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An electromagnetic wave of frequency  \dpi{100} f = 3.0\: M\! Hz passes from a vacuum into a dielectric medium with permittivity \dpi{100} \epsilon_r = 4.0

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 unchanged


Option: 4

wavelength and frequency both remain unchanged.


Answers (1)

best_answer

As discussed in

Wavelength of EM Wave -

\lambda =\frac{\lambda_{o}}{\mu}

- wherein

\lambda _{o} = Wavelength in vacuum

\mu = Refractive index of the medium

\mu = \sqrt{\frac{\epsilon }{\epsilon _{0}}}=\sqrt {\epsilon_r}=\sqrt{4}= 2

Since    \mu \: \alpha\: \frac{1}{\lambda }

\therefore \: \: \: \:W avelength\: is\: halved

Posted by

Ritika Jonwal

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