For the plane electromagnetic wave given by $$E = E_0 \sin(\omega t - kx)$$ and $$B = B_0 \sin(\omega t - kx)$$, the ratio of average electric energy density to average magnetic energy density is

We have a plane electromagnetic wave with $$E = E_0 \sin(\omega t - kx)$$ and $$B = B_0 \sin(\omega t - kx)$$, and we need to find the ratio of average electric energy density to average magnetic energy density.

The instantaneous electric energy density is $$u_E = \frac{1}{2}\varepsilon_0 E^2$$. Since $$E^2 = E_0^2 \sin^2(\omega t - kx)$$ and the time average of $$\sin^2$$ is $$\frac{1}{2}$$, we get:

$$\langle u_E \rangle = \frac{\varepsilon_0 E_0^2}{4}$$

Similarly, the instantaneous magnetic energy density is $$u_B = \frac{B^2}{2\mu_0}$$, so:

$$\langle u_B \rangle = \frac{B_0^2}{4\mu_0}$$

Now, for an electromagnetic wave in free space, the amplitudes are related by $$E_0 = c B_0$$, where $$c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$$. This gives $$\frac{E_0^2}{B_0^2} = c^2 = \frac{1}{\varepsilon_0 \mu_0}$$.

So the ratio of average energy densities is:

$$\frac{\langle u_E \rangle}{\langle u_B \rangle} = \frac{\varepsilon_0 E_0^2 / 4}{B_0^2 / (4\mu_0)} = \varepsilon_0 \mu_0 \cdot \frac{E_0^2}{B_0^2} = \varepsilon_0 \mu_0 \cdot \frac{1}{\varepsilon_0 \mu_0} = 1$$

This is a fundamental result: in an electromagnetic wave, the energy is shared equally between the electric and magnetic fields.

Hence, the correct answer is Option 4.