Seven years after Hertz, J.C. Bose produced electromagnetic waves of wavelength 25 mm to 5 mm in his laboratory. The highest frequency of these waves is (Take \(c=3\times10^{8}\ \mathrm{m\,s^{-1}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(6\times10^{10}\ \mathrm{Hz}\)
Since \( \nu\lambda=c \), the frequency is inversely proportional to the wavelength. The highest frequency belongs to the shortest wavelength, 5 mm.
\( \lambda=5\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} \nu &= \frac{c}{\lambda} \\ &= \frac{3\times10^{8}}{5\times10^{-3}} \\ &= 6\times10^{10}\ \mathrm{Hz} \end{aligned} \)
NCERT (2026-27):
Nature of electromagnetic waves – “… This relation is often written in terms of frequency, ν (=ω/2π) and wavelength, λ (=2π/k) as”
\( \displaystyle 2\pi\nu=c\left(\frac{2\pi}{\lambda}\right) \quad \text{or} \) \( \nu\lambda=c \)\( \qquad [8.9(b)] \)