Modern Physics 7 Question 21
22. An electromagnetic wave of intensity $50 Wm^{-2}$ enters in a medium of refractive index ’ $n$ ’ without any loss. The ratio of the magnitudes of electric fields and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by
(a) $\frac{1}{\sqrt{n}}, \sqrt{n}$
(b) $(\sqrt{n}, \sqrt{n})$
(c) $\frac{1}{\sqrt{n}}, \frac{1}{\sqrt{n}}$
(d) $\sqrt{n}, \frac{1}{\sqrt{n}}$
(Main 2019, 11 Jan I)
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Solution:
- In the free space, the speed of electromagnetic wave is given as,
$$ c=\frac{1}{\sqrt{\mu _0 \varepsilon _0}}=\frac{E _0}{B _0} $$
where, $E _0$ and $B _0$ are the amplitudes of varying electric and magnetic fields, respectively.
Now, when it enters in a medium of refractive index ’ $n$ ‘, its speed is given as,
$$ v=\frac{1}{\sqrt{\mu \varepsilon}}=\frac{1}{\sqrt{K \varepsilon _0 \mu}}=\frac{E}{B} $$
where, $K$ is dielectric strength of the medium.
Using Eqs. (i) and (ii), we get
$$ \frac{v}{c}=\frac{1}{\sqrt{K}} $$
$\left(\because\right.$ For a transparent medium, $\left.\mu _0 \approx \mu\right)$
Also, refractive index of medium is ’ $n$ ’ and is given as
$$ \frac{c}{v}=n \text { or } \frac{v}{c}=\frac{1}{n} $$
$\therefore$ From Eqs. (iii) and (iv), we get
$$ n=\sqrt{K} \text { or } K=n^{2} $$
The intensity of a EM wave is given as,
$$ I=\frac{1}{2} \varepsilon _0 E _0^{2} c $$
and in the medium, it is given as $I^{\prime}=\frac{1}{2} K \varepsilon _0 E^{2} v$
It is given that, $I=I^{\prime}$
$\Rightarrow \quad \frac{1}{2} \varepsilon _0 E _0^{2} c=\frac{1}{2} K \varepsilon _0 E^{2} v$ or $\frac{E _0}{E} \quad{ }^{2}=\frac{K v}{c}$
From Eqs. (iv), (v) and (vi), we get
$$ \frac{E _0}{E}{ }^{2}=\frac{n^{2}}{n}=n \quad \text { or } \quad E _0 / E=\sqrt{n} $$
Similarly, $\frac{1}{2} \cdot \frac{B _0^{2}}{\mu _0} c=\frac{1}{2} \cdot \frac{B^{2}}{\mu _0} v \Rightarrow \frac{B _0}{B}=\frac{1}{\sqrt{n}}$
Alternate method
We know that,
$$ \frac{E _0}{B _0}=c \text { air/vacuum } \quad \frac{E}{B} \underset{\text { medium }}{ }=v $$
$$ \begin{array}{rlrl} \text { Also, } & n & =\frac{c}{v} \\ & & \frac{E _0 / B _0}{E / B} & =\frac{c}{v}=n \\ \Rightarrow & \frac{E _0 / E}{B _0 / B} & =n \end{array} $$
This is possible only if $\frac{E _0}{E}=\sqrt{n}$ and $\frac{B _0}{B}=\frac{1}{\sqrt{n}}$.
