Interference

Interference

Steady Wave

There are two forms of steady wave. One is steady vecter wave, like the EM wave:

\[ \begin{cases} \vec{E}(P, t) = \vec{E}_0(P) \cos (\omega t - \phi(P)) \\ \vec{H}(P, t) = \vec{H}_0(P) \cos (\omega t - \phi(P)) \end{cases} \]

The other is steady scalar wave:

\[ \begin{gathered} U(P, t) = A(P) \cos (\omega t - \phi(P)) \\\\ \tilde{U}(P, t) = A(P) e^{i(\phi(P) - \omega t)} = A(P) e^{i\phi(P)} e^{-i\omega t} \\\\ \tilde{U}(P) = A(P) e^{i\phi(P)} \end{gathered} \]

For steady plane wave, we have

\[ \begin{gathered} \begin{cases} A(P) = \text{constant} \\ \phi(P) = \vec{k} \bullet \vec{r} = k_x x + k_y y + k_z z + \phi_0 \end{cases} \\\\ \tilde{U}(P) = A e^{i\phi(P)} = A e^{i(\vec{k} \bullet \vec{r} + \phi_0)} = A e^{i(k_x x + k_y y + k_z z + \phi_0)} \end{gathered} \]

For steady spherical wave, we have

\[ \begin{gathered} \begin{cases} A(P) = \dfrac{a}{r} \\ \phi(P) = kr + \phi_0 \end{cases} \\\\ \tilde{U}(P) = \frac{a}{r} e^{i(kr + \phi_0)} \end{gathered} \]

The wave intensity is \(\(I(P) = [A(P)]^2 = \tilde{U}^*(P) \cdot \tilde{U}(P)\)\)

Wave Superposition and Interference

The wave superposition principle is that, for scalar wave, the resultant wave is the sum of all the waves; For vector wave, the resultant wave is the vector sum of all the waves.

When two waves interfere, the intensity of the resultant wave is

\[ \begin{aligned} I(P, t) &= \tilde{U}^*(P, t) \cdot \tilde{U}(P, t) \\ &= \left( A_1 e^{-i\phi_1} e^{i\omega_1 t} + A_2 e^{-i\phi_2} e^{i\omega_2 t} \right) \left( A_1 e^{i\phi_1} e^{-i\omega_1 t} + A_2 e^{i\phi_2} e^{-i\omega_2 t} \right) \\ &= A_1^2 + A_2^2 + 2A_1A_2 \cos [(\phi_1 - \phi_2) - (\omega_1 - \omega_2)t] \\ &= I_1 + I_2 + 2\sqrt{I_1I_2} \cos [(\phi_1 - \phi_2) - (\omega_1 - \omega_2)t] \end{aligned} \]

So if \(\delta(P) = \phi_1 - \phi_2\) is not steady, or \(\omega_1 \neq \omega_2\), no interference will occur. If \(\delta\) changes with time, we call it incohrent.

For two spherical plane waves, if \(\phi_1 = \phi_2\), then \(\(\delta(P) = \frac{2\pi}{\lambda} (r_1 - r_2)\)\)

If \(\Delta L = r_1 - r_2 = m\lambda\), \(I(P)\) is maximum; If \(\Delta L = r_1 - r_2 = (m + \dfrac{1}{2})\lambda\), \(I(P)\) is minimum.

Young’s Double Slit Quantitative

In Young’s double slit expriment, we have \(\(\Delta L = d\sin\theta, \quad \sin\theta = \frac{y}{L}\)\)

So constructive interference happens at \(y = \dfrac{m\lambda L}{d}\).

Destructive interference happens at \(y = \dfrac{(m + \dfrac{1}{2})\lambda L}{d}\).

Thin Film Interference

If \(n_1 > n_2\), no phase change upon reflection.

If \(n_1 < n_2\), phase change of \(180^{\circ}\) upon reflection, equivalent to the wave shifting by \(\lambda/2\).

If the thickness is the same at every point, fringes of equal inclination will appear:

\[ \begin{aligned} \Delta L &= (ARC) - (AB) \\ &= n \frac{2h}{\cos i} - n_1 \cdot 2h\tan i \cdot \sin i_1 \\ &= 2h (\frac{n}{\cos i} - \frac{n_1\sin i_1 \sin i}{\cos i}) \\ &= 2nh (\frac{1}{\cos i} - \frac{\sin^2 i}{\cos i}) \\ &= 2nh\cos i \end{aligned} \]

Consider the half-wave loss, \(\Delta L = 2nh\cos i + \lambda/2\).

\[ \cos i_{m+1} - \cos i_m = -\sin i (i_{m+1} - i_m) = \frac{\lambda}{2nh} \\\\ \Delta r_m = r_{m+1} - r_m = f(i_{m+1} - i_m) = - \frac{f\lambda}{2nh\sin i_m} \]

So the fringes of equal inclination are internally sparse and externally dense. Moreover, \(n\) or \(h\) is bigger, \(\Delta r\) is smaller.

---

If the thickness is not the same at every point, fringes of equal thickness will appear. The difference of optical path is:

\[ \begin{aligned} \Delta L &= (QABP) - (QP) \\ &= (QA) - (QP) + (ABP) \\ &= -n AP \sin i + 2(AB) \\ &= -n \cdot (2h\tan i) \cdot \sin i + \frac{2nh}{\cos i} \\ &= 2nh (\frac{1}{\cos i} - \frac{\sin^2 i}{\cos i}) \\ &= 2nh \cos i \end{aligned} \]

For the air film, the gap between fringes is \(\(\Delta x = \frac{\Delta h}{\tan\theta} = \frac{\lambda}{2\theta}\)\)

For the Newton's ring, the radis of its minimum fringes is

\[ h_m = R - \sqrt{R^2 - r_m^2} \approx \frac{1}{2} \frac{r_m^2}{R} = \frac{1}{2} m\lambda \\\\ \Rightarrow r_m = \sqrt{m\lambda R} \]

It can be seen that the fringes of the Newton's ring are also internally sparse and externally dense.