Question

If ${\epsilon }_0$ and ${\mu }_0$ are respectively the electric permittivity and magnetic permeability of free space, $\epsilon$ and $\mu$ are the corresponding quantities in a medium, the index of refraction of the medium is :

• A. $\sqrt{\frac{{\epsilon }_0{\mu }_0}{\epsilon \mu }}$
• B. $\sqrt{\frac{\epsilon \mu }{{\epsilon }_0{\mu }_0}}$
• C. $\sqrt{\frac{{\epsilon }_0\mu }{\epsilon {\mu }_0}}$
• D. $\sqrt{\frac{\epsilon }{{\epsilon }_0}}$

Answer 1

Answer: (B)

Refractive index of medium is given by $n=\sqrt{{\mu }_r{\epsilon }_r}$ Here, $\mu ={\mu }_0{\mu }_r$ $\Rightarrow$ ${\mu }_r=\frac{\mu }{{\mu }_0}$ and $\epsilon ={\epsilon }_0{\epsilon }_r$ $\Rightarrow$ ${\epsilon }_r=\frac{\epsilon }{{\epsilon }_0}$ $\therefore$ $n=\sqrt{\frac{\mu }{{\mu }_0}.\frac{\epsilon }{{\epsilon }_0}}\sqrt{\frac{\epsilon \mu }{{\epsilon }_0{\mu }_0}}$ NOTE: The above expression can be written as $n=\sqrt{\frac{\epsilon \mu }{{\epsilon }_0{\mu }_0}}=\frac{C_{vacuum}}{C_{medium}}$ as $c_v=\frac{1}{\sqrt{{\epsilon }_0{\mu }_0}}$ = speed of light in vacuum $c_m=\frac{1}{\sqrt{\epsilon \mu }}$ = speed of light in medium