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An Electromagnetic Wave of frequency \mathrm{v}=3.0 \mathrm{MHz} passes from vacuum into a dielectric medium with permittivity \varepsilon=4.0. Them

Option: 1

Wavelength is doubled and the frequency remains unchanged


Option: 2

Wavelength is doubled and frequency becomes half


Option: 3

Wavelength is halved and frequency remains unchange


Option: 4

Wavelength and frequency both become unchanged


Answers (1)

best_answer

In vacuum, \mathrm{\varepsilon_0=1 }
In medium,\mathrm{ \varepsilon=4 }
So, refractive index
\mathrm{\begin{aligned} \mu & =\sqrt{\varepsilon / \varepsilon_0}=\sqrt{4 / 1}=2 \\ \lambda^{\prime} & =\frac{\lambda}{\mu}=\frac{\lambda}{2} \\ v & =\frac{c}{\mu}=\frac{c}{2} \end{aligned} }
wavelength
and wave velocity v= \mathrm{\frac{c}{\mu}=\frac{c}{2} }
Hence, it is clear that wavelength and velocity will become half but frequency remains uncharged when the wave is passing through any medium.

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Rishi

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