Physics Wave & Modern Optics 100% Free Open Access
Chapter 1 • Theory & Derivations

Interference by Division of Wavefront

Huygens' principle, wave superposition in complex notation, Young's double slit, fringe width & geometry, Fresnel's biprism, and Lloyd's mirror.

§1.1Huygens-Fresnel Principle, Wave Superposition and Complex Notation

The wave nature of light rests on the fundamental principle that light propagation is governed by harmonic scalar wave equations derived directly from Maxwell's electrodynamics: $$\nabla^2 \psi - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} = 0$$ where $\psi(\mathbf{r}, t)$ represents any scalar component of the optical electric field $\mathbf{E}(\mathbf{r}, t)$.

1. Complex Exponential Wave Representation

In monochromatic optical analysis, real oscillatory fields are most efficiently represented using Euler's complex notation: $$\psi(\mathbf{r}, t) = \operatorname{Re} \left\{ \tilde{E}(\mathbf{r}) e^{-i\omega t} \right\} = \operatorname{Re} \left\{ E_0 e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)} \right\}$$ where $\tilde{E}(\mathbf{r}) = E_0 e^{i(\mathbf{k}\cdot\mathbf{r} + \phi)}$ is the complex amplitude (phasor), $k = \frac{2\pi}{\lambda} = \frac{\omega}{c}$ is the wave number, $\omega$ is the angular optical frequency, and $\phi$ is the initial phase constant.

2. Linear Superposition of Two Coherent Optical Fields

Consider two monochromatic electromagnetic waves of identical angular frequency $\omega$ intersecting at an observation point $P(\mathbf{r})$. The individual electric field scalar amplitudes are: $$E_1 = E_{01} e^{i(\mathbf{k}_1 \cdot \mathbf{r} - \omega t + \phi_1)}, \quad E_2 = E_{02} e^{i(\mathbf{k}_2 \cdot \mathbf{r} - \omega t + \phi_2)}$$ By the principle of linear superposition in linear dielectric media, the total electric field at $P$ is the algebraic sum of the individual field vectors: $$E_{\text{total}} = E_1 + E_2 = \left( E_{01} e^{i\delta_1} + E_{02} e^{i\delta_2} \right) e^{-i\omega t}$$ where $\delta_1 = \mathbf{k}_1 \cdot \mathbf{r} + \phi_1$ and $\delta_2 = \mathbf{k}_2 \cdot \mathbf{r} + \phi_2$.

3. Observable Optical Intensity and the Interference Term

Optical detectors (photodiodes, CCD sensors, human eye) cannot track instantaneous optical oscillations ($~10^{14}-10^{15} \text{ Hz}$). Instead, detectors measure the time-averaged irradiance (optical intensity) $I$: $$I = \langle |E_{\text{total}}|^2 \rangle = \frac{1}{2} E_{\text{total}} E_{\text{total}}^*$$ Substituting the complex superposition: $$I = \frac{1}{2} \left( E_{01} e^{i\delta_1} + E_{02} e^{i\delta_2} \right) \left( E_{01} e^{-i\delta_1} + E_{02} e^{-i\delta_2} \right)$$ $$I = \frac{1}{2} \left[ E_{01}^2 + E_{02}^2 + E_{01}E_{02} \left( e^{i(\delta_1 - \delta_2)} + e^{-i(\delta_1 - \delta_2)} \right) \right]$$ Using Euler's identity $e^{i\delta} + e^{-i\delta} = 2\cos \delta$, and defining the phase difference $\delta = \delta_2 - \delta_1$: $$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \delta$$ where $I_1 = \frac{1}{2} E_{01}^2$ and $I_2 = \frac{1}{2} E_{02}^2$ are the intensities of the independent beams. The term $J_{12} = 2\sqrt{I_1 I_2} \cos \delta$ is the **Interference Term**.

4. Conditions for Sustained, High-Contrast Interference

For stable, observable interference fringes to persist in space and time:
  1. Monochromaticity & Frequency Matching: The interfering beams must possess identical or nearly identical frequencies ($\omega_1 = \omega_2$). If frequencies differ by $\Delta \omega$, the cross term oscillates at $\cos(\Delta \omega \cdot t)$ and time-averages to zero.
  2. Constant Phase Relationship (Coherence): The relative phase difference $\delta$ must remain strictly invariant over the observation time interval $T_{\text{obs}} \gg \tau_c$, where $\tau_c$ is the coherence time of the source.
  3. Parallel Polarization Vectors: If the electric fields are orthogonal ($\mathbf{E}_1 \perp \mathbf{E}_2$), their scalar product vanishes: $\mathbf{E}_1 \cdot \mathbf{E}_2 = 0$, giving $I = I_1 + I_2$ with zero interference modulation (Fresnel-Arago Law 1).
  4. Equal Amplitudes ($I_1 \approx I_2$): When $I_1 = I_2 = I_0$, the fringe visibility (contrast) reaches its theoretical maximum of unity: $$\mathcal{V} = \frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}} = \frac{4I_0 - 0}{4I_0 + 0} = 1$$ $$I(\delta) = 4I_0 \cos^2\left(\frac{\delta}{2}\right)$$

§1.2Young’s Double-Slit Experiment, Hyperbolic Fringes and Intensity Distribution

In 1801, Thomas Young provided the definitive experimental proof of the wave nature of light by dividing a primary wavefront into two secondary coherent wavelets using two closely spaced narrow slits $S_1$ and $S_2$.

1. Geometric Path Difference Derivation

Let two parallel slits separated by center-to-center distance $d$ illuminate a screen placed at distance $D$, where $D \gg d$. Let the origin $O$ be the center of the screen, and let $P$ be a point on the screen at distance $y$ from $O$. The physical paths traveled by the two wavelets from slits $S_1(0, d/2)$ and $S_2(0, -d/2)$ to $P(D, y)$ are: $$r_1 = \sqrt{D^2 + \left(y - \frac{d}{2}\right)^2} = D \left[ 1 + \frac{\left(y - d/2\right)^2}{D^2} \right]^{1/2}$$ $$r_2 = \sqrt{D^2 + \left(y + \frac{d}{2}\right)^2} = D \left[ 1 + \frac{\left(y + d/2\right)^2}{D^2} \right]^{1/2}$$ Applying the binomial expansion $(1 + u)^{1/2} = 1 + \frac{1}{2}u - \dots$ for $y, d \ll D$: $$r_1 \approx D + \frac{(y - d/2)^2}{2D}, \quad r_2 \approx D + \frac{(y + d/2)^2}{2D}$$ The optical path difference $\Delta = r_2 - r_1$ is: $$\Delta = \frac{(y + d/2)^2 - (y - d/2)^2}{2D} = \frac{2yd}{2D} = \frac{y d}{D}$$ In angular coordinates where $\theta$ is the angle subtended at the slit midpoint: $\sin \theta \approx \tan \theta = \frac{y}{D}$, yielding: $$\Delta = d \sin \theta$$

2. Constructive and Destructive Interference Conditions

The corresponding optical phase difference is: $$\delta = \frac{2\pi}{\lambda} \Delta = \frac{2\pi d y}{\lambda D}$$
  • Bright Fringes (Intensity Maxima): Occur when the path difference is an integer multiple of the wavelength: $$\Delta = m \lambda \implies y_m = m \frac{\lambda D}{d}, \quad m \in \{0, \pm 1, \pm 2, \dots\}$$
  • Dark Fringes (Intensity Minima): Occur when the path difference is a half-integral multiple of the wavelength: $$\Delta = \left(m + \frac{1}{2}\right) \lambda \implies y_m' = \left(m + \frac{1}{2}\right) \frac{\lambda D}{d}, \quad m \in \{0, \pm 1, \pm 2, \dots\}$$

3. Linear Fringe Width (Fringe Spacing) $\beta$

The distance between any two consecutive bright or dark fringes is constant: $$\beta = y_{m+1} - y_m = \frac{(m+1)\lambda D}{d} - \frac{m\lambda D}{d} = \frac{\lambda D}{d}$$ This equation provides a direct, high-precision laboratory method for measuring the optical wavelength $\lambda = \frac{\beta d}{D}$.

4. 3D Spatial Fringe Shape: Hyperboloids of Revolution

The locus of all points in 3D space with a constant path difference from two point sources $S_1$ and $S_2$ is defined by: $$|r_2 - r_1| = \text{constant} = m\lambda$$ By classical analytic geometry, this is the definition of a **hyperboloid of two sheets** having $S_1$ and $S_2$ as foci. When intercepted by a flat planar screen placed perpendicular to the central axis at $x = D$, the intersection of these hyperboloids with the plane $x = D$ produces narrow hyperbolic curves. Near the central axis ($y, z \ll D$), the vertices of these hyperbolas have extremely small curvature, appearing to high precision as straight, equispaced parallel interference fringes.

§1.3Fresnel’s Biprism: Virtual Coherent Sources and Wavelength Determination

Augustin-Jean Fresnel designed the biprism to overcome the criticism that Young's fringes were merely edge-diffraction effects produced by the slit edges. The Fresnel biprism produces two mutually coherent virtual sources purely by refraction without any aperture edges between the two beams.

1. Optical Construction and Refraction by Biprism

A Fresnel biprism consists of two acute prisms joined at their bases, with an obtuse angle of approximately $179^{\circ}$ and two very small refracting angles $\alpha \approx 30' \approx 0.5^{\circ}$. A narrow monochromatic slit $S$ illuminated by wavelength $\lambda$ is placed at distance $u$ in front of the flat face of the biprism. Light passing through the upper half is deviated downwards by angle $\delta$, while light passing through the lower half is deviated upwards by the identical angle $\delta$. For a thin prism of refractive index $n$ and refracting angle $\alpha$, the angle of minimum deviation is: $$\delta = (n - 1)\alpha$$

2. Separation Between Virtual Coherent Sources $d$

Due to refraction, the light appears to diverge from two virtual point sources $S_1$ and $S_2$ located in the plane of the original slit $S$: $$d = 2 u \delta = 2 u (n - 1) \alpha$$ Because both $S_1$ and $S_2$ originate from the same primary wavefront of slit $S$, they maintain strict mutual phase coherence.

3. Fringe Width and Screen Separation

Let the distance from the slit $S$ to the micrometer eyepiece (screen) be $D$. The fringe width on the observation plane is given by the standard interference relation: $$\beta = \frac{\lambda D}{d} = \frac{\lambda D}{2 u (n - 1) \alpha}$$

4. The Displacement Method for Direct Measurement of $d$

Direct physical measurement of the virtual distance $d$ (which is on the order of $0.5 - 2 \text{ mm}$) introduces significant experimental error. Fresnel solved this by inserting a convex lens between the biprism and the eyepiece. For a fixed distance $D > 4f$, there exist two conjugate positions of the convex lens that produce sharp real images of $S_1$ and $S_2$ in the focal plane of the micrometer eyepiece:
  • Position 1 (Magnified image separation $d_1$): $m_1 = \frac{v_1}{u_1} = \frac{d_1}{d}$
  • Position 2 (Diminished image separation $d_2$): $m_2 = \frac{v_2}{u_2} = \frac{d_2}{d}$
By the principle of optical reversibility, $u_1 = v_2$ and $v_1 = u_2$, therefore: $$m_1 \cdot m_2 = \frac{d_1}{d} \cdot \frac{d_2}{d} = 1 \implies d^2 = d_1 d_2 \implies d = \sqrt{d_1 d_2}$$ Consequently, the optical wavelength is determined with exceptional accuracy without needing to know the refractive index $n$ or biprism angle $\alpha$: $$\lambda = \frac{\beta d}{D} = \frac{\beta \sqrt{d_1 d_2}}{D}$$

§1.4Lloyd’s Mirror: Grazing Incidence and the Fundamental Half-Wave (\pi) Phase Shift

In 1834, Humphrey Lloyd developed a single-reflector interference configuration that definitively confirmed the electromagnetic boundary condition predicting a $\pi$ phase shift upon external reflection.

1. Experimental Geometry and Ray Tracing

A monochromatic primary point source $S_1$ of wavelength $\lambda$ is placed at a very small height $h$ above the plane of an optical flat front-surface mirror of length $L$. A screen is placed at distance $D$ perpendicular to the mirror plane. Light from $S_1$ propagates to the screen via two paths:
  1. Direct Wave: Propagates directly from $S_1$ to the screen at height $y$.
  2. Reflected Wave: Strikes the mirror at grazing incidence and reflects to the screen, appearing to originate from the virtual mirror image $S_2$ located at depth $h$ below the mirror surface.
The effective distance between the two interfering coherent sources is: $$d = 2h$$

2. The Crucial Half-Wave Phase Discontinuity ($\pi$ Phase Jump)

From Maxwell's electromagnetic boundary conditions (Fresnel reflection equations), when an optical wave traveling in an optically rarer medium ($n_1 = 1$) reflects at the boundary of a denser medium ($n_2 > 1$) at grazing incidence (angle of incidence $\theta_i \to 90^{\circ}$): $$r_{\perp} = \frac{\cos \theta_i - \sqrt{n^2 - \sin^2 \theta_i}}{\cos \theta_i + \sqrt{n^2 - \sin^2 \theta_i}} \xrightarrow{\theta_i \to 90^{\circ}} -1 = e^{i\pi}$$ This reflection introduces an abrupt, non-geometric phase discontinuity of exactly $\pi$ radians (equivalent to an optical path penalty of $\frac{\lambda}{2}$). The net optical path difference between the reflected and direct waves arriving at height $y$ is: $$\Delta_{\text{net}} = (r_2 - r_1) + \frac{\lambda}{2} = \frac{y d}{D} + \frac{\lambda}{2}$$

3. Inversion of Interference Conditions

Setting $\Delta_{\text{net}}$ equal to integral and half-integral multiples of $\lambda$:
  • Dark Fringes (Destructive Interference): $$\frac{y d}{D} + \frac{\lambda}{2} = \left(m + \frac{1}{2}\right)\lambda \implies y_m = m \frac{\lambda D}{d}, \quad m \in \{0, 1, 2, \dots\}$$
  • Bright Fringes (Constructive Interference): $$\frac{y d}{D} + \frac{\lambda}{2} = m\lambda \implies y_m' = \left(m - \frac{1}{2}\right) \frac{\lambda D}{d}, \quad m \in \{1, 2, 3, \dots\}$$

4. The Vanishing Central Fringe at the Mirror Edge

At the point of grazing contact with the mirror surface ($y = 0$), the geometric path difference vanishes: $r_2 - r_1 = 0$. In standard Young's double-slit interference, $y = 0$ corresponds to the central **bright** maximum. However, in Lloyd's mirror, due to the $-\pi$ phase jump on reflection: $$\Delta_{\text{net}}(y=0) = 0 + \frac{\lambda}{2} = \frac{\lambda}{2} \implies I(y=0) = 0$$ Thus, the central fringe in Lloyd's mirror is **strictly dark**. When white light is used, the central fringe is completely achromatic and jet black, unambiguously proving that reflection from a denser medium induces a phase change of $\pi$ radians.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
MediumExample 1.1: Quantitative Precision Measurement in Fresnel Biprism
In a Fresnel biprism experiment with sodium light of wavelength $\lambda = 589.3\text{ nm}$, the distance between the primary slit and the micrometer eyepiece is $D = 1.20\text{ m}$. Using a convex lens placed at two conjugate positions, the separations between the magnified and diminished images of the virtual sources are measured to be $d_1 = 4.05\text{ mm}$ and $d_2 = 2.45\text{ mm}$ respectively. Calculate: (a) the separation $d$ between the virtual coherent sources, (b) the fringe width $\beta$ observed on the micrometer scale, and (c) the number of bright fringes observed across a $15\text{ mm}$ field of view.
Step 1: Calculate the virtual source separation d using the conjugate displacement formula
$$d = \sqrt{d_1 d_2} = \sqrt{(4.05 \times 10^{-3} \text{ m}) \times (2.45 \times 10^{-3} \text{ m})} = \sqrt{9.9225 \times 10^{-6} \text{ m}^2} \approx 3.150 \times 10^{-3} \text{ m} = 3.150 \text{ mm}$$

The lens magnification at conjugate positions satisfies $m_1 m_2 = 1$, making the geometric mean of the two image separations equal to the true separation of the virtual sources.

Step 2: Determine the linear fringe width beta
$$\beta = \frac{\lambda D}{d} = \frac{(589.3 \times 10^{-9} \text{ m}) \times (1.20 \text{ m})}{3.150 \times 10^{-3} \text{ m}} = \frac{7.0716 \times 10^{-7}}{3.150 \times 10^{-3}} \approx 2.245 \times 10^{-4} \text{ m} = 0.2245 \text{ mm}$$

Each fringe pair occupies exactly $0.2245\text{ mm}$ on the eyepiece focal plane.

Step 3: Calculate the total number of fringes across the field of view
$$N = \frac{W}{\beta} = \frac{15.0 \text{ mm}}{0.2245 \text{ mm}} \approx 66.82 \implies N = 66 \text{ complete bright fringes}$$

Over a $1.5\text{ cm}$ field of view, 66 bright interference bands are resolved.

HardExample 1.2: Shift of Fringes by Introduction of Thin Transparent Mica Sheet
A thin transparent sheet of mica of refractive index $\mu = 1.58$ is inserted into the path of one of the interfering beams in a double-slit experiment illuminated by $\lambda = 550\text{ nm}$. The central zero-order bright fringe shifts across the screen by a distance equal to the spacing of 14 bright fringes. (a) Derive the general expression for the fringe shift $y_0$. (b) Calculate the precise thickness $t$ of the mica sheet.
Step 1: Derive the optical path difference introduced by a dielectric plate of thickness t
$$\text{Path in air} = t, \quad \text{Optical path in medium} = \mu t$$ $$\Delta_{\text{extra}} = \mu t - t = (\mu - 1)t$$ $$\text{Total path difference at screen position } y: \quad \Delta = \frac{y d}{D} - (\mu - 1)t$$

The medium slows the phase velocity to $c/\mu$, adding an optical distance $(\mu - 1)t$ to the traversed arm.

Step 2: Relate the central fringe position to the fringe count shift n
$$\text{At the shifted central fringe } (\Delta = 0): \quad \frac{y_0 d}{D} = (\mu - 1)t \implies y_0 = \frac{D}{d} (\mu - 1)t$$ $$\text{Since fringe width } \beta = \frac{\lambda D}{d}, \quad y_0 = n \beta = n \frac{\lambda D}{d} \implies n \lambda = (\mu - 1)t$$

The number of shifted fringes $n$ depends purely on the extra optical path divided by the wavelength.

Step 3: Solve for thickness t with numerical parameters
$$t = \frac{n \lambda}{\mu - 1} = \frac{14 \times (550 \times 10^{-9} \text{ m})}{1.58 - 1} = \frac{7.70 \times 10^{-6} \text{ m}}{0.58} \approx 1.328 \times 10^{-5} \text{ m} = 13.28 \ \mu\text{m}$$

The mica sheet has a physical thickness of $13.28$ micrometers.