Thin-film and related interference ideas

Two coherent rays, one film. Light incident nearly normally on a thin film of thickness $t$ and refractive index $n$ (film in air, so $n_{\text{air}}\approx1$ on both sides) partially reflects at the top surface and partially transmits, reflects off the bottom surface, and emerges parallel to the first ray. The second ray has travelled an extra optical path inside the film. At near-normal incidence,

where: $t$ is the film thickness, $n$ is the film's refractive index, and the factor of $2$ counts the round trip down and back up through the film. (For oblique incidence at refraction angle $\theta_r$ the more general form is $2nt\cos\theta_r$, but this article stays with the near-normal case.)

The $\lambda/2$ reflection rule. A wave reflecting off a boundary where the refractive index increases — rarer medium to denser medium — flips phase by $\pi$, equivalent to an extra path of $\lambda/2$. Reflecting off a denser-to-rarer boundary produces no such shift. For an air-bounded film (soap film, air wedge), the top-surface reflection is air-to-film (rarer to denser, so it gets the shift) and the bottom-surface reflection is film-to-air (denser to rarer, no shift). Exactly one of the two rays picks up the extra $\lambda/2$, so the effective path difference between them is

Bright and dark conditions (reflected light). Setting $\Delta$ equal to a whole number of wavelengths gives constructive interference (bright film), and setting it equal to a half-integer number gives destructive interference (dark film):

Because of the built-in half-wave shift, these conditions are the opposite of the naive $2nt=m\lambda$-is-bright guess. One striking consequence: as $t\to0$, the path difference collapses to just $\lambda/2$ — permanently the dark condition — which is why the thinnest patch of a soap film, moments before it bursts, looks black rather than bright.

Newton's rings. Rest a plano-convex lens of large radius of curvature $R$ on a flat glass plate. A thin wedge of air is trapped between them, with thickness $t$ growing away from the contact point; for a sphere touching a plane, $t \approx r^2/(2R)$ at radial distance $r$. This is again an air-bounded film (glass-air-glass, one shift), so the dark-ring condition $2t=m\lambda$ turns, after substituting $t\approx r^2/2R$, into

The centre ($r=0$, where $t=0$) is dark, matching the thin-film limit above, and successive rings crowd closer together as $m$ grows because $r_m$ scales as $\sqrt m$, not linearly.

A Newton's rings apparatus uses sodium light, $\lambda = 589\ \text{nm}$. The radius of the 5th dark ring is measured as $r_5 = 3.0\ \text{mm}$. Find the lens's radius of curvature $R$, and predict the radius of the 10th dark ring.

Step 1. Apply $r_m=\sqrt{m\lambda R}$ to the 5th ring and solve for $R$: $r_5^2 = 5\lambda R \Rightarrow R = \dfrac{r_5^2}{5\lambda} = \dfrac{(3.0\times10^{-3})^2}{5\times(589\times10^{-9})} = \dfrac{9.0\times10^{-6}}{2.945\times10^{-6}} \approx 3.06\ \text{m}$.

Step 2. Use the same $R$ to predict the 10th ring: $r_{10} = \sqrt{10\lambda R} = \sqrt{10\times589\times10^{-9}\times3.06} = \sqrt{1.802\times10^{-5}} \approx 4.24\times10^{-3}\ \text{m} = 4.24\ \text{mm}$.

Step 3. Sanity check with the $\sqrt m$ scaling directly, without recomputing $R$: $r_{10}/r_5 = \sqrt{10/5} = \sqrt2 \approx 1.414$, and indeed $4.24/3.0 \approx 1.41$. This confirms the ring radii really do grow as $\sqrt m$, not proportionally to $m$. The lens has radius of curvature $R \approx 3.06\ \text{m}$, and the 10th dark ring has radius $r_{10} \approx 4.24\ \text{mm}$.

• Writing $2nt=m\lambda$ for "bright" without checking how many reflections carry the $\lambda/2$ shift — an air-bounded film has exactly one shift, which flips which condition ($m\lambda$ or $(m-\tfrac12)\lambda$) means bright versus dark.
• Using thickness $t$ directly in the path-difference formula instead of the optical path $nt$ — the wavelength inside the film is $\lambda/n$, so the phase accumulated depends on $n$, not just on the physical thickness.
• Treating Newton's rings as evenly spaced like Young's double-slit fringes — because $r_m\propto\sqrt m$, the gap between consecutive rings shrinks as $m$ increases, unlike the constant fringe spacing of two-slit interference.
• Forgetting that the centre of the Newton's-rings pattern is dark, not bright — since $t\to0$ there, the effective path difference reduces to just the $\lambda/2$ reflection term, the same reasoning that makes a vanishingly thin soap film look black.

Ask the tutor: if you want to work through the reflection-shift bookkeeping for a different arrangement, such as an anti-reflection coating on glass or an oil film on water, ask the AI tutor to trace each interface step by step.