The dielectric constant of air is 1.006. The speed of an electromagnetic wave travelling in air is \( a\times10^{8}\,\mathrm{m\,s^{-1}} \), where a is about
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3
In a medium of permittivity \( \varepsilon \) and permeability \( \mu \), an electromagnetic wave travels with speed \( v=1/\sqrt{\mu\varepsilon} \). For air, \( \mu\approx\mu_0 \) and \( \varepsilon=K\varepsilon_0 \), where \( K \) is the dielectric constant. Since \( 1/\sqrt{\mu_0\varepsilon_0} \) is the speed of light in vacuum, \( c=3\times10^{8}\,\mathrm{m\,s^{-1}} \), the speed in air is:
\( \begin{aligned} v &= \frac{1}{\sqrt{\mu_0K\varepsilon_0}} \\ &= \frac{c}{\sqrt{K}} \\ &= \frac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{\sqrt{1.006}} \\ &\approx 2.99\times10^{8}\,\mathrm{m\,s^{-1}} \end{aligned} \)
Therefore, \( a\approx3 \).
NCERT (2026-27):
Nature of Electromagnetic Waves – “… in a material medium of permittivity ε and magnetic permeability μ, the velocity of light becomes,
\( \displaystyle v=\frac{1}{\sqrt{\mu\varepsilon}} \)\( \qquad (8.11) \)
Thus, the velocity of light depends on electric and magnetic properties of the medium.”