The resultant intensity of two interfering waves

In summary,The resultant field is given by$$\begin{aligned}\overrightarrow{E}=\overrightarrow{E}_{1}+\overrightarrow{E}_{2}=\widehat{i} E_{0}\cos \omega t+\widehat{j}E_{0}\cos \left( \omega t+\phi \right) \\\Rightarrow E^{2}=E_{0}^{2}\cos ^{2}\omega t+E_{0}^{2}\cos ^{2}\left( \omega t+\phi \right) \\\Rightarrow E^{2
  • #1
Dom Tesilbirth
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Homework Statement
Consider the interference of two coherent electromagnetic waves whose electric field vectors are given by ##\overrightarrow{E}_{1}=\widehat{i}E_{0}\cos \omega t## and ##\overrightarrow{E}_{2}=\widehat{j}E_{0}\cos \left( \omega t+\phi \right),## where ##\phi## is the phase difference. The intensity of the resulting wave is given by ##\dfrac{\varepsilon _{0}}{2}\langle E^{2}\rangle##, where ##\langle E^{2}\rangle## in the time average of ##E^{2}.## What is the total intensity?
Relevant Equations
$$I=\dfrac{\varepsilon _{0}}{2}\langle E^{2}\rangle$$
My Try:

The resultant field is given by
$$\begin{aligned}\overrightarrow{E}=\overrightarrow{E}_{1}+\overrightarrow{E}_{2}=\widehat{i} E_{0}\cos \omega t+\widehat{j}E_{0}\cos \left( \omega t+\phi \right) \\
\Rightarrow E^{2}=E_{0}^{2}\cos ^{2}\omega t+E_{0}^{2}\cos ^{2}\left( \omega t+\phi \right) \\
\Rightarrow E^{2}=E_{0}^{2}\left\{ \cos ^{2}\omega t+\cos ^{2}\left( \omega t+\phi \right) \right\} \\
\Rightarrow \langle E^{2}\rangle =E_{0}^{2}\dfrac{\left[ \int ^{t}_{0}\left\{ \cos ^{2}\omega t+\cos ^{2}\left( \omega t+\phi \right) \right\} dt\right] }{\int ^{t}_{0}dt}\\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{t}\times \int ^{t}_{0}\left[ \dfrac{1}{2}\left\{ \left( 1+\cos \left( 2\omega t\right) \right) +\left( 1+\cos \left( 2\left( \omega t+\phi \right) \right) \right) \right\} \right] \\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{2t}\times \int ^{t}_{0}\left( 2+\cos 2\omega t+\cos 2\left( \omega t+\phi \right) \right) dt\\
\Rightarrow \langle E^{2}\rangle =\dfrac{E_{0}^{2}}{2t}\left[ 2t+\dfrac{\sin \omega t}{\omega }+\dfrac{\sin 2\left( \omega t+\phi \right) }{2\omega }\right] \end{aligned}$$
$$\therefore I=\dfrac{\varepsilon _{0}}{2}\langle E^{2}\rangle$$
$$\Rightarrow I=\dfrac{\varepsilon _{0}}{2}\times \dfrac{E_{0}^{2}}{2t}\left[ 2t+\dfrac{\sin \omega t}{\omega }+\dfrac{\sin ^{2}( \omega t+\phi) }{2\omega }\right]$$
But the answer is suppose to be ##\varepsilon _{0}E_{0}^{2}.## How do I get this answer?
 
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  • #2
Intensity is the power per unit area. The power from the two waves just add, irrespective of phase.
Eo means the peak electric field, so to find power we need the RMS electric field, which is Eo/√2 .
Intensity of wave 1 = ε (Eo/√2)^2 = ε Eo^2/2
Intensity of wave 2 also = ε Eo^2/2
Total intensity =2 x ε Eo^2/2 = ε Eo^2
When you calculated total intensity by adding the fields I think you might have missed the fact that Eo is the peak rather than the RMS field.
 
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  • #3
Dom Tesilbirth said:
Homework Statement:: Consider the of two coherent electromagnetic waves whose electric field vectors are given by ##\overrightarrow{E}_{1}=\widehat{i}E_{0}\cos \omega t## and ##\overrightarrow{E}_{2}=\widehat{j}E_{0}\cos \left( \omega t+\phi \right),## where ##\phi## is the phase difference. The intensity of the resulting wave is given by ##\dfrac{\varepsilon _{0}}{2}\langle E^{2}\rangle##, where ##\langle E^{2}\rangle## in the time average of ##E^{2}.## What is the total intensity?
The problem statement is wrong. The factor of 1/2 comes from the averaging process.
$$I = \langle \epsilon_0 E^2 \rangle = \frac 12 \epsilon_0 E_0^2.$$
 
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  • #4
Dom Tesilbirth said:
But the answer is suppose to be ##\varepsilon _{0}E_{0}^{2}.## How do I get this answer?
Some points to consider….

1. The intensity of a simple, sinusoidal, EM wave is the magnitude of its Poynting vector. In terms of electric field amplitude (E₀), it is ½cε₀E₀². So unless you are using natural units (taking c=1, which you haven't stated) something is wrong!

2.Presumably both waves are meant to be traveling in the same direction (say z) so the equations should be:, for example:

##\overrightarrow{E_1}(z,t)=\widehat{i}E_{0}\cos( \omega t - kz)##
##\overrightarrow{E_2}(z,t)=\widehat{j}E_{0}\cos (\omega t - kz + \phi)##

(This makes the resultant wave elliptically polarised.)

3. Since you have 2 equal amplitude waves traveling in the same direction, then, from consideration of conservation of energy, the total intensity is immediately 2 x ½cε₀E₀² = cε₀E₀². But I guess you are trying to do a formal derivation.

4. You may have made algebra errors after your integration (from t=0 to t). Check.

5. In this context, ‘mean’ values must be taken over an integer number of cycles. (The contribution from a fraction of a cycle must be ignored, which also makes sense when considering averages over long periods). The simplest approach is to take the integral over 1 cycle: t = 0 to t = T (where T = ##\frac {2 \pi}{\omega}##). And, of course, the integral of sin(anything) or cos(anything) over a full cycle is zero, which can save work and algebra errors.
 
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  • #5
Steve4Physics said:
2.Presumably both waves are meant to be traveling in the same direction (say z) so the equations should be:, for example:

##\overrightarrow{E_1}(z,t)=\widehat{i}E_{0}\cos( \omega t - kz)##
##\overrightarrow{E_2}(z,t)=\widehat{j}E_{0}\cos (\omega t - kz + \phi)##
The problem statement probably should have read "Consider the sum of two coherent electromagnetic waves whose electric field vectors at a certain point are given by..."

Using your equations, setting ##z = 0## yields the equations in the problem statement. Setting ##z## equal to some other constant should yield the same result in the end for ##I##, although the algebra will be messier.
 
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  • #6
jtbell said:
The problem statement probably should have read "Consider the sum of two coherent electromagnetic waves whose electric field vectors at a certain point are given by..."
I wish I could edit my question again. The actual question reads, “Consider the interference of two coherent electromagnetic waves...” Thank you for your reply.
 
  • #7
Fixed it for you.
 
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  • #8
jtbell said:
Using your equations, setting ##z = 0## yields the equations in the problem statement. Setting ##z## equal to some other constant should yield the same result in the end for ##I##, although the algebra will be messier.
I agree. I was (deliberately) leaving it for the OP to deduce this for themselves.
 
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1. What is meant by "interfering waves"?

Interfering waves refer to the phenomenon where two or more waves meet and interact with each other, resulting in a change in the overall intensity of the waves.

2. How is the resultant intensity of two interfering waves calculated?

The resultant intensity can be calculated by adding the individual intensities of the waves and taking into account the phase difference between them. The intensity can also be affected by factors such as the wavelength and amplitude of the waves.

3. What is the difference between constructive and destructive interference?

Constructive interference occurs when two waves with the same frequency and amplitude meet and combine, resulting in a larger overall intensity. Destructive interference, on the other hand, occurs when two waves with opposite phases cancel each other out, resulting in a decrease in intensity.

4. Can the resultant intensity of interfering waves be negative?

No, the resultant intensity of interfering waves cannot be negative. Intensity is a measure of the energy carried by a wave per unit area, and it cannot be negative. However, it can be zero in the case of complete destructive interference.

5. How does the distance between the two interfering waves affect the resultant intensity?

The distance between the two interfering waves can affect the resultant intensity in two ways. Firstly, as the distance increases, the intensity decreases due to the spreading out of the waves. Secondly, the distance can also affect the phase difference between the waves, which in turn can impact the overall intensity.

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