Question

A light wave is incident normally on a glass slab of refractive index $1.5$. If $4\% $ of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be:
A) 24 V/m
B) 30 V/m
C) 6 V/m
D) 10 V/m

Answer 1

Hint: To solve this question, we have used the relation of intensity of Electromagnetic waves with the electric field. We can easily calculate the value of amplitude of the electric field for the propagating in the glass medium using this formula by comparing the intensity of the incident wave and the reflected wave.

Formulae used:
$I = \dfrac{1}{2}\varepsilon {E^2}v$
Here $I$ is the intensity of the wave in air , $\varepsilon $ is the permittivity of medium, $E$ is the amplitude of electric field in the medium and $v$ is the speed of the wave in the medium.

Complete step by step answer:
As the light ray is coming from the air, its intensity will be
$I = \dfrac{1}{2}{\varepsilon _0}{E_0}^2c$........................(1)
where $I$ is the intensity of the wave in air , ${\varepsilon _0}$is the permittivity of air, ${E_0}$ is the amplitude of electric field in air and $c$ is the speed of the wave in the air.
In the glass medium, the intensity of the light ray will be,
${I'} = \dfrac{1}{2}\varepsilon {E^2}v$.........................(2)
Here ${I'}$ is the intensity of the wave in air , $\varepsilon$ is the permittivity of medium, $E$ is the amplitude of electric field in the medium and $v$ is the speed of the wave in the medium.
We know that
$\varepsilon = {\varepsilon _0}{\varepsilon _r}$
Where ${\varepsilon _0}$ is the permittivity of air, $\varepsilon $ is the permittivity of medium and ${\varepsilon _r}$ is the relative permittivity of medium.
$ \Rightarrow {I'} = \dfrac{1}{2}{\varepsilon _0}{\varepsilon _r}{E^2}v$...........................(3)
Also in the question it's given that $4\% $of the light is reflected. So remaining intensity is absorbed by the slab.
$ \Rightarrow = \dfrac{{96}}{{100}}I$........................(4)
Substituting the values of ${I'}$ and $I$ in equation 4, we get
$ \Rightarrow \dfrac{1}{2}{\varepsilon _0}{\varepsilon _r}{E^2}v = \dfrac{{96}}{{100}} \times \dfrac{1}{2}{\varepsilon _0}{E_0}^2c$
$ \Rightarrow {\varepsilon _r}{E^2}v = \dfrac{{96}}{{100}} \times {E_0}^2c$
$ \Rightarrow E = \sqrt {\left( {\dfrac{{96}}{{100}} \times {E_0}^2\dfrac{c}{{{\varepsilon _r}v}}} \right)} $
Putting the value of ${E_0} = 30$, $\dfrac{c}{v} = \mu $ and $\sqrt {{\varepsilon _r}} = \mu $ we get
$ \Rightarrow$ E = 24 V/m.

So option (A) is the answer.

Note: While calculating the value of the amplitude of the electric field we should be careful about the values of the permittivity. The permittivity value of the medium in which the wave is travelling should be taken otherwise we can get incorrect answers. Also the refractive index of the medium in which the wave is travelling should be taken.