An electromagnetic wave of frequency 3.0 MHz | vedclass

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(A) Given,frequency of the electromagnetic $(EM)$ wave,$f = 3 \ MHz = 3 	imes 10^6 \ Hz$. Permittivity of the non-magnetic medium,$\epsilon = 16 \eps

Vedclass · Accepted Answer

$-75 \ m$

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(A) Given,frequency of the electromagnetic $(EM)$ wave,$f = 3 \ MHz = 3 	imes 10^6 \ Hz$. Permittivity of the non-magnetic medium,$\epsilon = 16 \epsilon_0$. The wavelength of the $EM$ wave in vacuum is $\lambda = \frac{c}{f} = \frac{3 	imes 10^8 \ m/s}{3 	imes 10^6 \ Hz} = 100 \ m$. The refractive index of the medium is $n = \sqrt{\epsilon_r \mu_r}$. Since the medium is non-magnetic,$\mu_r = 1$. Thus,$n = \sqrt{\frac{\epsilon}{\epsilon_0}} = \sqrt{16} = 4$. The wavelength in the medium is $\lambda' = \frac{\lambda}{n} = \frac{100 \ m}{4} = 25 \ m$. The change in wavelength is $\Delta \lambda = \lambda' - \lambda = 25 \ m - 100 \ m = -75 \ m$.

An electromagnetic wave of frequency $3.0 \ MHz$ passes from vacuum into a non-magnetic medium with permittivity $\epsilon = 16 \epsilon_0$,where $\epsilon_0$ is the free space permittivity. The change in wavelength is

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