Radiant power of a series of light waves

In summary, E is a quantity that represents the superposition of multiple reflections of a wave, and is given in watts. Units check out nicely when integrated, but when squared the answer is a huge number in watts^2. To get the absolute value of the intensity, you need to know the amplitude of the incident wave.
  • #1
Ngineer
64
1
Hello everyone,

I need to calculate the radiant power of an interference pattern of a series of light wave reflections. I need a value in Watts that would plug in nicely into a photodetector's responsivity function (given in Amps/Watts) and thus giving me an estimation of the output current.

I thought the formula would look something like this:

upload_2016-1-18_15-28-4.png

T is the period (all waves have the same period but are at different phases.)
E|x=0 is the superposition of all the waves (interference pattern) evaluated at x=0

To me, this formula makes the most sense, as it gives a a result in Watts.
However, my supervisor suggests that I square that value. Doesn't this give a result in Watts2?

Thank you for your help!
 
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  • #2
What is E and what are the units for this quantity?
 
  • #3
Hi nasu,

E is the sum (superposition) of all the reflections, in other words, it is the interference of multiple light waves having the same frequency/wavelength but differing in phase.

Since it represents the superposition of multiple light waves, my assumption was that it is given in watts (radiant power).
 
  • #4
If E were indeed the radiant power and you want the radiant power, what is the point of integrating?
Where do you get this E from? You measure it? You calculate it theoretically? How do you represent one of the individual waves in the superposition? What equation?

Usually "E" is used to represent the electric field associated with the wave. I suppose your supervisor assumed that this is the case.
The intensity of the wave is proportional to the square of the amplitude of the wave (electric filed in case of EM wave).
 
  • #5
Hi nasu,

If E were indeed the radiant power and you want the radiant power, what is the point of integrating?
Integrating over one period and then dividing by the period to get the average radiant power.

Where do you get this E from? You measure it? You calculate it theoretically?
Theoretically.
upload_2016-1-18_18-35-7.png

R1 is the reflection coefficient of a mirror.
ΔΦ is the phase shift corresponding to a distance between two mirrors, given by 2kd.
d1,d2 are phase shifts inherent to the reflection process. I'm assuming them to be zero for now and thus I get:
upload_2016-1-18_18-33-6.png


upload_2016-1-18_18-46-11.png


How do you represent one of the individual waves in the superposition? What equation?
upload_2016-1-18_18-38-10.png


Usually "E" is used to represent the electric field associated with the wave. I suppose your supervisor assumed that this is the case.
The intensity of the wave is proportional to the square of the amplitude of the wave (electric filed in case of EM wave).

That might have been the case. However, when I evaluate my formula (without the square) units check out nicely but I get the answer in a huge square root. With the square,the answer is a nice polynomial of R1, but again units will not checkout.
 
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  • #6
According to your formula, E has no dimensions. The reflection coefficients are just ratios between amplitudes or intensities. This is, unless you use a different definition of these Ri. So you won't get neither watt nor watt^2, no matter what you integrate (E or E^2).
In order to find either amplitude or intensity in absolute values you need to know the amplitude of intensity or the incident wave. What you have there, assuming it is correct, is just a relative quantity, the fraction of the incident wave that is reflected and so on. I am not sure what your system is, it seems that you have multiple reflections.
It looks like the sum of relative amplitudes, in the approximation of plane waves. If you multiply this by the amplitude of the incident wave you will get a total amplitude. To get the intensity you need to take the square of the absolute value and multiply by whatever constant is required by the system of units you are using.

In SI, the average intensity is given by
## I=\frac{\epsilon_0 c E^2}{2 } ##
where E is the amplitude of the wave (maximum electric field)., c is the speed of light and epsilon_0 is permitivity of vacuum. This include the average over a period of the squared amplitude (see the 1/2 factor).This will give you intensity, i.e. watts per unit area. If you want total power you need to multiply by the cross sectional area of the beam. But you assume plane waves, not a finite beam.
 
Last edited:
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  • #7
Thank you. I actually forgot to mention that I have a quantity Plaser in mW corresponding to the radiant power of the input beam multiplied by the whole definition.

Thanks again.
 

Related to Radiant power of a series of light waves

What is radiant power?

Radiant power refers to the amount of energy that is transmitted through electromagnetic radiation, such as light waves, per unit time. It is often measured in watts (W).

How is the radiant power of a series of light waves calculated?

The radiant power of a series of light waves can be calculated by multiplying the intensity of the light (measured in watts per square meter) by the surface area that the light is being emitted on.

What factors can affect the radiant power of a series of light waves?

The radiant power of a series of light waves can be affected by factors such as the intensity and wavelength of the light, the distance the light travels, and the medium through which the light is passing.

What are some common applications of measuring radiant power?

Measuring radiant power is important in a variety of fields, including astronomy, physics, and engineering. It can be used to determine the energy output of stars, the efficiency of solar panels, and the power consumption of electronic devices.

How does the radiant power of light waves impact our daily lives?

The radiant power of light waves is crucial for our daily lives as it allows us to see and communicate through various technologies such as light bulbs, lasers, and fiber optics. It also plays a role in many natural processes, such as photosynthesis in plants.

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