# An electromagnetic wave of intensity

Question:

An electromagnetic wave of intensity $50 \mathrm{Wm}^{-2}$ enters in a medium of refractive index ' $\mathrm{n}$ ' 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:

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

2. (2) $(\sqrt{\mathrm{n}}, \sqrt{\mathrm{n}})$

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

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

Correct Option: , 3

Solution:

(3) The speed of electromagnetic wave in free space is given by

$\mathrm{C}=\frac{1}{\sqrt{\mu_{0} \in_{0}}}$       ....(1)

In medium, $\mathrm{v}=\frac{1}{\sqrt{\mathrm{k} \in_{0} \mu_{0}}} \ldots$ (ii)

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

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

$\frac{1}{2} \in_{0} \mathrm{E}_{0}^{2} \mathrm{C}=$ intensity $=\frac{1}{2} \in_{0} \mathrm{kE}^{2} \mathrm{v}$

$\therefore \mathrm{E}_{0}^{2} \mathrm{C}=\mathrm{kE}^{2} \mathrm{v}$

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

similarly

$\frac{B_{0}^{2} C}{2 \mu_{0}}=\frac{B^{2} v}{2 \mu_{0}} \Rightarrow \frac{B_{0}}{B}=\frac{1}{\sqrt{n}}$