Question

For an EM Wave, E = ${E_0}\sin 12 \times {10^6}\left[ {Z - 2 \times {{10}^8}t} \right]\dfrac{N}{C}$ in a medium, then what is its refractive index?
A) $\dfrac{2}{3}$
B) $\dfrac{3}{2}$
C) $\dfrac{4}{3}$
D) $\dfrac{5}{3}$

Answer 1

The equation of an EM Wave in a medium is $E = {E_0}\sin \left( {kz - wt} \right)$. The velocity of light in the medium is given by v = $\dfrac{w}{k}$. The refractive index of a medium is the ratio of the speed of light in vacuum with the speed of light in the medium.

Complete step-by-step solution:
The equation of the given EM Wave is $E = {E_0}\sin 12 \times {10^6}\left[ {Z - 2 \times {{10}^8}t} \right]\dfrac{N}{C}$. We first simplify the the given equation, we obtain,
$E = {E_0}\sin \left[ {12 \times {{10}^6}Z - 24 \times {{10}^{14}}t} \right]\dfrac{N}{C}$ …equation (1)
We now compare equation (1) with the general equation of the EM Wave, which is, $E = {E_0}\sin \left( {kz - wt} \right)\dfrac{N}{C}$, we observe that,
$k = 12 \times {10^6}$ …equation (2)
$w = 24 \times {10^{14}}$ …equation (3)
The velocity of light in the given medium is the ratio of w with k. We represent the velocity of light in the medium as v. On substituting the values obtained in equation (2) and (3), we get,
$v = \dfrac{w}{k} = \dfrac{{24 \times {{10}^{14}}}}{{12 \times {{10}^6}}} = 2 \times {10^8}$ m/s
The velocity of light in vacuum is $c = 3 \times {10^8}$ m/s. The refractive index, $\mu$, of the medium can be found out by dividing the velocity of light in the vacuum by the velocity of light in the medium. We get,
$\mu = \dfrac{c}{v} = \dfrac{{3 \times {{10}^8}}}{{2 \times {{10}^8}}} = \dfrac{3}{2}$
Hence, the refractive index of the medium is $\dfrac{3}{2}$.

Therefore, the correct answer of the question is option B.

Note: The speed of the EM wave in a medium depends on the refractive index of the medium. The speed of EM Wave in a medium is inversely proportional to the refractive index of the medium. The higher the refractive index of the medium, slower the speed of light in the medium.