Electromagnetic Waves

JEE Main level, single correct, medium

110 of 165

Question 110 of 165

An electromagnetic wave of intensity 50Wm−250 \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 :

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Answer: (C) (n,1n)\left(\sqrt{n}, \frac{1}{\sqrt{n}}\right)

The speed of electromagnetic wave in free space is given by

C=1μ0ϵ0\mathrm{C}=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}

In medium, v=1k∈0μ0…\mathbf{v}=\frac{1}{\sqrt{\mathrm{k} \in_{0} \mu_{0}}} \ldots (ii)

Dividing equation (i) by(ii), we get ∴CV=k=n\therefore \frac{\mathrm{C}}{\mathrm{V}}=\sqrt{\mathrm{k}}=\mathrm{n}

12ϵ0E02C=\frac{1}{2} \epsilon_{0} \mathrm{E}_{0}^{2} \mathrm{C}= intensity =12ϵ0kE2v=\frac{1}{2} \epsilon_{0} \mathrm{kE}^{2} \mathrm{v}

∴E02C=kE2v\therefore \mathrm{E}_{0}^{2} \mathrm{C}=\mathrm{kE}^{2} \mathrm{v}

⇒E02E2=kVC=n2n⇒E0E=n\Rightarrow \frac{E_{0}^{2}}{E^{2}}=\frac{k V}{C}=\frac{n^{2}}{n} \Rightarrow \frac{E_{0}}{E}=\sqrt{n}

similarly

B02C2μ0=B2v2μ0⇒B0 B=1n\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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