Exercises — Thin film interference — reflected and transmitted

Top interface: air ($n=1.00$) → soap ($n=1.33$). This is rarer→denser, so it flips ($+\lambda /2$). Bottom interface: soap ($1.33$) → air ($1.00$). Denser→rarer, so no flip. Total = one flip (odd). Odd ⇒ the standard "constructive = whole wavelength" rule swaps. So for reflection:

It is the air (vacuum) wavelength. The factor $n$ in $2nt$ has already converted the geometric path into an optical path, i.e. it already accounts for the shorter wavelength ${\lambda }_{film}=\lambda /n$ inside the film. Writing ${\lambda }_{film}$ on the right would count the slow-down twice. Keep $\lambda$ = air value.

One flip (from Ex 1.1) ⇒ bright is the half-integer condition $2nt=(m+\frac{1}{2})\lambda$. Smallest $t$ ⇒ smallest allowed $m=0$: $t=\frac{\lambda }{4n}=\frac{650}{4(1.33)}=\boxed{122.2\text{ nm}}.$ Why $m=0$? $m=0$ gives the least $t>0$; negative $m$ would give negative thickness.

Dark (one flip) is $2nt=m\lambda$. $m=0$ gives $t=0$ (the film-about-to-pop black spot — real but trivial). Smallest non-zero ⇒ $m=1$: $t=\frac{\lambda }{2n}=\frac{650}{2(1.33)}=\boxed{244.4\text{ nm}}.$ Note this is exactly twice the first bright thickness — bright and dark alternate every $\lambda /4n$ in thickness.

Flip count: air($1.00$)→oil($1.40$) = rarer→denser = flip. oil($1.40$)→water($1.33$) = denser→rarer = no flip. Net = one flip (odd) ⇒ bright is half-integer. Optical path: $2nt\cos {\theta }_r=2(1.40)(450)(0.90)=1134\text{ nm}=1.89\lambda .$ Bright wants $(m+\frac{1}{2})$: nearest is $1.5$ or $2.5$ — both far. Dark wants integer: nearest is $2.0$, and $1.89$ is only $0.11$ off. So the reflection is close to dark (nearly a minimum).

One flip ⇒ dark reflection is $2nt=m\lambda$, so $\lambda =\frac{2nt}{m}=\frac{2(1.33)(300)}{m}=\frac{798}{m}$ nm.

• $m=1:798$ nm (infrared — not visible).
• $m=2:399$ nm (violet edge — barely visible). The strongest visible dark line is at $\approx \boxed{399\text{ nm}}$ (violet). So the film reflects white minus violet — it looks slightly yellow-green.

Flip count: air($1.00$)→coating($1.38$) = flip. coating($1.38$)→glass($1.52$) = rarer→denser = flip too. Two flips (even) ⇒ the standard rule holds: dark reflection needs the half-integer condition $2n_{c}t=\left(m+\frac{1}{2}\right)\lambda .$ Thinnest ⇒ $m=0$: $t=\frac{\lambda }{4n_c}=\frac{550}{4(1.38)}=\boxed{99.6\text{ nm}}.$ Steel-man check: it looks like Ex 2.1's formula but the physics is opposite — there one flip made bright the half-integer; here two flips make dark the half-integer. Same algebra, flipped meaning. This is why the coating destroys reflection (and by Conservation of energy, boosts transmission).

By Conservation of energy, transmission is the photographic negative of reflection, so if reflection is near dark, transmission is near bright. Check directly: transmitted rays carry no net $\lambda /2$ (the internal reflection that feeds the transmitted interference has no flip in this ordering), so transmitted bright is the integer condition $2nt\cos {\theta }_r=m\lambda$. We found $2nt\cos {\theta }_r=1.89\lambda \approx 2\lambda$ (integer!). So transmission is indeed near bright — consistent. ✓

As $t\to 0$, the geometric-optical path $2nt\cos {\theta }_r\to 0$. But one flip contributes an extra $\lambda /2$. So the total path difference $\Delta \to 0+\frac{\lambda }{2}=\frac{\lambda }{2}$ — exactly half a wavelength, the perfect destructive condition. The two rays return in perfect anti-step, cancel, and the film is black, even with zero thickness. Formally: dark needs $2nt=m\lambda$; at $t=0$ that is $m=0$, satisfied. So $t\to 0$ is dark. There is no free parameter to fumble — the flip alone forces it. (This is the soap bubble's black cap just before it pops.)

Bright (one flip): $2nt=(m+\frac{1}{2})\lambda$, i.e. $t=\frac{(m+\frac{1}{2})\lambda }{2n}$. Need integers $m_1,m_2$ with $\left(m_1+\frac{1}{2}\right){\lambda }_1=\left(m_2+\frac{1}{2}\right){\lambda }_2(=2nt).$ So $\frac{m_1+\frac{1}{2}}{m_2+\frac{1}{2}}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{640}{480}=\frac{4}{3}$. Write $m_1+\frac{1}{2}=\frac{2m_1+1}{2}$, similarly for $m_2$. Then $\frac{2m_1+1}{2m_2+1}=\frac{4}{3}$. Both $2m_i+1$ are odd; we need odd:odd $=4:3$. Scale $4:3$ by odd $k$: $4k:3k$ — need both odd, but $4k$ is always even. So $4:3$ can't be two odds directly; multiply by $3$: $12:9$ (even:odd, no). By $5$: $20:15$ (no). No solution has $2m_1+1$ even, so exact simultaneous bright is impossible — the closest is where both are near half-integers. Nearest practical thickness: the shared optical path $2nt$ must be a common near-value of $(m_1+\frac{1}{2})480$ and $(m_2+\frac{1}{2})640$. Try $2nt=1680$ nm: $1680/480=3.5=(3+\frac{1}{2})$ ✓ exact bright for $480$; $1680/640=2.625$ — not half-integer. Try $2nt=1440$: $1440/480=3.0$ (dark for 480). The genuinely best small compromise where both are exactly bright does not exist; the honest answer is: no finite $t$ makes both exactly bright — a clean mastery-level "prove impossibility" result. (If forced to pick, $t=\frac{3.5\times 480}{2(1.33)}=632$ nm makes 480 exactly bright and 640 only mildly off.)

2.1 $t=122.2$ nm · 2.2 $t=244.4$ nm · 3.1 path $=1134$ nm $=1.89\lambda$ (near dark) · 3.2 $\lambda \approx 399$ nm · 4.1 $t=99.6$ nm · 4.2 near bright ($1.89\lambda \approx 2\lambda$) · 5.2 no exact simultaneous solution.