To solve this problem, we need to understand the contributions of the electric and magnetic fields to the intensity of an electromagnetic wave.$\\$The intensity $(I)$ of an electromagnetic wave is given by the Poynting vector, which is the cross product of the electric field $(E)$ and the magnetic field $(B): \\ I = \frac{1}{\mu_0} E \times B \\$The magnitude of the Poynting vector is $I = \frac{1}{\mu_0} EB \sin \theta,$ where $\theta$ is the angle between $E$ and $B. \\$For electromagnetic waves, $E$ and $B$ are perpendicular, so $\sin \theta = 1. \\$Thus, the intensity is $I = \frac{1}{\mu_0} EB. \\$The speed of light $(c)$ in a vacuum is related to $E$ and $B$ by the equation:$\\ c = \frac{E}{B}. \\$Substituting $B = \frac{E}{c}$ into the intensity equation, we get:$\\ I = \frac{1}{\mu_0} E \left(\frac{E}{c}\right) = \frac{E^2}{\mu_0 c}. \\$Similarly, the intensity can also be expressed in terms of $B: \\ I = \frac{1}{\mu_0} \left(\frac{B}{c}\right) B = \frac{B^2 c}{\mu_0}. \\$Thus, the contributions to the intensity from the electric and magnetic fields are:$\\ I_E = \frac{E^2}{\mu_0 c}, \quad I_B = \frac{B^2 c}{\mu_0}. \\$Using $c = \frac{E}{B},$ we can express $B = \frac{E}{c}. \\$Substituting $B = \frac{E}{c}$ into $I_B,$ we get:$\\ I_B = \frac{\left(\frac{E}{c}\right)^2 c}{\mu_0} = \frac{E^2}{\mu_0 c}. \\$Thus, the contributions from the electric and magnetic fields are equal:$\\ I_E = I_B. \\$Therefore, the ratio of contributions made by the electric field and magnetic field components to the intensity is $1 : 1. \\$This corresponds to Option 1.