Q. If $\varepsilon_{0}\ and\ \mu_{0}$ are the electric permittivity and magnetic permeability of free space and $\varepsilon\ and\ \mu$ are the corresponding quantities in the medium, the index of refraction of the medium in terms of above parameter is
A) $\frac{\varepsilon\mu}{\varepsilon_{0}\mu_{0}}$
B) $\left( \frac{\varepsilon\mu}{\varepsilon_{0}\mu_{0}}\binom{\ }{\ } \right)^{1/2}$
C) $\left( \frac{\varepsilon_{0}\mu_{0}}{\varepsilon\mu} \right)$
D) $\left( \frac{\varepsilon_{0}\mu_{0}}{\varepsilon\mu} \right)^{1/2}\ $
Explanation
(b) Velocity of light in vacuum
$c\ = \ \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}$
velocity of light in medium
$v\ = \ \frac{1}{\sqrt{\mu\ \varepsilon}}$
$\therefore\ \ \ \ \ \ \ \mu = \frac{c}{v} = \left( \frac{\mu\varepsilon}{{\mu_{0}\varepsilon}_{0}\ } \right)^{1/2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.