Question

An EM wave from air enters a medium. The electric fields are ${\vec E_1} = {E_{01}}\hat x\cos [2\pi \nu (\dfrac{z}{c} - t)]$ in air and ${\vec E_2} = {E_{02}}\hat x\cos [k(2z - ct)]$ in medium, where the wave number $k$and frequency $\nu$ refer to their values in air. The medium is non-magnetic. If ${\varepsilon _{r1}}$and ${\varepsilon _{r2}}$refer to relative permittivity of air and medium respectively, which of the following options is correct?
A) $\dfrac{{{\varepsilon _{r1}}}}{{{\varepsilon _{r2}}}} = \dfrac{1}{4}$
B) $\dfrac{{{\varepsilon _{r1}}}}{{{\varepsilon _{r2}}}} = \dfrac{1}{2}$
C) $\dfrac{{{\varepsilon _{r1}}}}{{{\varepsilon _{r2}}}} = 4$
D) $\dfrac{{{\varepsilon _{r1}}}}{{{\varepsilon _{r2}}}} = 2$

Answer 1

We can easily solve the question if we know the standard equation of a wave in any medium i.e. ${\vec E_{}} = {E_0}\cos [kz - \omega t)]$ . We can compare the values given in the question with the one in the standard equation and find out the velocity of the waves in air and in the given medium. We can easily use the velocities to find the ratio of relative permittivity of air to that of the given medium if we remember Snell’s law.

Formula Used:
$v = \dfrac{\omega }{k}$
Where $v$is the linear velocity of the wave, $\omega$ is the angular velocity of the wave and $k$is the wavenumber of the wave.
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{{v_2}}}{{{v_1}}}$
This is Snell's law. Here ${n_1}$ and ${n_{_2}}$are the refractive index of the mediums 1 and 2 respectively and ${v_1}$and ${v_2}$ are the speed of the light in the medium 1 and 2 respectively.

Complete step by step answer:
The standard equation of any wave in any medium is given by;
${\vec E_{}} = {E_0}\cos [kz - \omega t)]$
By comparing the equation of the wave in air i.e., ${\vec E_1} = {E_{01}}\hat x\cos [2\pi \nu (\dfrac{z}{c} - t)]$ with the standard equation of waves, we get
$k = \dfrac{{2\pi \nu }}{c}$ and $\omega = 2\pi \nu$.
Where $k$ is the wave number of the wave, $\nu$is the frequency of the wave, $c$is the speed of light and $\omega$ is the angular velocity of the wave.
$\Rightarrow {v_{air}} = \dfrac{\omega }{k} = c$
By comparing the equation of the wave in the given medium ${\vec E_2} = {E_{02}}\hat x\cos [k(2z - ct)]$ with the standard equation of waves, we get
$k = 2k$ And $\omega = kc$
$\Rightarrow {v_{medium}} = \dfrac{\omega }{k} = \dfrac{c}{2}$
Using Snell’s law,
$\dfrac{{{n_{air}}}}{{{n_{medium}}}} = \dfrac{{{v_{medium}}}}{{{v_{air}}}}$
Where ${n_{air}}$ is the refractive index of air, ${n_{medium}}$ is the refractive index of given medium, ${v_{air}}$is the speed of wave in air and ${v_{medium}}$is the speed is wave in the given medium.
$\Rightarrow \dfrac{{{n_{air}}}}{{{n_{medium}}}} = \dfrac{1}{2}$
We know that the permittivity of a medium $\varepsilon$is directly proportional to the square of refractive index of the medium${n^2}$,
$\dfrac{{{\varepsilon _{r1}}}}{{{\varepsilon _{r2}}}} = \dfrac{{{n_1}^2}}{{{n_2}^2}}$
$\Rightarrow \dfrac{{{\varepsilon _{r1}}}}{{{\varepsilon _{r2}}}} = \dfrac{1}{4}$

Hence, the correct option is (A).

Note: While comparing any wave equation with the standard wave equation we should be careful with the angular velocity part and the wave number part. The comparison should be done only after proper rearrangement. If proper arrangement will not be there we can get errors in the result. Also we should be careful while writing the velocity and refractive index equation.