Interference¶

The superposition of Waves¶

\[\frac{1}{v^2}\frac{\partial^2 \psi }{\partial t^2}=\frac{\partial^2 \psi }{\partial x^2}+\frac{\partial^2 \psi }{\partial y^2}+\frac{\partial^2 \psi }{\partial z^2} =\nabla^2\psi\]

This equation is linear, \(\psi(\vec{r},t)\) and its derivatives appear only to the first power. Consequently, if \(\psi_i(\vec{r},t)\) are solutions,

\[\psi(\vec{r},t)=\sum_{i=1}^n C_i \psi_i(\vec{r},t)\]

will be a solution as well.

Adding two waves that have the same frequncy amd wavelength.

\[E_1=E_{01}\cos\left( \alpha_1-\omega t \right)\\ E_2=E_{02}\cos\left( \alpha_2-\omega t \right)\]

where \(\alpha_i=kx_i+\phi_i\) with \(x_i\)being the distance from the source \(s_i\) of the waveto the point of observation.

1. Algebreic Method¶

The linear combination of the waves is

\[\begin{align*} E\equiv E_0\cos\left( \alpha-\omega t \right) = E_1+E_2 \end{align*}\]

where (notice \(\cos\omega t\) and \(\sin\omega t\) are linearly independent)

\[\begin{align*} E_0\cos \alpha &=E_{01}\cos\alpha_1+E_{02}\cos\alpha_2\\ E_0\sin \alpha &=E_{01}\sin\alpha_1+E_{02}\sin\alpha_2 \end{align*} \]

square the two equations and add them up:

Amplitude:

\[\boxed{\begin{align*} E_0^2=E_{01}^2+E_{02}^2+2E_{01}E_{02}\cos\left( \alpha_2-\alpha_1 \right) \end{align*}}\]

interference term (干涉项):

\[2E_{01}E_{02}\cos\left( \alpha_2-\alpha_1 \right)\]

phase difference:

\[\begin{align*} \delta \equiv \left( \alpha_2-\alpha_1 \right)=\frac{2\pi}{\lambda}(x_2-x_1)+(\phi_2-\phi_1) \end{align*}\]

interfere constructively: when \(\left( \alpha_2-\alpha_1 \right)=2m\pi\) for integer \(m\)
interfere destructively: when \(\left( \alpha_2-\alpha_1 \right)=(2m+1)\pi\) for integer \(m\)

2. The Complex Method¶

\[\begin{align*} \tilde{E}_1=&E_{01}e^{i(\alpha_1-\omega_1 t)} =E_{01}e^{i\alpha_1}e^{-i\omega_1 t}\\ &E_1 = Re(\tilde{E_1}) \end{align*}\]

\[\begin{align*} \tilde{E}={E_{0}e^{i\alpha}}e^{-i\omega t}=\left[ {E_{01}e^{i\alpha_1}+E_{02}e^{i\alpha_2}} \right]e^{-i\omega t} \end{align*}\]

where \(E_{0}e^{i\alpha}\) is known as the complex amplitude of the composite wave. Since \(E_0^2=\left( E_0e^{i\alpha} \right)^* \cdot \left(E_0e^{i\alpha}\right)\), we find

\[\begin{align*} E_0^2=&\left[ E_{01}e^{-i\alpha_1}+E_{02}e^{-i\alpha_2} \right]\left[ E_{01}e^{i\alpha_1}+E_{02}e^{i\alpha_2} \right]\\ =&E_{01}^2+E_{02}^2+E_{01}E_{02}\left[ e^{i(\alpha_2-\alpha_1)} + e^{-i(\alpha_2-\alpha_1)} \right]\\ =&E_{01}^2+E_{02}^2+2E_{01}E_{02}\cos(\alpha_2-\alpha_1) \end{align*}\]

3. Phasor Addition¶

We represent a wave, which has an amplitude and a phase, to a vector, know as a phasor , in a two-dimensional plane, such that

\[ \begin{align*} E_i=&E_{0i}\cos(\alpha_i-\omega t)\\ =&\left(E_{0i}\cos\alpha_i\right)\cos\omega t+\left(E_{0i}\sin\alpha_i\right)\sin\omega t\\ \Longrightarrow\vec{E}_i=&\left(E_{0i}\cos\alpha_i\right)\hat{x}+\left(E_{0i}\sin\alpha_i\right)\hat{y} \end{align*} \]

The algebraic sum, \(E=E_1+E_2\) is the projection on the x axis of the corresponding phasor sum.

If we denote the phase delay \(\phi=\alpha_2-\alpha_1\) as shown in the phasor diagram, the law of cosines(余弦定理) applied to the triangle of sides \(\vec{E}_{01}\), \(\vec{E}_{02}\) and \(\vec{E}_{0}\) yields

\[ \begin{align*} E_0^2=E_{01}^2+E_{02}^2+E_{01}E_{02}\cos(\alpha_2-\alpha_1) \end{align*} \]

Summary

The equivalence of the following formulas - Algebraic Method: \(\cos\omega t\) and \(\sin\omega t\) are linearly independent

\[\begin{align*} &E_0\cos\alpha\cos\omega t+E_0\sin\alpha\sin\omega t= \\ &E_{01}\cos\alpha_1\cos\omega t+E_{01}\sin\alpha_1\sin\omega t+\\ &E_{02}\cos\alpha_2\cos\omega t+E_{02}\sin\alpha_2\sin\omega t\end{align*}\]

Complex Method: The real part and the imaginary part are independent

\[ E_0e^{i\alpha}e^{-i\omega t}=\left[E_{01}e^{i\alpha_1}+E_{02}e^{i\alpha_2}\right]e^{-i\omega t} \]

Phasor Additon: The vectors that are pertincular to each other are independent

\[\begin{align*} &E_0\cos\alpha\hat{x}+E_0\sin\alpha\hat{y}=\\ &E_{01}\cos\alpha_1\hat{x}+E_{01}\sin\alpha_1\hat{y}+\\ &E_{02}\cos\alpha_2\hat{x}+E_{02}\sin\alpha_2\hat{y} \end{align*}\]

Young's Double-Slit Interference Experiment¶

Natural Light¶

Naturual light is incoherent light

The light from two fine incandescent wires wouldn't interfere, because the light is emitted by vast numbers of atoms in the wires, acting randomly and independently for extremely short times. The light is said to be incoherent (不相干, 离散).

Conditions for Interference¶

Condition for Interference

Two beams must have (nearly) the same frequency \(\omega\) . therwise, the phase difference is time-dependent. During the detection interval, the interference pattern will be averaged away.
The clearest pattern (with maximum contrast) exists when interfering waves have (nearly) equal amplitude.
Initial phase difference can exist between sources, as long as it remains constant; the two sources are said to be coherent (相干, 一致) .

Young's Interference Experiment¶

The waves are in phase (相同) when they pass through the two slits because there they are just portions of the same incident wave. However, once they have passed the slits, the two waves must travel different distances to reach point \(P\) on screen \(C\).

The path length difference is, when \(d\ll D\),

\[ \begin{align*} \Delta L=d\sin\theta \end{align*} \]

The electric field components of these waves at point \(P\) are not in phase and vary with time as

\[ \begin{align*} E_1=E_0\cos(kr_1-\omega t)=E_0\cos(kL+\beta -\omega t)\\ E_2=E_0\cos(kr_2-\omega t)=E_0\cos(kL-\beta -\omega t) \end{align*} \]

where the phase difference \(\left[ L=\frac{r_1+r_2}{2}=\sqrt{D^2+y^2} \right]\)

\[ \begin{align*} \delta_2=2\beta=k\Delta L=\frac{2\pi d}{\lambda}\sin\theta \end{align*} \]

The total intensity is thus given by

\[ \begin{align*} &I=E_{0}^2+E_{0}^2+E_{0}E_{0}\cos(\delta_2)\\ &= 2E_0^2\left[ 1+\cos(2\beta) \right]\\ &=I_{max}\cos^2\beta \end{align*} \]

\[\beta = \frac{\pi \Delta L}{\lambda}\]

a bright fringe appears when

\[\begin{align*} \Delta L=d\sin\theta=m\lambda, m\in \mathbb{Z} \end{align*}\]

a dark fringe appears when

\[\begin{align*} \Delta L=d\sin\theta = \left(m+\frac{1}{2}\right)\lambda, m\in \mathbb{Z} \end{align*}\]

We can then find the angle \(\theta\) to any fringe and thus use the values of \(m\) to label the fringes.

Interference from Thin Films¶

Reflection Phase Change (相变)¶

Reflection at an interface can cause a phase change (相变), depending on the indexes of refraction on the two sides of the interface.
There is no phase change for refraction(折射).

Reflection	Refletion Phase Shift
Off Lower Indexes	0
Off Higher Indexes	0.5 wavelength

\(r_1\) has an additional reflection phase shift 0.5 wavelength. \(n_2>n_{air}\)
There is no such shift for \(r_2\). \(n_2 > n_{air}\)

Interference¶

The path difference of \(r_1\) and \(r_1\): \(\Delta L = 2L\)
The wavelength in the medium \((r_2)\):

\[\begin{align*} &\lambda_2 =\frac{v_2}{f}=\frac{c}{n_2 f}=\frac{\lambda}{n_2}\\ &k_2 = \frac{2\pi}{\lambda_2} = n_2 k \end{align*}\]

the reflected \(r_1\) and \(r_2\) ，and their superposition are：

\[\begin{align*} \vec{E}_1=&E_0\cos\left(k\left(x+\frac{1}{2}\lambda\right)-\omega t\right)\\ =&E_0\cos(\pi+kx-\omega t)\\ \vec{E}_2=&E_0\cos(k_2 2L+kx-wt)\\ \therefore \, \vec{E}^2 =&\left( \vec{E}_1+\vec{E}_2 \right)^2=2E_0^2\left(1+\cos(k_2 2L-\pi) \right) \end{align*}\]

The maximum interference and the bright rigion appear if

\[\begin{align*} (k_2 2L-\pi)&=2m\pi\\ 2L=\left(m+\frac{1}{2}\right)\lambda_2&=\left(m+\frac{1}{2}\right)\frac{\lambda}{n_2}, m\in \mathbb{Z} \end{align*}\]

The minimum interference and the dark region appear if

\[\begin{align*} (k_2 2L-\pi)&=(2m+1)\pi\\ 2L=m\lambda_2&=m\frac{\lambda}{n_2}, m\in \mathbb{Z} \end{align*}\]