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Step-by-Step Solution
Step 1: Identify the given electric fields
In air, the electric field is given by
$ \overrightarrow{E_1} = E_{01}\,\hat{x}\,\cos \left[2 \pi \nu \left(\frac{z}{c} - t\right)\right]. $
In the medium, the electric field is given by
$ \overrightarrow{E_2} = E_{02}\,\hat{x}\,\cos \left[k \left(2z - c t\right)\right]. $
Here, $k$ and $\nu$ are the wave number and frequency as they would be measured in air. We also note that the medium is non-magnetic, i.e., its permeability $ \mu $ is the same as in air.
Step 2: Determine the wave velocity in air
From the argument of the cosine function in air,
$ 2 \pi \nu \left(\frac{z}{c} - t\right), $
the wave velocity in air is $c$, which is consistent with the known speed of light in vacuum/air.
Also, the speed of an electromagnetic wave in air can be expressed as
$
c = \frac{1}{\sqrt{\mu\,\varepsilon_{r_1}\,\varepsilon_0}}.
$
(Here, $ \varepsilon_{r_1} $ is the relative permittivity of air and $ \varepsilon_0 $ is the permittivity of free space.)
Step 3: Determine the wave velocity in the medium
From the argument of the cosine function in the medium,
$ k \left(2z - c t\right), $
notice the factor of 2 with $z$ and 1 with $ct$. We can rewrite this term to see how $z$ and $t$ change with respect to the phase:
$ k \left(2z - ct\right) = 2\,k \left(z - \frac{c}{2}t\right). $
Hence, the effective speed of the wave in the medium appears to be $ \frac{c}{2} $. In a non-magnetic medium, the speed of an electromagnetic wave is
$
\frac{c}{2} = \frac{1}{\sqrt{\mu\,\varepsilon_{r_2}\,\varepsilon_0}}.
$
(Here, $\varepsilon_{r_2}$ is the relative permittivity of the medium.)
Step 4: Relate the two velocities to the relative permittivities
We have the two key expressions:
1. $ c = \frac{1}{\sqrt{\mu\,\varepsilon_{r_1}\,\varepsilon_0}}, $
2. $ \frac{c}{2} = \frac{1}{\sqrt{\mu\,\varepsilon_{r_2}\,\varepsilon_0}}. $
Divide the first by the second:
$
\frac{c}{\,c/2\,}
= \frac{\frac{1}{\sqrt{\mu\,\varepsilon_{r_1}\,\varepsilon_0}}}{\frac{1}{\sqrt{\mu\,\varepsilon_{r_2}\,\varepsilon_0}}}
\quad \Longrightarrow \quad
2 = \sqrt{\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}}}.
$
Therefore,
$
\sqrt{\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}}} = 2
\quad \Longrightarrow \quad
\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}} = 4
\quad \Longrightarrow \quad
\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac{1}{4}.
$
Step 5: State the final result
Thus, the correct ratio of the relative permittivities is
$
\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac{1}{4}.
$
Hence, the correct option is
$
\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac{1}{4}.
$
