An electromagnetic wave of frequency $3 \mathrm{~MHz}$ passes from vacuum into a dielectric medium $(\mu_r=1)$ of relative permittivity 4.0. Then,

The frequency of an electromagnetic wave is determined by the source and remains constant when the wave travels from one medium to another. Thus, the frequency remains $3 \text{ MHz}$.

The speed of an electromagnetic wave in vacuum is $c = \dfrac{1}{\sqrt{\mu_0 \epsilon_0}}$. In a dielectric medium with relative permittivity $\epsilon_r$ and relative permeability $\mu_r$, the speed is $v = \dfrac{1}{\sqrt{\mu_0 \mu_r \epsilon_0 \epsilon_r}} = \dfrac{c}{\sqrt{\mu_r \epsilon_r}}$.

Given $\mu_r = 1$ and $\epsilon_r = 4.0$, the speed in the medium is $v = \dfrac{c}{\sqrt{1 \times 4.0}} = \dfrac{c}{2}$.

The wavelength in a medium is given by $\lambda' = \dfrac{v}{f}$. Since $v = \dfrac{c}{2}$ and $f$ is constant, $\lambda' = \dfrac{c/2}{f} = \dfrac{1}{2} \left( \dfrac{c}{f} \right) = \dfrac{\lambda}{2}$.

Therefore, the wavelength is halved and the frequency remains unchanged.

Answer: Wavelength is halved and frequency remains unchanged