Question

An electromagnetic wave of intensity $50W{m}^2$ enters in a medium of refractive index ‘n’ without any loss. The ratio of the magnitudes of the electric fields, and the ratio of the magnitudes of the magnetic fields of the wave before and after entering into the medium are respectively, given by:
(A) $\left({\displaystyle \frac{1}{\sqrt{n}}},{\displaystyle \frac{1}{\sqrt{n}}}\right)$
(B) $\left(\sqrt{n},{\displaystyle \frac{1}{\sqrt{n}}}\right)$
(C) $\left(\sqrt{n},\sqrt{n}\right)$
(D) $\left({\displaystyle \frac{1}{\sqrt{n}}},\sqrt{n}\right)$

Answer 1

Hint:In order to solve this question,we are going to use the relations for the speed of light in vacuum and the speed of light in a medium of dielectric constant k and how they are related to electric field and magnetic field.After putting all the values of all the quantities that is given in question we can arrive at correct answer.

Complete step by step answer:
We know that speed of light in vacuum $c={\displaystyle \frac{1}{\sqrt{{\mu }_0\times {\in }_0}}}$ , where $${\in }_0$$ is the absolute electrical permittivity of the free space and $${\mu }_0$$ is the absolute permeability of free space.Velocity of light(v) in a medium is given by
$v={\displaystyle \frac{1}{\sqrt{k{\mu }_0\times {\in }_0}}}$,
where k is the dielectric constant of the medium.
Also the refractive index of a medium is defined as the velocity of light in vacuum to the velocity of light in the medium.
$${\displaystyle \frac{c}{v}}=n$$
Dividing above equations we get,
${\displaystyle \frac{c}{v}}={\displaystyle \frac{{\displaystyle \frac{1}{\sqrt{{\mu }_0\times {\in }_0}}}}{{\displaystyle \frac{1}{\sqrt{k{\mu }_0\times {\in }_0}}}}}\Rightarrow n=\sqrt{k}$
We know that for a transparent medium $${\mu }_r\approx {\mu }_0$$
Intensity of electromagnetic waves in vacuum = ${\displaystyle \frac{1}{2}}{\in }_0{E}_0^{2}c$
Intensity of electromagnetic waves in a medium = ${\displaystyle \frac{1}{2}}{\in }_{0}k{E}^{2}v$
Equating both the equations we get, $E_0^{2}c=k{E}^{2}v$
$\Rightarrow {\displaystyle \frac{E_0^2}{E^2}}={\displaystyle \frac{kv}{c}}={\displaystyle \frac{n^2}{n}}\Rightarrow {\displaystyle \frac{E_0}{E}}=\sqrt{n}$
Similarly, we can find the speed of light in the magnetic field region.
$$\Rightarrow {\displaystyle \frac{B_0^{2}c}{2{\mu }_0}}={\displaystyle \frac{B^{2}v}{2{\mu }_0}}\Rightarrow {\displaystyle \frac{B_0}{B}}={\displaystyle \frac{1}{\sqrt{n}}}$$

Hence,option (B) is the correct answer.

Note:The product of absolute permittivity and absolute permeability of space is related to the velocity of light. This relation forms the basis of the theory of relativity. . It is the maximum speed in space by which a body can travel. No body in space can travel with a velocity greater than the velocity of light. Speed of light always decreases when it enters a medium other than the vacuum. This ratio of the speed of light in vacuum to the speed of light in that medium is known as the refractive index of that medium.