An electromagnetic wave of frequency v = 3 MHz passes from vacuum into a dielectric medium with permittivity 4. What happens to the wavelength?


Frequency, v = 3 MHz

Permittivity, ε = 4

During refraction, the frequency remains constant

\(v_{med}=\frac{1}{\sqrt{\mu _{0}\epsilon _{0}4}}=\frac{c}{2}\) \(\frac{\lambda _{med}}{\lambda _{air}}=\frac{v_{med}}{v_{air}}=\frac{\frac{c}{2}}{c}=\frac{1}{2}\)

Therefore, the wavelength is halved and frequency remains constant.

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