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(C) The intensity of an electromagnetic wave is given by $I = U_{av} c$.In terms of the electric field, the average energy density is $U_{av} = \frac{

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$1 : 1$

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(C) The intensity of an electromagnetic wave is given by $I = U_{av} c$.In terms of the electric field, the average energy density is $U_{av} = \frac{1}{2} \varepsilon_{0} E_{0}^{2}$.In terms of the magnetic field, the average energy density is $U_{av} = \frac{1}{2} \frac{B_{0}^{2}}{\mu_{0}}$.Since the relationship between the electric field amplitude $E_{0}$ and magnetic field amplitude $B_{0}$ is $E_{0} = c B_{0}$, we can substitute this into the electric field energy density expression:$U_{av} (	ext{electric}) = \frac{1}{2} \varepsilon_{0} (c B_{0})^{2} = \frac{1}{2} \varepsilon_{0} c^{2} B_{0}^{2}$.Using the relation $c^{2} = \frac{1}{\mu_{0} \varepsilon_{0}}$, we get:$U_{av} (	ext{electric}) = \frac{1}{2} \varepsilon_{0} \left( \frac{1}{\mu_{0} \varepsilon_{0}} ight) B_{0}^{2} = \frac{1}{2} \frac{B_{0}^{2}}{\mu_{0}} = U_{av} (	ext{magnetic})$.Thus, the contributions from both fields are equal, and the ratio is $1 : 1$.

The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is

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