Intensity of light

Question

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?

Answer 1

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

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

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 $\sin^2(\mathrm{anything})$ is a half, so we get:

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

and the average energy transport is independent 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.

Answer 2

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 $$n_\nu = \frac{P \cdot \Delta t}{ E_{\nu}}.$$

Answer 3

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..?

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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...