In an electromagnetic wave, the ratio of energy densities of electric and magnetic fields is _______.

The correct option is 1 : 1.

To understand why this is the correct ratio, we can look at the expressions for the energy density of the electric field and the magnetic field in an electromagnetic wave.

The energy density associated with the electric field ($u_E$) is given by:

$u_E=\frac{1}{2}{\epsilon }_0{E}^2$

where ${\epsilon }_0$ is the permittivity of free space and $E$ is the electric field strength.

Similarly, the energy density associated with the magnetic field ($u_B$) is given by:

$u_B=\frac{B^2}{2{\mu }_0}$

where ${\mu }_0$ is the permeability of free space and $B$ is the magnetic field strength.

For an electromagnetic wave propagating in vacuum, the relationship between the electric field and magnetic field amplitudes at any point is given by:

$E=cB$

where $c$ is the speed of light in vacuum. The speed of light is related to the electromagnetic properties of free space by the relation:

$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\Rightarrow c^2=\frac{1}{{\mu }_0{\epsilon }_0}$

Now, let's substitute $E=cB$ into the expression for the electric energy density:

$u_E=\frac{1}{2}{\epsilon }_0{\left(cB\right)}^2=\frac{1}{2}{\epsilon }_0{c}^2{B}^2$

Substituting $c^2=\frac{1}{{\mu }_0{\epsilon }_0}$ into this equation gives:

$u_E=\frac{1}{2}{\epsilon }_0(\frac{1}{{\mu }_0{\epsilon }_0}{B}^2$

Simplifying the expression, the permittivity ${\epsilon }_0$ terms cancel out:

$u_E=\frac{B^2}{2{\mu }_0}$

This shows that the electric energy density is exactly equal to the magnetic energy density:

$u_E=u_B$

Therefore, the ratio of the energy density of the electric field to the magnetic field is:

$\frac{u_E}{u_B}$ = 1

Hence, the ratio is 1 : 1.