Q161 / 180 · KCET 2015 · Physics

If \( \varepsilon_{0} \) and \( \mu_{0} \) are the permittivity and permeability of free space and \( \varepsilon \) and \( \mu \) are the
corresponding quantities for a medium, then refractive index of the medium is

  • A
    \( \sqrt{\frac{\mu_{0} \varepsilon_{0}}{\mu \varepsilon}} \)
  • Correct answer
    B
    \( \sqrt{\frac{\mu \varepsilon}{\mu_{0} \varepsilon_{0}}} \)
  • C
    \( 11 \)
  • D
    Insufficient information

Solution

Given $\varepsilon_{0}=$ permittivity of free space $; \mu_{0}=$ permeabilityof free space,$\varepsilon=$ permittivity of medium; $\mu=$ permeability of medium Refractive index of medium, $n=\frac{C}{v} \rightarrow(1)$ We know $c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\Rightarrow v=\frac{1}{\sqrt{\mu \varepsilon}}$ Therefore, $n=\frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu_{0} \varepsilon_{0}}}=\sqrt{\frac{\mu \varepsilon}{\mu_{0} \varepsilon_{0}}}[$ from Eq. (1)]

Hence, the correct answer is option: 'B'.

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