Magnitude and average value of energy intensity of light beam

In summary: Ax was a typo, it should have been the unit vector ##\hat{x}##. And I'll keep that in mind. Thanks again for all your help.In summary, the electric and magnetic fields in an electromagnetic wave vary sinusoidally and reach their extrema at the same points in time. The electric field vector varies only in the x-y plane while the magnetic field vector varies only in the z-y plane. The magnitude of the Poynting vector, which represents the rate of energy transfer in the wave, fluctuates at twice the frequency of the wave and is proportional to the product of the maximum electric and magnetic field strengths. The average value of the Poynting vector is half of the instantaneous peak value.
  • #1
dl447342
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Homework Statement
Explain why the magnitude of ##\vec{S}## , denoted ##S##, fluctuates at twice the frequency of the wave. Explain why the average value of ##S## is half the instantaneous peak value.
Relevant Equations
I know ##\vec{S} = \frac{1}{\mu_0} \vec{E}\times \vec{B},## where ##\mu_0## is the magnetic permeability of the vacuum, and the other two variables represent the electric field vector and magnetic field vector. Also, ##c = \lambda f,## where ##c## is the speed of light, ##\lambda## is the wavelength, and ##f## is the frequency, and the momentum of a photon is ##\frac{h}{\lambda}##, where ##h## is Planck's constant.
I tried using the equations above, but I wasn't really able to come up with an intuitive explanation. From my understanding, the electric field vector only varies in the x-y plane while the magnetic field vector only varies in the z-y plane. Also, both vary sinusoidally and both reach extrema (peaks or troughs) at the same points in time.
 
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  • #2
Herer is the short answer $$2\cos^2 \theta =1+ \cos 2\theta$$
 
  • #3
hutchphd said:
Herer is the short answer $$2\cos^2 \theta =1+ \cos 2\theta$$
Why is that the answer? I know that's true from basic trignometric identities, but what is ##\theta##?

Could you provide some web links so I can learn more about this concept?
 
  • #4
This is because E and B both contain ##\cos ## and so the cross-product contains a ##\cos^2##.
##\theta=2\pi\frac {(x-ct)}\lambda##
Does that help?
 
  • #5
dl447342 said:
Homework Statement:: Explain why the magnitude of ##\vec{S}## , denoted ##S##, fluctuates at twice the frequency of the wave. Explain why the average value of ##S## is half the instantaneous peak value.
The statements you are asked to explain apply to sinusoidal waves. They may apply to some other waveforms too, but I guess whoever set the question intended you to assume that the field strengths vary as, say, sin(ωt ) (or as sin(ωt + φ) if you want to be a bit more general).

Ideally, there should have been some mention of 'sinusoidal' in the question,
 
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  • #6
Steve4Physics said:
The statements you are asked to explain apply to sinusoidal waves. They may apply to some other waveforms too, but I guess whoever set the question intended you to assume that the field strengths vary as, say, sin(ωt ) (or as sin(ωt + φ) if you want to be a bit more general).

Ideally, there should have been some mention of 'sinusoidal' in the question,
Okay. I think the waves have particular formulas that would make the computation of the magnitude of ##\vec{S}## easier.
 
  • #7
So can you write an explicit equation for the magnitude of S ??
 
  • #8
hutchphd said:
So can you write an explicit equation for the magnitude of S ??
The formula for ##\vec{E}## is ##\vec{E} = \hat{j} E_{max} \cos(kx - \omega t)## and the formula for ##\vec{B}## is ##\vec{B} = \hat{k} B_{max} \cos(kx - \omega t)##, where ##\omega## is the angular frequency and ##k## is the wave number, equal to ##2\pi## divided by the wavelength, ##t## represents time and ##x## represents the positive with respect to the origin along the x-axis.
Using the right-hand rule, where ##\vec{B}## is assumed to be along the z-axis while ##\vec{E}## is assumed to be on the y-axis), I get that ##\vec{S}## should point in the direction of the positive x-axis. So for an explicit equation, I was thinking of something like ## \frac{E_{max}B_{max}}{2 \mu_0} (1+\cos 2(kx - \omega t))\hat{i}## where ##\hat{i}## is the unit vector in the direction of the x-axis.

##\omega = 2\pi f## where f is the frequency of "each electromagnetic wave" and so the magnitude of ##S## fluctuates at twice this value with respect to time as it has a factor of ##2## in ##\cos 2(kx - \omega t)##.
 
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  • #9
Haven't checked the details but looks good. The working you hand-in should show the use of the trig' identity you used: cos²x = (1+cos(2x))/2.

Really there's no need to consider the vector aspects for the question as posed. The directions don't affect the answers.

To answer the part about the average being half the peak, you might want to consider S expressed in terms of cos²(...). Do you know what the graph y = cos²x looks like?
 
  • #10
This looks good except you have the ##\vec B ## in the ##\hat x## direction. It should be in ##\hat z ## then ##\hat y \times \hat z=\hat x## which is what you want for ##\hat S##. Also what is the "ax" the expressions?
In general I prefer ##\hat x, \hat y, \hat z ## to i,j,k mostly because k is the wavevector also
 
  • #11
hutchphd said:
This looks good except you have the ##\vec B ## in the ##\hat x## direction. It should be in ##\hat z ## then ##\hat y \times \hat z=\hat x## which is what you want for ##\hat S##. Also what is the "ax" the expressions?
In general I prefer ##\hat x, \hat y, \hat z ## to i,j,k mostly because k is the wavevector also
Thanks. I fixed it.
 
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FAQ: Magnitude and average value of energy intensity of light beam

What is the difference between magnitude and average value of energy intensity of light beam?

The magnitude of energy intensity of a light beam refers to the maximum value of energy per unit area at any given point in the beam. It is a measure of the strength or brightness of the beam. On the other hand, the average value of energy intensity is the average of the energy per unit area over the entire beam, and it takes into account the variations in intensity throughout the beam.

How is the magnitude of energy intensity of a light beam measured?

The magnitude of energy intensity of a light beam is typically measured using a photometer, which is a device that measures the amount of light energy per unit area. The photometer is placed at a specific distance from the light source and measures the intensity of the light at that point.

What factors affect the magnitude and average value of energy intensity of a light beam?

The magnitude and average value of energy intensity of a light beam can be affected by several factors, including the power of the light source, the distance from the light source, the wavelength of the light, and any obstructions or filters in the path of the light beam.

How does the magnitude and average value of energy intensity of a light beam impact its applications?

The magnitude and average value of energy intensity of a light beam are important considerations in various applications, such as laser technology, medical imaging, and optical communications. A higher magnitude of energy intensity is often desirable for more powerful and efficient performance, while a consistent average value is important for accurate measurements and data transmission.

Can the magnitude and average value of energy intensity of a light beam be manipulated?

Yes, the magnitude and average value of energy intensity of a light beam can be manipulated through various means, such as using lenses, filters, or mirrors to control the direction and concentration of the beam. Additionally, the power and wavelength of the light source can also be adjusted to change the magnitude and average value of energy intensity.

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