The electric field of an electromagnetic wave in vacuum is \(E_y=20\sin(200\pi x-\omega t)\ \mathrm{V\,m^{-1}}\), where \(x\) is in metres and \(t\) in seconds. The frequency of the wave is (Take \(c=3\times10^{8}\ \mathrm{m\,s^{-1}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3\times10^{10}\ \mathrm{Hz}\)
The number multiplying \( x \) is the wave number \( k \). It gives the wavelength through \( k=2\pi/\lambda \). The frequency then follows from \( \nu\lambda=c \).
\( k=200\pi\ \mathrm{m^{-1}} \)
\( \begin{aligned} \lambda &= \frac{2\pi}{k} \\ &= \frac{2\pi}{200\pi} \\ &= 0.01\ \mathrm{m} \end{aligned} \)
\( \begin{aligned} \nu &= \frac{c}{\lambda} \\ &= \frac{3\times10^{8}}{0.01} \\ &= 3\times10^{10}\ \mathrm{Hz} \end{aligned} \)
NCERT (2026-27):
Nature of electromagnetic waves – “… Here k is related to the wave length λ of the wave by the usual equation”
\( \displaystyle k=\frac{2\pi}{\lambda} \)\( \qquad (8.8) \)
Nature of electromagnetic waves – “… This relation is often written in terms of frequency, ν (=ω/2π) and wavelength, λ (=2π/k) as”
\( \displaystyle 2\pi\nu=c\left(\frac{2\pi}{\lambda}\right) \quad \text{or} \) \( \nu\lambda=c \)\( \qquad [8.9(b)] \)