An electromagnetic wave of frequency nu = 3.0 | vedclass

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(C) The refractive index $n$ of a medium is given by $n = \sqrt{\frac{\mu\varepsilon}{\mu_0\varepsilon_0}}$.Assuming the medium is non-magnetic,$\mu =

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Wavelength is halved and frequency remains unchanged

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(C) The refractive index $n$ of a medium is given by $n = \sqrt{\frac{\mu\varepsilon}{\mu_0\varepsilon_0}}$.Assuming the medium is non-magnetic,$\mu = \mu_0$,so $n = \sqrt{\frac{\varepsilon}{\varepsilon_0}}$.Given the relative permittivity (dielectric constant) $K = \frac{\varepsilon}{\varepsilon_0} = 4.0$,we have $n = \sqrt{4.0} = 2$.The frequency $
u$ of an electromagnetic wave depends only on the source and remains unchanged when passing through different media.The speed of the wave in the medium is $v = \frac{c}{n} = \frac{c}{2}$.Since $v = 
u\lambda$,the new wavelength $\lambda' = \frac{v}{
u} = \frac{c/2}{
u} = \frac{\lambda}{2}$.Thus,the wavelength is halved and the frequency remains unchanged.

An electromagnetic wave of frequency $\nu = 3.0\,MHz$ passes from vacuum into a dielectric medium with permittivity $\varepsilon = 4.0\varepsilon_0$. Then

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