There are four media in which light is traveling. The speed of light in each medium along with relative permeability of the medium are given as below. Arrange the relative permittivity of each of the medium in increasing order.
(A) for Medium I, \( V_{I} = 2\times10^{8}\,\mathrm{m/s} \), \( \mu_{r} = 1 \)
(B) for Medium II, \( V_{II} = 1.5\times10^{8}\,\mathrm{m/s} \), \( \mu_{r} = 1 \)
(C) for Medium III, \( V_{III} = 3\times10^{8}\,\mathrm{m/s} \), \( \mu_{r} = 1 \)
(D) for Medium IV, \( V_{IV} = 2\times10^{8}\,\mathrm{m/s} \), \( \mu_{r} = 2 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C), (D), (A), (B)
The speed of light in a medium is set by its permittivity and permeability.
\( \displaystyle v = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{\sqrt{\mu_{r}\varepsilon_{r}}} \;\Rightarrow\; \varepsilon_{r} = \frac{1}{\mu_{r}}\left(\frac{c}{v}\right)^{2} \)
Taking \( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \):
(A) for Medium I, \( V_{I} = 2\times10^{8}\,\mathrm{m/s} \), \( \mu_{r} = 1 \) — \( \displaystyle \varepsilon_{r} = \left(\frac{3}{2}\right)^{2} = 2.25 \)
(B) for Medium II, \( V_{II} = 1.5\times10^{8}\,\mathrm{m/s} \), \( \mu_{r} = 1 \) — \( \displaystyle \varepsilon_{r} = \left(\frac{3}{1.5}\right)^{2} = 4 \)
(C) for Medium III, \( V_{III} = 3\times10^{8}\,\mathrm{m/s} \), \( \mu_{r} = 1 \) — \( \displaystyle \varepsilon_{r} = \left(\frac{3}{3}\right)^{2} = 1 \)
(D) for Medium IV, \( V_{IV} = 2\times10^{8}\,\mathrm{m/s} \), \( \mu_{r} = 2 \) — \( \displaystyle \varepsilon_{r} = \frac{1}{2}\left(\frac{3}{2}\right)^{2} = 1.125 \)
In increasing order, the relative permittivity is \( \varepsilon_{C} \lt \varepsilon_{D} \lt \varepsilon_{A} \lt \varepsilon_{B} \), that is (C), (D), (A), (B).