A radio wave has a wavelength of 3.0 m in vacuum. It enters a non-magnetic medium \((\mu_r=1)\) of relative permittivity 4. The wavelength of the wave in the medium is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.5 m
The frequency of a wave is set by its source. So it does not change when the wave enters the medium. The speed changes, and the wavelength changes with it.
In a medium, \( v=1/\sqrt{\mu\varepsilon} \). With \( \mu=\mu_r\mu_0 \) and \( \varepsilon=\varepsilon_r\varepsilon_0 \), this gives \( v=c/\sqrt{\mu_r\varepsilon_r} \).
\( \lambda_0=3.0\ \mathrm{m} \)
\( \mu_r=1 \)
\( \varepsilon_r=4 \)
\( \begin{aligned} v &= \frac{c}{\sqrt{\mu_r\varepsilon_r}} \\ &= \frac{c}{\sqrt{1\times4}} \\ &= \frac{c}{2} \end{aligned} \)
\( \begin{aligned} \lambda &= \frac{v}{\nu} \\ &= \frac{c}{2\nu} \\ &= \frac{\lambda_0}{2} \\ &= 1.5\ \mathrm{m} \end{aligned} \)
NCERT (2026-27):
Sources of electromagnetic waves – “… Consider a charge oscillating with some frequency. … The oscillating electric and magnetic fields thus regenerate each other, so to speak, as the wave propagates through the space. The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. …”
Nature of electromagnetic waves – “… These replace ε0 and μ0 in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity ε and magnetic permeability μ, the velocity of light becomes,”
\( \displaystyle v=\frac{1}{\sqrt{\mu\varepsilon}} \)\( \qquad (8.11) \)