On putting in the measured values of \( \mu_0 \) and \( \varepsilon_0 \), Maxwell’s expression \( 1/\sqrt{\mu_0\varepsilon_0} \) for the speed of electromagnetic waves in vacuum gives about \( 3\times10^{8}\,\mathrm{m\,s^{-1}} \). Which of the following was concluded from this value?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → light is an electromagnetic wave
The value \( 3\times10^{8}\,\mathrm{m\,s^{-1}} \) is the speed of light found from optical measurements. Maxwell’s equations describe waves of coupled electric and magnetic fields. These waves travel in vacuum with the speed of light, so light itself was taken to be an electromagnetic wave. These waves need no material medium, and in vacuum they all travel with the same speed \( c \), whatever their wavelength.
NCERT (2026-27):
Introduction – “The most important prediction to emerge from Maxwell’s equations is the existence of electromagnetic waves, which are (coupled) time-varying electric and magnetic fields that propagate in space. The speed of the waves, according to these equations, turned out to be very close to the speed of light( 3 ×108 m/s), obtained from optical measurements. This led to the remarkable conclusion that light is an electromagnetic wave. Maxwell’s work thus unified the domain of electricity, magnetism and light.”