Waves interference in terms of fields and intensity

Question

let's suppose that two electromagnetic waves (let's focus for instance on their electric fields) with same amplitude A and frequency, but with a difference $\Delta\Phi$: $$E_{1}=A\cdot \sin\left(kz-\omega t\right)$$ $$E_{2}=A\cdot \sin\left(kz-\omega t+\Delta\Phi\right)$$

Now, I've seen two kinds of analysis of this phenomenon:

1) Thinking in terms of electric field (or magnetic field)

The total electric field $E=E_{1}+E_{2}$ may be expressed in this way: $$E=2A\cdot \sin\left(kz-\omega t+\frac{\Delta\Phi}{2}\right)\cos\left(\frac{\Delta\Phi}{2}\right)$$

It's simply an oscillating electric field with same frequency and with amplitude equal to:$$2A\cdot \cos\left(\frac{\Delta\Phi}{2}\right)$$

This term allows us to define constructive and destructive interference. (Reference)

In this situation, I'd say that there will be:

• Constructive Interference ($A<E_{max}\leq2A$ that means $A<2A\cdot \cos\left(\frac{\Delta\Phi}{2}\right)\leq2A$ that means $\frac{1}{2}<\cos\left(\frac{\Delta\Phi}{2}\right)\leq1$) if and only if $$-60°<\frac{\Delta\Phi}{2}\leq0°$$ that means $$-120°<\Delta\Phi\leq0°$$

where there is completely constructive interference $2A\cdot \cos\left(\frac{\Delta\Phi}{2}\right)=2A$ in the specific case $\Delta\Phi=0°$ (in-phase waves).

• Destructive Interference ($0<E_{max}\leq A$ that means $0<2A\cdot \cos\left(\frac{\Delta\Phi}{2}\right)\leq A$ that means $0<\cos\left(\frac{\Delta\Phi}{2}\right)\leq \frac{1}{2}$) if and only if $$60°<\frac{\Delta\Phi}{2}\leq 90°$$ that means $$120°<\Delta\Phi\leq180°$$

where there is completely destructive interference $2A\cdot \cos\left(\frac{\Delta\Phi}{2}\right)=0$ in the specific case $\Delta\Phi=180°$ (out-of-phase waves).

2) Thinking in terms of wave intensity (Reference)

In this case we get$$I=2I_{0}\cdot\left(1+\cos\left(\Delta\Phi\right)\right)$$, where $$I_0$$ is the mean intensity of the single electromagnetic waves ($I_0 = \frac{(E_1)^2}{Z_0} =\frac{(E_2)^2}{Z_0} $).

In this case, I'd say that there will be:

• Constructive Interference ($2I_0 < I \leq 4I_0$ that means $0 < \cos(\Delta\Phi) \leq 1$) if $$ 0° < \Delta\Phi < 90°$$.

where there is completely constructive interference in the specific case $\Delta\Phi=0°$ (in-phase waves).

• Destructive Interference ($0 \leq I < 2I_0$ that means $-1 \leq \cos(\Delta\Phi) < 0$) if $$ 180° < \Delta\Phi < 270°$$.

where there is completely destructive interference in the specific case $\Delta\Phi=180°$ (out-of-phase waves).

Conclusions

• I) Only the definition of completely constructive and completely destructive interferences coincide in both analysis. Which is the physical meaning of the fact that partial constructive/destructive interference are achieved by different phase shifts if we think in terms of fields or intensity?

• II) Which kind of analysis is more useful in practice and why?

Answer 1

In wave mechanics, you have to use overlapping of fields not the intensity. This leads to all phenomena of interference and diffraction.

But for case of electromagnetic wave, the obsevations are generaly accumulated of many beams from both radiation sources. To maintain a constant phase difference between the two radiation sources is certainly not easy, and which is called `coherence`. For two coherent radiation sources, the combination radiation is the overlapping of two fields. The constructive case renders an amplitude of `2A`, thus an intensity of 4 times the origin intensity. ( Of course, there should also appear spatial variation to have other places become darker to keep energy conservation. That mean the phase difference should depends on space position, which is known as optical path differnece.)

In the other hand, if two radiation sources are not in coherence (called inconherence). The phase difference would not be a constant, but a function of time and space. The phsae difference appears more like a random phase. For two inconherent sources, the summing over all possible phases will effectively cancels the cross term between two fields in the expression of the intensity. The result will be equal to sum of two intensities. For example, two light bulbs in a room, the total intensity is basically the sum of the intensity of each bulb.

Answer 2

If we consider the interference pattern on a ccd camera due to the superposition of two coherent waves, both descriptions are equivalent. To see this, let's start with two $sin$-waves, as you did. To describe the interference patterns, we should

1. separate the spatial and the temporal components $$ \sin(kz - \omega t ) + \sin(kz + \omega t + \varphi) = 2 \sin(kz + \varphi/2)\cdot \cos(\omega t + \varphi/2) $$
2. square this expression, because the ccd camera records the intensity and not the electric field,
3. use the time average $ 1/T \int_0^T dt \;\cos^2(\omega t) = 1/2 $, because the frequency is so large that the ccd camera is unable to resolve the oscillation.

Hence, we end up with a term $I \propto 2\sin^2(kz + \varphi/2)$. Using trigonometric relationships we can write $ 2\sin^2(kz + \varphi/2) = 1 - \cos(2 k z + \varphi) $. Hence, the only difference between the two descriptions are

• the minus sign in front of the $\cos$-term, which is only a matter of how we define the phase $\Delta \phi = 2 k z + \varphi - \pi$,
• and a factor of two, which is probably due to the fact that you didn't consider the time average -- which is the standard phasor procedure.

Answer 3

Both methods are equivalent because $I = 2 I_0 (1+\cos (\Delta \Phi))=4I_0\cos^2\left ( \dfrac {\Delta \Phi}{2}\right)$ where $I_0 \propto A^2$.
The second representation is equivalent to your $2A \cos\left(\dfrac{\Delta\Phi}{2}\right)$ squared.

So the interference pattern for both representations is a set of $\cos^2$fringes.

Answer 4

Might be worth noting that individual intensities of Electromagnetic waves don't add in the general case.

Intensity is proportional to the square of the magnitude of the wave.

If $k$ is the proportional constant, then

$I_1=kE_1^2$

$I_2=kE_2^2$

But the electric fields are vector quantities.

$\vec{E_{tot}}=\vec{E_1}+\vec{E_2}$

So:

$I_{tot}=k(\vec{E_1}+\vec{E_2})^2=k(E_1^2+E_2^2+2\vec{E_1}\cdot\vec{E_2})=I_1+I_2+$ (cross term)

Both intensity and field treatments should give you the same results. The field is the fundamental entity.