Question
The energy contained in a small volume through which an electromagnetic wave is passing oscillates with:
  1. Zero frequency.
  2. The frequency of the wave.
  3. Half the frequency of the wave.
  4. Double the frequency of the wave.

Answer

  1. Double the frequency of the wave.
Explanation:
The energy per unit volume of an electromagnetic wave,
$\text{u}=\frac{1}{2}\in_0\text{E}^2+\frac{\text{B}^2}{2\mu_0}$
The energy of the given volume can be calculated by multiplying the volume with the above expression.
$\text{U}=\text{u}\times\text{V}=\Big(\frac{1}{2}\in_0\text{E}^2+\frac{\text{B}^2}{2\mu_0}\Big)\times\text{V}\ ....(\text{i})$
Let the direction of propagation of the electromagnetic wave be along the z-axls. Then, the electric and magnetic fields at a particular point are given by,
$\text{E}_\text{x}=\text{E}_0\sin(\text{kz}-\omega\text{t})$
$\text{B}_\text{y}=\text{B}_0\sin(\text{kz}-\omega\text{t})$
Substituting the values of electric and magnetic fields in (1) we get,
$\text{U}=\Big(\frac{1}{2}\in_0\big(\text{E}_0^2\sin^2(\text{kz}-\omega\text{t})+\frac{\text{B}^2_0\sin^2(\text{kz}-\omega\text{t})}{2\mu_0}\Big)\times\text{V}$
$\text{U}=\Big(\in_0\text{E}^2_0\frac{(1-\cos(2\text{kz}-2\omega\text{t}))}{4}+\frac{\text{B}_0^2(1-\cos(2\text{kz}-2\omega\text{t}))}{4\mu_0}\Big)\times\text{V}$
From the above expression, it can be easily understood that the energy of the electric and magnetic fields have angular frequency $2\omega$ Thus. the frequency of the energy of the electromagnetic wave will also be double.

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