Why is contribution to intensity of light equal for both magnetic and electric field? In chapter on waves, my sir taught us that intensity of wave is directly proportional to the square of amplitude of wave. So when we were asked what is contribution of electric field and magnetic field to intensity of light, I used the relation that amplitude of electric field is $c$ times amplitude of magnetic field. When I used this I got the contributions of magnetic field to electric field to be in ratio of $1:c^2$. But this was wrong and the answer was $1:1$. When I asked my sir the reason for this, he said both cases are different, but got angry when I asked him to elaborate. Could anyone please explain this to me?
 A: The energy per unit volume, $U_\text{B}$, in a magnetic field is given by
$$U_\text{B}=\frac{1}{2\mu_0} B^2.$$
The energy per unit volume, $U_\text{E}$ in an electric field is
$$U_\text{E}=\frac12\epsilon_0 E^2.$$
As you say, in an e-m wave,
$$E=cB.$$
Therefore
$$U_\text{E}=\frac12 \epsilon_0 c^2B^2=\frac{1}{2 \mu_0} B^2=U_B.$$
Here we have used the relationship $\mu_0 \epsilon_0=\frac{1}{c^2}$.
The mean rate of flow of energy in a plane wave per unit area of wavefront is then
$$\bar I=c\left[\frac{1}{2\mu_0}\overline{B^2} +\frac12\epsilon_0 \overline{E^2} \right].$$
The two terms in the square brackets are equal, that is the electric field energy and magnetic field energy contribute equally to the wave energy per unit area, that is to the wave intensity.
A: In addition to the other answers, here is actually a simple argument (partially hinted at in Philip's answer):
The EM wave is fundamentally an oscillation between electric and magnetic fields. In other words, the energy goes back-and-forth between the electric and the magnetic component. Therefore, the electric and magnetic energies necessarily have to be equal !
How this translates to electric field and magnetic field values depends on the unit system used.
A: The ratio of the energy densities can't be $1/c^2$, because their ratio has to be unitless.
The easiest way to see that the energy densities must be equal is by considering the case where two wave packets collide head-on, requiring that energy be conserved regardless of the polarization. This argument goes through regardless of what system of units you use.
