SciLearn: wave interference

This post deals with the interference of light.

Light is a strange thing. It's made of photons, yet most of the time, it behaves as a wave. Now I won't go into the quantum mechanical details of light; I'm just going to talk about one particular nature of light: interference, which occurs whenever light behaves as a wave. I'll provide a systematic summary of wave interference and slit interference patterns.

The first step to interference is to begin with the basic equations of wave motion. Let's pick a simple sine wave in 1 dimensions:

d(x, t) = a sin ϕ = a sin[kx - ωt + ϕ₀]

where:
d = displacement;
a = amplitude;
k = angular wavenumber;
x = position;
ω = angular frequency;
t = time;
ϕ = phase;
ϕ₀ = initial phase.

When two waves interfere, their displacement superpose on each other, so they can simply be added to produce the resultant wave (for simplicity, amplitude of both waves are equal):

$\begin{align*} D (x, t) &= a \sin \phi_1 + a \sin \phi_2 \\ &= 2 a \cos \frac{\phi_1 - \phi_2}{2} \sin \frac{\phi_1 + \phi_2}{2} \\ &= 2 a \cos \frac{\Delta \phi}{2} \sin \frac{\sum \phi}{2} \\ &= 2 a \cos \frac{\Delta \phi}{2} \sin \langle\phi\rangle \end{align*}$

where
ϕ₁ = k₁x₁ - ω₁t + ϕ_0,1;
ϕ₂ = k₂x₂ - ω₂t + ϕ_0,2;
Δϕ = ϕ₁ - ϕ₂;
Σϕ = ϕ₁ + ϕ₂.
⟨ϕ⟩ = Σϕ / 2
The above equation involved the use of a "sum-to-product" trigonometric identity.

There are special scenarios under which the above equation simplifies:

Standing wave:
x₁ = x = x₂
k₁ = k = k₂
-ω₁ = ω = ω₂
ϕ_0,1 = 0 = ϕ_0,2
The displacement equation is:

$\begin{align*} & \quad \: D (x, t) \\ &= 2 a \cos \frac{\Delta \phi}{2} \sin \frac{\sum \phi}{2} \\ &= 2 a \cos \frac{\left(k x + \omega t\right) - \left(k x - \omega t\right)}{2} \\ & \quad \: \times \sin \frac{\left(k x + \omega t\right) + \left(k x - \omega t\right)}{2} \\ &= 2 a \cos \omega t \sin kx \end{align*}$
The boundary conditions for a standing wave is that the ends, at x = 0 and x = L, are fixed discontinuities (i.e. the ends of the standing wave cannot move):

D(0, t) = D(L, t) = 0 (for any t).
Solving this equation yields the allowable wavenumbers and wavelengths (λ) for standing waves:

$\begin{align*} D (L, t) &= 0 \\ A \cos \omega t \sin kL &= 0 \\ \sin kL &= 0 \\ kL &= n \pi, \quad n \in \mathbb{Z} \\ k &= \frac{n \pi}{L} \quad \Leftrightarrow \quad \lambda = \frac{2 L}{n} \end{align*}$
Coherent interference:
k₁ = k = k₂
ω₁ = ω = ω₂
The displacement equation is:

$\begin{align*} & \quad \: D (x, t) \\ &= A \cos \frac{\Delta \phi}{2} \sin \langle \phi \rangle \\ &= A \cos \frac{\left(k x_1 - \omega t + \phi_{0,1}\right) - \left(k x_2 - \omega t + \phi_{0,2}\right)}{2} \\ & \quad \: \times \sin \frac{\left(k x_1 - \omega t + \phi_{0,1}\right) + \left(k x_2 - \omega t + \phi_{0,2}\right)}{2} \\ &= A \cos \frac{k \left(x_1 - x_2\right) + \left(\phi_{0,1} - \phi_{0,2}\right)}{2} \\ & \quad \: \times \sin \left(k \frac{x_1 + x_2}{2} - \omega t + \frac{\phi_{0,1} + \phi_{0,2}}{2}\right) \\ &= A \cos \frac{k \Delta x + \Delta \phi_0}{2} \sin \left(k_2 \langle x \rangle - \omega t + \langle \phi_0 \rangle\right) \end{align*}$
The condition for destructive interference is:

$\begin{align*} D (x, t) &= 0 \\ 2 a \cos \frac{\Delta \phi}{2} \sin \langle \phi \rangle &= 0 \\ \cos \frac{\Delta \phi}{2} &= 0 \\ \frac{\Delta \phi}{2} &= n \pi + \frac{\pi}{2}, \quad n \in \mathbb{Z} \\ \Delta \phi &= \left(2 n + 1 \right)\pi \end{align*}$
Similarly, for constructive interference:

$\begin{align*} D (x, t) &= \pm 2a \sin \langle \phi \rangle \\ 2 a \cos \frac{\Delta \phi}{2} \sin \langle \phi \rangle &= \pm 2a \sin \langle \phi \rangle \\ \cos \frac{\Delta \phi}{2} &= \pm 1 \\ \frac{\Delta \phi}{2} &= n \pi, \quad n \in \mathbb{Z} \\ \Delta \phi &= 2 n\pi \end{align*}$
Beat interference:
x₁ = x = x₂
k₁ = k = k₂
ϕ_0,1 = ϕ₀ = ϕ_0,2
The displacement equation is:

$\begin{align*} & \quad \: D (x, t) \\ &= A \cos \frac{\Delta \phi}{2} \sin \langle \phi \rangle \\ &= A \cos \frac{\left(k x - \omega_1 t + \phi_0\right) - \left(k x - \omega_2 t + \phi_0\right)}{2} \\ & \quad \: \times \sin \frac{\left(k x - \omega_1 t + \phi_0\right) + \left(k x - \omega_2 t + \phi_0\right)}{2} \\ &= A \cos \left(\frac{\Delta \omega}{2} t\right) \sin \left(k x - \langle\omega\rangle t + \phi_0\right) \\ &= A \cos \omega_\text{mod} t \sin \left(k x - \langle\omega\rangle t + \phi_0\right) \end{align*}$
where ω_mod is the modulation angular frequency.

Now, at last, I can show the general equations (approximations) for multiple-slit interference.

For single-slits, the minima angles (θ_min) can be found using:
a sin θ_min = n λ where n is a positive integer and a is the slit width.

For double-slits, multiple-slits, or diffraction gratings, the primary maxima angles can be found using:
d sin θ_max = n λ where n is zero or a positive integer and d is the slit separation.

The two results actually apply to any slit interference: θ_min will be the primary minima angles and θ_max will be the primary maxima angles. The shape of the intensity plot will always have an envelope curve that is shaped like the intensity curve of a single-slit diffraction, and the minima of this envelope curve is the "primary minima".

So that was my brief summary of light interference.