeh rarer→denser hai, isliye flip hota hai ($+\lambda /2$). Bottom interface: soap ($1.33$) → air ($1.00$). Denser→rarer, isliye koi flip nahi. Total = ek flip (odd). Odd ⇒ standard "constructive = whole wavelength" rule swap ho jaata hai. Toh reflection ke liye:

Yeh air (vacuum) wavelength hai. $2nt$ mein factor $n$ ne already geometric path ko optical path mein convert kar diya hai, yaani iske andar film mein shorter wavelength ${\lambda }_{film}=\lambda /n$ already account ho chuki hai. Right side par ${\lambda }_{film}$ likhne se slow-down dobara count ho jaata. $\lambda$ = air value hi rakhni hai.

Ek flip (Ex 1.1 se) ⇒ bright half-integer condition hai $2nt=(m+\frac{1}{2})\lambda$. Sabse chota $t$ ⇒ sabse chota allowed $m=0$: $t=\frac{\lambda }{4n}=\frac{650}{4(1.33)}=\boxed{122.2\text{ nm}}.$ $m=0$ kyun? $m=0$ se sabse kam $t>0$ milta hai; negative $m$ se negative thickness aayegi.

Dark (ek flip) hai $2nt=m\lambda$. $m=0$ se $t=0$ milta hai (film-about-to-pop black spot — real hai par trivial). Sabse chota non-zero ⇒ $m=1$: $t=\frac{\lambda }{2n}=\frac{650}{2(1.33)}=\boxed{244.4\text{ nm}}.$ Note karo yeh pehli bright thickness se exactly double hai — bright aur dark thickness mein $\lambda /4n$ ke gap par alternate karte hain.

Flip count: air($1.00$)→oil($1.40$) = rarer→denser = flip. oil($1.40$)→water($1.33$) = denser→rarer = koi flip nahi. Net = ek flip (odd) ⇒ bright half-integer hai. Optical path: $2nt\cos {\theta }_r=2(1.40)(450)(0.90)=1134\text{ nm}=1.89\lambda .$ Bright chahiye $(m+\frac{1}{2})$: nearest $1.5$ ya $2.5$ hai — dono kaafi door hain. Dark chahiye integer: nearest $2.0$ hai, aur $1.89$ sirf $0.11$ off hai. Toh reflection dark ke kaafi paas hai (nearly a minimum).

Ek flip ⇒ dark reflection hai $2nt=m\lambda$, toh $\lambda =\frac{2nt}{m}=\frac{2(1.33)(300)}{m}=\frac{798}{m}$ nm.

• $m=1:798$ nm (infrared — visible nahi).
• $m=2:399$ nm (violet edge — barely visible). Sabse strong visible dark line $\approx \boxed{399\text{ nm}}$ par hai (violet). Toh film white minus violet reflect karti hai — thodi yellow-green dikhti hai.

Flip count: air($1.00$)→coating($1.38$) = flip. coating($1.38$)→glass($1.52$) = rarer→denser = flip bhi. Do flips (even) ⇒ standard rule laagoo hota hai: dark reflection ke liye half-integer condition chahiye $2n_{c}t=\left(m+\frac{1}{2}\right)\lambda .$ Sabse patla ⇒ $m=0$: $t=\frac{\lambda }{4n_c}=\frac{550}{4(1.38)}=\boxed{99.6\text{ nm}}.$ Steel-man check: formula Ex 2.1 jaisa lagta hai par physics ulti hai — wahan ek flip ne bright ko half-integer banaya; yahan do flips dark ko half-integer banate hain. Same algebra, ulta matlab. Isliye coating reflection destroy karti hai (aur Conservation of energy se transmission boost hoti hai).

Conservation of energy se, transmission reflection ka photographic negative hota hai, toh agar reflection dark ke paas hai, transmission bright ke paas hogi. Seedha check karo: transmitted rays mein net $\lambda /2$ nahi hota (is ordering mein internal reflection jo transmitted interference feed karti hai mein koi flip nahi hota), toh transmitted bright integer condition hai $2nt\cos {\theta }_r=m\lambda$. Humne paya tha $2nt\cos {\theta }_r=1.89\lambda \approx 2\lambda$ (integer!). Toh transmission sach mein bright ke paas hai — consistent hai. ✓

Jab $t\to 0$, geometric-optical path $2nt\cos {\theta }_r\to 0$. Par ek flip extra $\lambda /2$ contribute karta hai. Toh total path difference $\Delta \to 0+\frac{\lambda }{2}=\frac{\lambda }{2}$ — exactly half a wavelength, perfect destructive condition. Dono rays perfect anti-step mein wapas aate hain, cancel ho jaate hain, aur film black hoti hai, zero thickness hone par bhi. Formally: dark ke liye chahiye $2nt=m\lambda$; $t=0$ par yeh $m=0$ hai, satisfied. Toh $t\to 0$ dark hai. Koi free parameter nahi hai galti karne ke liye — flip akela isko force karta hai. (Yeh soap bubble ka black cap hai jo pop hone se theek pehle dikhta hai.)

Bright (ek flip): $2nt=(m+\frac{1}{2})\lambda$, yaani $t=\frac{(m+\frac{1}{2})\lambda }{2n}$. Integers $m_1,m_2$ chahiye jaise $\left(m_1+\frac{1}{2}\right){\lambda }_1=\left(m_2+\frac{1}{2}\right){\lambda }_2(=2nt).$ Toh $\frac{m_1+\frac{1}{2}}{m_2+\frac{1}{2}}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{640}{480}=\frac{4}{3}$. Likho $m_1+\frac{1}{2}=\frac{2m_1+1}{2}$, similarly $m_2$ ke liye. Tab $\frac{2m_1+1}{2m_2+1}=\frac{4}{3}$. Dono $2m_i+1$ odd hain; humein odd:odd $=4:3$ chahiye. $4:3$ ko odd $k$ se scale karo: $4k:3k$ — dono odd chahiye, par $4k$ hamesha even hota hai. Toh $4:3$ seedhe do odds nahi ban sakta; $3$ se multiply karo: $12:9$ (even:odd, nahi). $5$ se: $20:15$ (nahi). Koi solution nahi jisme $2m_1+1$ even ho, toh exact simultaneous bright impossible hai — sabse paas woh hai jahan dono half-integers ke kaafi paas hon. Sabse practical thickness: shared optical path $2nt$ ko $(m_1+\frac{1}{2})480$ aur $(m_2+\frac{1}{2})640$ ka common near-value hona chahiye. Try karo $2nt=1680$ nm: $1680/480=3.5=(3+\frac{1}{2})$ ✓ $480$ ke liye exactly bright; $1680/640=2.625$ — half-integer nahi. Try karo $2nt=1440$: $1440/480=3.0$ (480 ke liye dark). Jo genuinely best small compromise ho jahan dono exactly bright hon, woh exist nahi karta; honest answer yeh hai: koi finite $t$ nahi jo dono ko exactly bright banaye — yeh ek clean mastery-level "prove impossibility" result hai. (Agar force karna pade, toh $t=\frac{3.5\times 480}{2(1.33)}=632$ nm se 480 exactly bright hoti hai aur 640 sirf thodi si off hoti hai.)

2.1 $t=122.2$ nm · 2.2 $t=244.4$ nm · 3.1 path $=1134$ nm $=1.89\lambda$ (dark ke paas) · 3.2 $\lambda \approx 399$ nm · 4.1 $t=99.6$ nm · 4.2 bright ke paas ($1.89\lambda \approx 2\lambda$) · 5.2 koi exact simultaneous solution nahi.