Interference by Division of Amplitude & Interferometry

Sir George Gabriel Stokes utilized the principle of optical reversibility—which states that in the absence of absorption or dissipative losses, any optical ray tracing can be reversed precisely along its trajectory—to establish fundamental thermodynamic relations governing amplitude reflection and transmission coefficients.

1. Ray Decomposition at a Planar Dielectric Interface

Consider an incident plane wave of unit amplitude $E_0 = 1$ propagating from medium 1 ($n_1$) toward an interface with medium 2 ($n_2$). At the interface:

• A reflected ray of amplitude $r$ is generated in medium 1.
• A transmitted (refracted) ray of amplitude $t$ enters medium 2.

Now reverse the two rays simultaneously:

1. Reversing the reflected ray $r$: upon reaching the boundary, it produces a reflected component $r \cdot r = r^2$ into medium 1, and a transmitted component $r \cdot t$ into medium 2.
2. Reversing the transmitted ray $t$: upon reaching the boundary from medium 2 (where reflection coefficient is $r'$ and transmission coefficient is $t'$), it produces a transmitted component $t \cdot t'$ back into medium 1, and a reflected component $t \cdot r'$ in medium 2.

2. Application of the Principle of Reversibility

For the reversed system to perfectly reconstitute the original incident ray of unit amplitude traveling in medium 1, and produce zero net field propagating backwards into medium 2: $$\text{In medium 1:} \quad r^2 + t t' = 1$$ $$\text{In medium 2:} \quad r t + t r' = 0 \implies t (r + r') = 0$$ Since the transmission coefficient $t \neq 0$: $$r' = -r = r e^{\pm i\pi}$$ $$t t' = 1 - r^2$$

3. Physical Significance

The relation $r' = -r$ proves conclusively that the amplitude reflection coefficient experienced by a wave incident from the rarer side ($r$) is equal in magnitude but exactly opposite in sign (differing by a phase angle of $\pi$ radians) to the reflection coefficient experienced by a wave incident from the denser side ($r'$). This phase inversion accounts for the loss of half a wavelength ($\lambda/2$) whenever light reflects externally from a dielectric medium of higher refractive index ($n_2 > n_1$).

When a beam of light strikes a thin transparent dielectric film of thickness $t$ and refractive index $\mu$, the beam is split into multiple reflected and transmitted components at the upper and lower surfaces, generating division-of-amplitude interference.

1. Derivation of the Cosine Law of Path Difference

Consider a plane wave incident at angle $i$ on a plane-parallel film of thickness $t$ and refractive index $\mu$. The angle of refraction inside the film is $r$, governed by Snell's law: $\sin i = \mu \sin r$. The optical path difference between the ray reflected from the top surface and the ray reflected from the bottom surface is: $$\Delta_{\text{geom}} = \mu (AB + BC) - AD$$ From the film geometry: $$AB = BC = \frac{t}{\cos r}$$ $$AC = 2 t \tan r, \quad AD = AC \sin i = (2 t \tan r)(\mu \sin r) = 2 \mu t \frac{\sin^2 r}{\cos r}$$ Substituting into the path difference: $$\Delta_{\text{geom}} = \mu \left( \frac{2t}{\cos r} \right) - 2 \mu t \frac{\sin^2 r}{\cos r} = \frac{2 \mu t}{\cos r} (1 - \sin^2 r) = 2 \mu t \cos r$$

2. Phase Change on Reflection and Net Path Difference

By Stokes' relations, the external reflection at the top surface (air $\to$ film) introduces an abrupt phase change of $\pi$, equivalent to an optical path difference of $\frac{\lambda}{2}$. The internal reflection at the bottom surface (film $\to$ air) undergoes no phase change. The net optical path difference for **reflected light** is: $$\Delta_{\text{net}} = 2 \mu t \cos r - \frac{\lambda}{2}$$

• Constructive Interference (Bright Film): $$2 \mu t \cos r = \left(m + \frac{1}{2}\right) \lambda, \quad m \in \{0, 1, 2, \dots\}$$
• Destructive Interference (Dark Film): $$2 \mu t \cos r = m \lambda, \quad m \in \{1, 2, 3, \dots\}$$

For **transmitted light**, there is no phase jump, and the conditions are exactly complementary ($2\mu t\cos r = m\lambda$ for maximum transmission).

3. Wedge-Shaped Films (Fringes of Equal Thickness)

For a film bounded by two flat glass plates inclined at a minute angle $\theta \ll 1\text{ rad}$: $$t(x) = x \theta$$ Fringes form straight, equispaced parallel bands running parallel to the line of contact. The fringe width is: $$\beta = \frac{\lambda}{2 \mu \theta}$$

4. Engineering of Quarter-Wave Anti-Reflection (AR) Coatings

In precision optical instruments (cameras, telescopes, lasers), reflection losses from glass elements ($n_g \approx 1.5$) are eliminated by depositing a thin dielectric film of refractive index $n_c$ such that: $$n_1 < n_c < n_g \quad (\text{e.g., air } n_1=1, \text{ MgF}_2 \ n_c=1.38, \text{ glass } n_g=1.52)$$ Because $n_1 < n_c$ and $n_c < n_g$, both the ray reflected from the air-coating interface and the ray reflected from the coating-glass interface undergo a $\pi$ phase shift at reflection. To produce destructive interference between the two reflected rays at normal incidence ($r = 0, \cos r = 1$): $$2 n_c d = \frac{\lambda_0}{2} \implies d = \frac{\lambda_0}{4 n_c}$$ Furthermore, for complete cancellation of the reflected intensity, the two reflected amplitudes must be equal ($r_1 = r_2$): $$\frac{n_c - 1}{n_c + 1} = \frac{n_g - n_c}{n_g + n_c} \implies n_c = \sqrt{n_g}$$

Newton's rings provide a classic example of fringes of equal thickness formed by a variable-thickness air film enclosed between a spherical convex surface and an optical flat.

1. Geometry of the Enclosed Air Film

A plano-convex lens of large radius of curvature $R$ (typically $1 - 3\text{ m}$) is placed with its curved surface in contact with an optically flat glass plate. Let $r_n$ be the radius of the $n$-th circular ring, and let $t$ be the thickness of the air film at distance $r_n$ from the point of contact $O$. By the geometric theorem of intersecting chords for a sphere of radius $R$: $$r_n^2 = t(2R - t)$$ Since $t \ll R$, the term $t^2$ is negligible, yielding: $$t = \frac{r_n^2}{2R}$$

2. Condition for Interference in Reflected Light

For an air film ($\mu = 1$) at normal incidence ($\cos r = 1$), the optical path difference between the beam reflected from the lower curved surface of the lens and the beam reflected from the flat plate is: $$\Delta = 2t + \frac{\lambda}{2} = \frac{r_n^2}{R} + \frac{\lambda}{2}$$

• Dark Rings (Destructive Interference): $$\frac{r_n^2}{R} + \frac{\lambda}{2} = \left(n + \frac{1}{2}\right) \lambda \implies r_n^2 = n R \lambda$$ $$\text{Ring Diameter } D_n = 2r_n = 2\sqrt{n R \lambda} \implies D_n^2 = 4 n R \lambda, \quad n \in \{0, 1, 2, \dots\}$$
• Bright Rings (Constructive Interference): $$\frac{r_n^2}{R} + \frac{\lambda}{2} = n \lambda \implies r_n^2 = \left(n - \frac{1}{2}\right) R \lambda$$ $$D_n'^2 = 4\left(n - \frac{1}{2}\right) R \lambda = 2(2n - 1) R \lambda, \quad n \in \{1, 2, 3, \dots\}$$

3. Central Dark Spot and Spacing Law

At the center of contact ($n = 0, r = 0$), $t = 0$. The path difference is purely $\Delta = \frac{\lambda}{2}$, resulting in destructive interference. Therefore, the central spot in reflected light is **strictly dark**. The difference in squares of diameters of the $(n+p)$-th and $n$-th dark rings is: $$D_{n+p}^2 - D_n^2 = 4(n+p)R\lambda - 4nR\lambda = 4 p R \lambda$$ $$\lambda = \frac{D_{n+p}^2 - D_n^2}{4 p R}$$ This relation allows the measurement of optical wavelengths with parts-per-thousand precision, independent of the exact position of the zero-th center.

4. Measurement of Refractive Index of Liquids

When a liquid of unknown refractive index $\mu$ is introduced between the lens and the glass plate: $$(D_{n+p}^2 - D_n^2)_{\text{liquid}} = \frac{4 p R \lambda}{\mu}$$ Dividing the air measurement by the liquid measurement: $$\mu = \frac{(D_{n+p}^2 - D_n^2)_{\text{air}}}{(D_{n+p}^2 - D_n^2)_{\text{liquid}}}$$

Invented by Albert A. Michelson in 1881, the Michelson interferometer is the archetypal amplitude-splitting dual-beam optical instrument. It played a pivotal role in testing the luminiferous aether (Michelson-Morley experiment) and in defining the international standard meter in terms of optical wavelengths.

1. Optical Architecture and Ray Trajectory

The instrument consists of:

• A monochromatic extended light source $S$.
• A precision beam splitter plate $G_1$, lightly silvered or dielectrically coated on its back surface to achieve $50/50$ division of amplitude ($R = T = 0.5$).
• A compensating plate $G_2$ of identical thickness and material as $G_1$, oriented strictly parallel to $G_1$.
• A fixed reference mirror $M_1$ and a movable mirror $M_2$ mounted on a precision micrometer carriage.

An incident ray is divided at the semi-reflecting face of $G_1$:

1. Beam 1 (Transmitted): Propagates through $G_1$, traverses compensating plate $G_2$, reflects off mirror $M_2$, re-traverses $G_2$, and reflects off the semi-reflecting face of $G_1$ toward the detector.
2. Beam 2 (Reflected): Reflects off the back face of $G_1$, reflects off mirror $M_1$, traverses $G_1$, and reaches the detector.

The compensating plate $G_2$ ensures that both beams traverse identical thicknesses of dispersive glass ($3 \times$ through plate thickness), eliminating chromatic dispersion.

2. Mathematical Formulation of Circular Fringes

Let $M_1'$ be the virtual image of mirror $M_1$ formed by reflection in the beam splitter $G_1$. The interferometer is optically equivalent to an air film of thickness $d = |d_2 - d_1|$ bounded by parallel planes $M_2$ and $M_1'$. For an extended source, rays entering the detector at inclination angle $\theta$ experience an optical path difference: $$\Delta = 2 d \cos \theta$$ Taking into account the phase change of $\pi$ upon internal vs external reflection at the beam splitter: $$2 d \cos \theta = m \lambda \quad (\text{Bright fringes})$$ $$2 d \cos \theta = \left(m + \frac{1}{2}\right) \lambda \quad (\text{Dark fringes})$$ Since the path difference is circularly symmetric about the normal to the mirrors, the resulting fringes of equal inclination (Haidinger fringes) are **concentric circles**.

3. Fringe Shift and Mirror Displacement

When the movable mirror $M_2$ is translated by distance $\Delta d$, the optical path changes by $2\Delta d$. If $N$ fringes cross the crosshair of the viewing telescope: $$2 \Delta d = N \lambda \implies \Delta d = N \frac{\lambda}{2}$$ This provides sub-nanometer displacement resolution.

4. Measurement of Doublet Wavelength Separation (e.g., Sodium D-Lines)

When illuminated by a source containing two closely spaced wavelengths $\lambda_1$ and $\lambda_2$ (such as the sodium doublet $\lambda_1 = 589.6\text{ nm}, \lambda_2 = 589.0\text{ nm}$), each wavelength produces an independent fringe pattern. As mirror $M_2$ is moved, the fringes periodically go from maximum visibility (bright rings coincident) to minimum visibility (bright rings of $\lambda_1$ falling on dark rings of $\lambda_2$). If the mirror displacement between two consecutive positions of maximum darkness (minimum visibility) is $d_{\text{dis}}$: $$d_{\text{dis}} = \frac{\lambda_1 \lambda_2}{2(\lambda_1 - \lambda_2)} \approx \frac{\bar{\lambda}^2}{2 \Delta \lambda} \implies \Delta \lambda = \frac{\bar{\lambda}^2}{2 d_{\text{dis}}}$$

Unlike two-beam interferometers which produce broad sinusoidal fringes, multiple-beam interferometers utilize repeated reflections between two highly reflective parallel surfaces to produce exceptionally sharp, needle-like transmission fringes.

1. Airy’s Derivation of Multiple-Beam Superposition

Consider a plane wave of unit amplitude $E_0 = 1$ incident at angle $\theta$ upon an optical cavity (Fabry-Pérot etalon) of plate spacing $d$ and refractive index $n$, bounded by two parallel plates each having intensity reflectance $R = r^2$ and transmittance $T = t^2$. The phase difference between any two consecutively transmitted rays is: $$\delta = \frac{2\pi}{\lambda} (2 n d \cos \theta)$$ The total transmitted complex amplitude is the infinite geometric series: $$E_T = t t' e^{i\phi_0} \left[ 1 + r^2 e^{i\delta} + r^4 e^{i2\delta} + r^6 e^{i3\delta} + \dots \right]$$ Using $t t' = 1 - r^2 = 1 - R$ (for lossless coatings), the infinite sum evaluates to: $$E_T = \frac{1 - R}{1 - R e^{i\delta}}$$ The transmitted optical intensity is: $$I_T = |E_T|^2 = \frac{(1 - R)^2}{|1 - R(\cos \delta + i \sin \delta)|^2} = \frac{(1 - R)^2}{(1 - R \cos \delta)^2 + R^2 \sin^2 \delta}$$ Expanding the denominator: $$(1 - R \cos \delta)^2 + R^2 \sin^2 \delta = 1 - 2R\cos \delta + R^2 = (1 - R)^2 + 2R(1 - \cos \delta) = (1 - R)^2 + 4R \sin^2\left(\frac{\delta}{2}\right)$$ Dividing numerator and denominator by $(1 - R)^2$ yields the celebrated **Airy Formula**: $$I_T(\delta) = \frac{I_0}{1 + F \sin^2\left(\frac{\delta}{2}\right)}$$ where $F$ is the **Coefficient of Finesse**: $$F = \frac{4R}{(1 - R)^2}$$

2. Transmission Characteristics and Fringe Sharpness

• When $\delta = 2m\pi$ ($2 n d \cos \theta = m\lambda$): $\sin^2(\delta/2) = 0 \implies I_T = I_0$ ($100\%$ transmission).
• When $\delta = (2m+1)\pi$: $\sin^2(\delta/2) = 1 \implies I_T = \frac{I_0}{1 + F}$. For $R = 0.95$, $F = \frac{4(0.95)}{(0.05)^2} = 1520$, yielding $I_{\text{min}} = \frac{I_0}{1521} \approx 0.00066 I_0$.

3. Full Width at Half Maximum (FWHM) and Resolving Power

The fringe intensity drops to half its maximum ($I_T = I_0/2$) when: $$F \sin^2\left(\frac{\Delta \delta_{1/2}}{4}\right) = 1 \implies \frac{\Delta \delta_{1/2}}{2} \approx \frac{2}{\sqrt{F}} \implies \Delta \delta = \frac{4}{\sqrt{F}} = \frac{2(1 - R)}{\sqrt{R}}$$ The **Effective Finesse** $\mathcal{F}$ is defined as the ratio of the Free Spectral Range (fringe separation $2\pi$) to the half-width $\Delta \delta$: $$\mathcal{F} = \frac{2\pi}{\Delta \delta} = \frac{\pi \sqrt{F}}{2} = \frac{\pi \sqrt{R}}{1 - R}$$ For $R = 0.98$, $\mathcal{F} = \frac{\pi \sqrt{0.98}}{0.02} \approx 155$. The chromatic resolving power of the Fabry-Pérot etalon is: $$\frac{\lambda}{\Delta \lambda} = m \mathcal{F} = m \frac{\pi \sqrt{R}}{1 - R}$$ For order $m = 10^4$ and $\mathcal{F} = 150$, the resolving power exceeds $1.5 \times 10^6$, enabling the resolution of isotopic and hyperfine atomic spectral splittings.