An electromagnetic wave of intensity 50 Wm^ -2 enters in a medium...

An electromagnetic wave of intensity

50 \mathrm{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 :

Options

A $\left(\frac{1}{\sqrt{\mathrm{n}}}, \frac{1}{\sqrt{\mathrm{n}}}\right)$
B $(\sqrt{n}, \sqrt{n})$
C $\left(\sqrt{n}, \frac{1}{\sqrt{n}}\right)$
D $\left(\frac{1}{\sqrt{n}}, \sqrt{n}\right)$

Correct answer

C. $\left(\sqrt{n}, \frac{1}{\sqrt{n}}\right)$

Step-by-step solution

The speed of electromagnetic wave in free space is given by

\mathrm{C}=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}

In medium,

\mathbf{v}=\frac{1}{\sqrt{\mathrm{k} \in_{0} \mu_{0}}} \ldots

(ii) Dividing equation (i) by(ii), we get

\therefore \frac{\mathrm{C}}{\mathrm{V}}=\sqrt{\mathrm{k}}=\mathrm{n}

\frac{1}{2} \epsilon_{0} \mathrm{E}_{0}^{2} \mathrm{C}=

intensity

=\frac{1}{2} \epsilon_{0} \mathrm{kE}^{2} \mathrm{v}

\therefore \mathrm{E}_{0}^{2}

\mathrm{C}=\mathrm{kE}

^{2} \mathrm{v}

\Rightarrow \frac{E_{0}^{2}}{E^{2}}=\frac{k V}{C}=\frac{n^{2}}{n} \Rightarrow \frac{E_{0}}{E}=\sqrt{n}

similarly

\frac{\mathrm{B}_{0}^{2} \mathrm{C}}{2 \mu_{0}}=\frac{\mathrm{B}^{2} \mathrm{v}}{2 \mu_{0}} \Rightarrow \frac{\mathrm{B}_{0}}{\mathrm{~B}}=\frac{1}{\sqrt{\mathrm{n}}}

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