How to calculate the intensity of the interference of two waves in a given point? [closed]

Question

There are two different point sources which produce spherical waves with the same power, amplitude, ω, wavenumber and phase.

I can calculate the intensity of each wave in a point: $$ I_1 = P / (4 \pi r_1^2) $$ $$ I_2 = P / (4 \pi r_2^2) $$ But how can I calculate the intensity of the resulting wave in that point?

Answer 1

I found the solution, but forgot to add it here. But since a considerable number of people has seen the question, the answer could be useful.

Notation

• $r_1$: distance between the given point and the focus 1.
• $r_2$: distance between the given point and the focus 2.
• $A_1$: amplitude in the given point of the wave generated by focus 1.
• $A_2$: amplitude in the given point of the wave generated by focus 2.
• $A_r$: amplitude in the given point of the resulting wave.
• $I_1$: intensity in the given point of the wave generated by focus 1.
• $I_2$: intensity in the given point of the wave generated by focus 2.
• $I_r$: intensity in the given point of the resulting wave.
• $P$: power of the initial waves.
• $\varphi$: phase difference between the initial waves in the given point.
• $k$: wavenumber of the initial and resulting waves.
• $\omega$: angular frequency of the initial and resulting waves.
• $v$ : velocity of the initial and resulting waves.

Preface

I will use the formula

$$ I = \frac{1}{2}\rho v \omega^2 A^2 $$

The resulting wave will be a wave of amplitude $A_r$ with the same $k$ and $\omega$ than the initial waves. Moreover, $\rho$ and $v$ will also be the same because they only depend on the environment.

Then,

$$ \begin{cases} I_r = \frac{1}{2}\rho v \omega^2 A_r^2 \\ I_2 = \frac{1}{2}\rho v \omega^2 A_2^2 \\ \end{cases} \implies I_r = I_2 \left(\frac{A_r}{A_2}\right)^2 = \frac{P}{4\pi r_2^2} \left(\frac{A_r}{A_2}\right)^2 $$

In order to express $I_r$ in terms of $r_1$ and $r_2$ instead of $A_1$ and $A_2$, I will use that the amplitude of an spherical wave is inversely proportional to the distance to the focus. That is:

$$ \frac{A_1}{A_2} = \frac{r_2}{r_1} $$

Answer

In a point with destructive interference ($\varphi = \pi$)

The resulting amplitude will be the difference of amplitudes:

$$ A_r = |A_1 - A_2| $$

Then, the resulting intensity is

$$ I_r = \frac{P}{4\pi r_2^2} \left(\frac{A_r}{A_2}\right)^2 = \frac{P}{4\pi r_2^2} \left(\frac{A_1 - A_2}{A_2}\right)^2 = \frac{P}{4\pi r_2^2} \left(\frac{A_1}{A_2}-1\right)^2 = \frac{P}{4\pi r_2^2} \left(\frac{r_2}{r_1}-1\right)^2 = \frac{P}{4\pi} \left(\frac{r_2-r_1}{r_1 r_2}\right)^2 $$

In a point with constructive interference ($\varphi = 0$)

The resulting amplitude will be the sum of amplitudes:

$$ A_r = A_1 + A_2 $$

Then, the resulting intensity is

$$ I_r = \frac{P}{4\pi r_2^2} \left(\frac{A_r}{A_2}\right)^2 = \frac{P}{4\pi r_2^2} \left(\frac{A_1 + A_2}{A_2}\right)^2 = \frac{P}{4\pi r_2^2} \left(\frac{A_1}{A_2}+1\right)^2 = \frac{P}{4\pi r_2^2} \left(\frac{r_2}{r_1}+1\right)^2 = \frac{P}{4\pi} \left(\frac{r_1+r_2}{r_1 r_2}\right)^2 $$

In general

The resulting amplitude will be ($\varphi$ is the phase difference):

$$ A_r = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos{\varphi}} $$

Then, the resulting intensity is

$$ I_r = \frac{P}{4\pi r_2^2} \left(\frac{A_r}{A_2}\right)^2 = \frac{P}{4\pi r_2^2} \frac{A_1^2 + A_2^2 + 2 A_1 A_2 \cos{\varphi}}{A_2^2} = \frac{P}{4\pi r_2^2} \left(\left(\frac{A_1}{A_2}\right)^2 + 1 + 2 \frac{A_1}{A_2} \cos{\varphi}\right) = \frac{P}{4\pi r_2^2} \left(\left(\frac{r_2}{r_1}\right)^2 + 1 + 2 \frac{r_2}{r_1} \cos{\varphi}\right) = \frac{P}{4\pi} \frac{r_1^2 + r_2^2 + 2 r_1 r_2 \cos{\varphi}}{(r_1 r_2)^2} $$