An electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into a dielectric medium with relative permittivity $\in$= 4.0. Then

Frequency of EM wave depends on nature of source but not medium. In case of wavelength, velocity in vacuum are f and $\mathrm{\lambda }$, then
${\mathrm{C}}_0=\frac{1}{\sqrt{{\mathrm{\mu }}_0{\mathrm{ϵ}}_0}}=\mathrm{f}\cdot {\mathrm{\lambda }}_{\text{vacuum }}$
Velocity in medium
${\mathrm{C}}_{\mathrm{m}}=\frac{{\mathrm{c}}_{\mathrm{o}}}{\sqrt{{\mathrm{\mu }}_{\mathrm{R}}{\mathrm{ϵ}}_{\mathrm{R}}}}=\mathrm{f}.{\mathrm{\lambda }}_{\text{medium }}\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$
For dielectric medium, μ_r = 1
$\therefore {\mathrm{C}}_{\mathrm{m}}=\frac{{\mathrm{C}}_0}{\sqrt{1\times 4}}=\frac{{\mathrm{C}}_0}{2}.\because {\mathrm{ϵ}}_{\mathrm{R}}=4$
$\mathrm{C}\propto \mathrm{\lambda }$
$\begin{array}{l}\frac{{\mathrm{C}}_{\mathrm{m}}}{{\mathrm{C}}_0}=\frac{{\mathrm{\lambda }}_{\mathrm{m}}}{{\mathrm{\lambda }}_0}\\ \therefore {\mathrm{\lambda }}_{\mathrm{m}}=\frac{{\mathrm{C}}_{\mathrm{m}}}{{\mathrm{C}}_0}\times {\mathrm{\lambda }}_0=\frac{{\mathrm{C}}_0}{2\times {\mathrm{C}}_0},{\mathrm{\lambda }}_0=\frac{{\mathrm{\lambda }}_0}{2}\end{array}$
Hence, wavelength is halved and frequency does not change.