An electromagnetic wave of frequency \( \nu=3.0\,\mathrm{MHz} \) passes from vacuum into a dielectric medium of relative permittivity 4.0 (relative permeability 1). Then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → the wavelength is halved and the frequency remains unchanged
In a medium of permittivity \( \varepsilon \) and permeability \( \mu \) an electromagnetic wave travels with speed \( v=1/\sqrt{\mu\varepsilon} \). The frequency is set by the source, and it stays the same when the wave enters another medium. So the wavelength \( \lambda=v/\nu \) changes in the same ratio as the speed.
With relative permittivity 4.0 and relative permeability 1, \( \varepsilon=4\varepsilon_0 \) and \( \mu=\mu_0 \):
\( \begin{aligned} v &= \frac{1}{\sqrt{\mu\varepsilon}} \\ &= \frac{1}{\sqrt{4\mu_0\varepsilon_0}} \\ &= \frac{c}{2} \end{aligned} \)
With \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \), the wavelength in vacuum is:
\( \begin{aligned} \lambda_0 &= \frac{c}{\nu} \\ &= \frac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{3.0\times10^{6}\,\mathrm{Hz}} \\ &= 100\,\mathrm{m} \end{aligned} \)
In the dielectric it is:
\( \begin{aligned} \lambda &= \frac{v}{\nu} \\ &= \frac{1.5\times10^{8}\,\mathrm{m\,s^{-1}}}{3.0\times10^{6}\,\mathrm{Hz}} \\ &= 50\,\mathrm{m} \end{aligned} \)
From the above, the speed falls to \( c/2 \) while the frequency stays 3.0 MHz. So the wavelength falls from 100 m to 50 m. Hence, the wavelength is halved and the frequency remains unchanged.
NCERT (2026-27):
Nature of Electromagnetic Waves – “… in a material medium of permittivity ε and magnetic permeability μ, the velocity of light becomes,
\( \displaystyle v=\frac{1}{\sqrt{\mu\varepsilon}} \qquad (8.11) \)
Thus, the velocity of light depends on electric and magnetic properties of the medium.”
Refraction of a Plane Wave – “… when a wave gets refracted into a denser medium (v1 > v2) the wavelength and the speed of propagation decrease but the frequency ν(= v/λ) remains the same.”