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If  \mathrm{ \varepsilon_0 }  and  \mathrm{ \mu_0 }  are the electric permittivity and magnetic permeability of free space and  \mathrm{ \varepsilon \text { and } \mu } are the corresponding quantities in the medium, the index of refraction of the medium in terms of above
parameter is

Option: 1

\mathrm{\frac{\varepsilon \mu}{\varepsilon_0 \mu_0}}


Option: 2

\left(\frac{\varepsilon \mu}{\varepsilon_0 \mu_0}\right)^{1 / 2}


Option: 3

\left(\frac{\varepsilon_0 \mu_0}{\varepsilon \mu}\right)


Option: 4

\left(\frac{\varepsilon_0 \mu_0}{\varepsilon \mu}\right)^{1 / 2}


Answers (1)

best_answer

Velocity of light in vacuum
\mathrm{c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} }
velocity of light in medium
\mathrm{\begin{aligned} & v =\frac{1}{\sqrt{\mu \varepsilon}} \\ \therefore \quad & \mu =\frac{c}{v}=\left(\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}\right)^{1 / 2} \end{aligned} }

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manish painkra

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