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Question: Answered & Verified by Expert
If \(\varepsilon_0\) and \(\mu_0\) are the electric permittivity and magnetic permeability in vacuum, \(\varepsilon\) and \(\mu\) are corresponding quantities in medium, then refractive index of the medium is
PhysicsRay OpticsVITEEEVITEEE 2022
Options:
  • A \(\sqrt{\frac{\varepsilon}{\varepsilon_0}}\)
  • B \(\sqrt{\frac{\varepsilon_0 \mu}{\varepsilon \mu_0}}\)
  • C \(\sqrt{\frac{\varepsilon_0 \mu_0}{\varepsilon \mu}}\)
  • D \(\sqrt{\frac{\varepsilon \mu}{\varepsilon_0 \mu_0}}\)
Solution:
2474 Upvotes Verified Answer
The correct answer is: \(\sqrt{\frac{\varepsilon \mu}{\varepsilon_0 \mu_0}}\)
We know that velocity of electromagnetic wave in vacuum \(\left(\mathrm{v}_0\right)=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\) and velocity of electromagnetic wave in medium is
\(\text {(v) }=\frac{1}{\sqrt{\mu \varepsilon}} \text {. }\)
Therefore refractive index of the medium
\(\begin{aligned}
(\mu) & =\frac{\text { Vel. of E.M. wave in vacuum }\left(\mathrm{v}_0\right)}{\text { Vel. of E.M. wave in medium(v) }} \\
& =\frac{1 / \sqrt{\mu_0 \varepsilon_0}}{1 / \sqrt{\mu \varepsilon}}=\sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}}
\end{aligned}\)

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