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

 

Option: 1

\mathrm{2 \frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu_0 \varepsilon_0}}}


Option: 2

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


Option: 3

\mathrm{\frac{1}{2} \frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu_0 \varepsilon_0}}}


Option: 4

none of these


Answers (1)

best_answer

We know that the velocity of light in vacuum \mathrm{c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}}

And the velocity of light in a medium \mathrm{v=\frac{1}{\sqrt{\mu \varepsilon}}}

Also the refractive index

\mathrm{n=\frac{\text { Velocity of light in medium }}{\text { Velocity of light in vaccum }}=\frac{v}{c}=\frac{1 / \sqrt{\mu_0 \varepsilon_0}}{1 / \sqrt{\mu \varepsilon}}=\frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu_0 \varepsilon_0}}}

Posted by

Devendra Khairwa

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