Interference

When two or more waves overlap, the resultant displacement is the sum of the individual displacements. For two coherent waves with amplitudes $E_1$ and $E_2$:

$E = E_1 + E_2 = E_0 \cos(\mathbf{k}\cdot\mathbf{r} - \omega t + \phi_1) + E_0 \cos(\mathbf{k}\cdot\mathbf{r} - \omega t + \phi_2)$

The time-averaged intensity is:

$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\Delta\phi$

Where $\Delta\phi = \phi_2 - \phi_1$ is the phase difference.

Two slits separated by distance $d$ are illuminated by coherent light of wavelength $\lambda$. The Screen is at distance $L \gg d$.

Condition for bright fringes (constructive interference):

$d\sin\theta = m\lambda, \quad m = 0, \pm 1, \pm 2, \ldots$

Condition for dark fringes (destructive interference):

$d\sin\theta = \left(m + \frac{1}{2}\right)\lambda, \quad m = 0, \pm 1, \pm 2, \ldots$

Derivation. The path difference between the two slits is $\Delta = d\sin\theta$. Constructive Interference occurs when $\Delta = m\lambda$ (phase difference $2m\pi$), and destructive when $\Delta = (m + 1/2)\lambda$ (phase difference $(2m+1)\pi$). $\blacksquare$

The fringe spacing on the screen:

$\Delta y = \frac{\lambda L}{d}$

Light reflecting from a thin film of thickness $t$ and refractive index $n$ undergoes interference Between the wave reflected from the top surface and the wave reflected from the bottom surface.

Path difference: $2nt\cos\theta_t$ where $\theta_t$ is the angle of refraction inside the film.

A phase shift of $\pi$ occurs upon reflection from a medium of higher refractive index. The condition For constructive interference (bright reflection) is:

$2nt\cos\theta_t = \left(m + \frac{1}{2}\right)\lambda \quad \mathrm{(one\ phase\ shift)}$

$2nt\cos\theta_t = m\lambda \quad \mathrm{(zero\ or\ two\ phase\ shifts)}$

:::caution Common Pitfall Always count the number of $\pi$ phase shifts that occur upon reflection. A reflection from Low-to-high refractive index introduces a $\pi$ shift; high-to-low does not. For a soap film in Air, there is one $\pi$ shift (at the top surface). For a coating on glass ($n_{\mathrm{coat} \lt n_{\mathrm{glass}}}$), there is also one shift. The conditions for constructive and destructive interference swap depending on whether the total number of shifts is odd or even.

A beam splitter divides light into two beams that travel perpendicular paths and recombine. The path Difference $\Delta = 2(d_1 - d_2)$ determines the interference pattern.

Moving mirror $M_1$ by distance $\Delta d$ shifts the pattern by $m = 2\Delta d / \lambda$ fringes.

The Michelson interferometer is used for precise length measurements, the determination of Refractive indices, and was historically crucial in establishing the invariance of the speed of Light (Michelson-Morley experiment).

The fringe visibility (or contrast) quantifies the sharpness of interference fringes:

$\mathcal{V} = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$

For two-beam interference with intensities $I_1$, $I_2$ and degree of temporal coherence $|\gamma|$:

$\mathcal{V} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}\,|\gamma(\tau)|$

For equal intensities ($I_1 = I_2$): $\mathcal{V} = |\gamma(\tau)|$.

The coherence function decays with path difference. For a Gaussian spectral profile of width $\Delta\lambda$:

$|\gamma(\tau)| = \exp\!\left[-\pi\left(\frac{\Delta\lambda \cdot \Delta x}{\lambda^2}\right)^2\right]$

Where $\Delta x = c\tau$ is the path difference. Fringes become unresolvable when $\Delta x \approx \lambda^2/\Delta\lambda = L_c$The coherence length.

Implication for interferometry. To observe interference fringes, the path difference between The two arms must satisfy $\Delta x \ll L_c$. A sodium lamp ($\Delta\lambda \approx 0.6$ nm at $\lambda = 589$ nm) gives $L_c \approx 0.58$ mm. A HeNe laser ($\Delta\lambda \approx 10^{-6}$ nm) Gives $L_c \approx 350$ m — fringes are visible over enormous path differences.

A Fabry-Perot etalon consists of two parallel, partially reflecting surfaces separated by Distance $d$. Multiple reflections create many beams that interfere, producing sharp transmission Peaks.

For a lossless etalon with reflectance $R$ and transmittance $T = 1 - R$Illuminated at angle $\theta$ inside a medium of refractive index $n$:

$\frac{I_T}{I_0} = \frac{T^2}{(1 - R)^2 + 4R\sin^2(\delta/2)} = \frac{1}{1 + F\sin^2(\delta/2)}$

Where $\delta = (4\pi/\lambda)\,nd\cos\theta$ is the round-trip phase and $F = 4R/(1-R)^2$ is the coefficient of finesse.

Transmission maxima occur when $\delta = 2m\pi$ ($m = 0, 1, 2, \ldots$), i.e., $2nd\cos\theta = m\lambda$.

Finesse:

$\mathcal{F} = \frac{\pi\sqrt{F}}{2} = \frac{\pi\sqrt{R}}{1 - R}$

The finesse determines the sharpness of the peaks: higher $R$ gives sharper peaks.

Free spectral range (frequency spacing between adjacent peaks):

$\Delta\nu_{\mathrm{FSR} = \frac{c}{2nd}}$

Resolving power:

$\mathcal{R} = \frac{\lambda}{\delta\lambda} = m\mathcal{F}$

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