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If we have 2 beams of light with equal intensities, but with different frequencies, wouldn't the one with the higher frequency generate more power?

If so, how come the intensity, which is in $W/m^2$, of the higher frequency beam is not higher?

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What do you mean by "generate more power"? –  Ebenezer Sklivvze Jan 12 '13 at 10:58
    
FYI, intensity is a poor word to use nowadays, $\frac{W}{m^2}$ is either irradiance or radiant exitance, depending on whether it is coming or going respectively. Intensity has become ambiguous between closely related fields, so people are attempting to drop its usage –  daaxix Jan 12 '13 at 16:29

3 Answers 3

If you have some electromagnetic wave e.g. a plane ave:

$$ E = E_m sin(kx - \omega t) $$

then the energy transport is given by the Poynting vector. For the plane wave above the energy transport works out to be:

$$ S = \frac{1}{c\mu_0} E_m^2 sin^2(kx - \omega t) $$

To calculate average energy transport we note that the average value of sin$^2$(anything) is a half, so we get:

$$ S_{av} = \frac{1}{2c\mu_0} E_m^2 $$

and the average energy transport is independant of the frequency.

As others have suggested, you may be getting confused by the fact that the energy of a photon is proportional to its frequency, $E = h\nu$. For your higher frequency light the energy of the photons will indeed be greater, but there will be fewer photons so the average energy transport works out the same as the lower frequency light.

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The light intensity $I$ is the power $P$, that irradiates an area $A$, $I=P/A$, therefore the units $\mathrm{W}\mathrm{m}^{-2}$. This means, that if you have two beams with equal intensities, the area is the same and the power has to be the same as well.

You are probably confused by the relation between frequency and energy of a photon, $E_\nu = h\nu$. But this says just, what the energy of a single photon is. When you have two beams with equal intensities and different frequencies, the one with higher frequency just contains less photons and the overall power is the same.

The number of photons $n_\nu$ with frequency $\nu$ radiating on the surface in a given time intervall $\Delta t$ is $$n_\nu = \frac{P \cdot \Delta t}{ E_{\nu}}.$$

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Because the frequency and intensity are not related to that extent (by which you got yourself confused). Frequency gives the number of vibrations per second (and so, the unit $s^{-1}$). As @daaxix says, you have used intensity to mean irradiance. Irradiance is the amount of power the wave has delivered per unit area and hence, $W/m^2$. Both are quite similar in their definitions.

Or in other words, the amplitude of the wave determines the irradiance. When you change the frequency of the wave (the vibrations), it is not mandatory for the amplitude to change.

Think it this way. Consider an ordinary light source emitting light of wavelength ($550\ \text{nm}$). This would be green. It is just similar to a torch light. Consider a LASER emitting the same wavelength $550\ \text{nm}$. This is also green. What if you have a look at both lights..?


Now, We could take a look at the other way round. In terms of particles, EM waves are considered to have photons. If the photons have more energy, the frequency is more. If the number of photons emitted from the source is more, it has higher irradiance. That's all...

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I don't like the comparison of a laser (plane wave/ gauss wave) and an ordinary torch light (spherical wave). One could explain the difference in intensity just be geometrical considerations. –  elcojon Jan 12 '13 at 14:19
    
@elcojon: Hi Elcojon. I thought that we don't necessarily have to consider plane/spherical wave-fronts for differentiating a basic property, taking the OP into a whole new concept. Your quote: "I don't like the....." Whether you don't like it or not, I don't think it's wrong :-) –  Waffle's Crazy Peanut Jan 12 '13 at 14:40
    
@elcojon, if you add angle, which would differ between plane waves and spherical waves, then you have radiance, with units $\frac{W}{m^2 sr}$, where $sr$ is steradians... –  daaxix Jan 12 '13 at 16:32
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@CrazyBuddy you should change intensity to irradiance above, intensity has essentially been deprecated in the active physics community because there are ambiguous definitions across closely related fields... –  daaxix Jan 12 '13 at 16:35

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