The electric field vector of a plane electromagnetic wave oscillates sinusoidally at a frequency of \( 4.5 \times 10^{10}\,\mathrm{Hz} \) in a medium with permittivity \( \epsilon_0 \) and permeability \( \mu_0 \). The wavelength of the wave is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6.67 \times 10^{-3}\,\mathrm{m} \)
A medium with permittivity \( \epsilon_0 \) and permeability \( \mu_0 \) is free space, so the wave travels at the speed of light, \( c = 1/\sqrt{\mu_0\epsilon_0} = 3\times10^{8}\,\mathrm{m\,s^{-1}} \). The wavelength follows from \( c = f\lambda \).
\( \displaystyle \lambda = \frac{c}{f} = \frac{3\times10^{8}}{4.5\times10^{10}} = 6.67\times10^{-3}\,\mathrm{m} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
