Question

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. \[\sqrt {\dfrac{{{\mu _0}{\varepsilon _0}}}{{\mu e}}} \]
B. \[\sqrt {\dfrac{{\mu \varepsilon }}{{{\mu _0}{\varepsilon _0}}}} \]
C. \[1\]
D. Insufficient information

Answer 1

Hint: In the formation of the electric field, permittivity measures the obstruction created by the material, while permeability is the ability of the material to allow magnetic lines to conduct through it. The refractive index is a value determined from the ratio of the velocity of light in the vacuum to that of the higher density of the second medium.

Formula used:
Refractive index of a medium:
\[n = \dfrac{c}{v}\] …… (1)
Where,
\[n\] indicates the refractive of the medium.
\[c\] indicates speed of light in the air.
\[v\] indicates speed of light in that medium.

Complete step by step answer:
Let, the velocity of light in medium and air be \[v\]and \[c\],
We know the refractive index of medium from the equation (1),
\[n = \dfrac{c}{v}\]
Where,
\[c = \dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}\]
\[v = \dfrac{1}{{\sqrt {\mu \varepsilon } }}\]
So,
$n = \dfrac{{\dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}}}{{\dfrac{1}{{\sqrt {\varepsilon \mu } }}}} \\$
$\Rightarrow n = \sqrt {\dfrac{{\mu \varepsilon }}{{{\mu _0}{\varepsilon _0}}}} \\$

Hence, the required answer is \[\sqrt {\dfrac{{\mu \varepsilon }}{{{\mu _0}{\varepsilon _0}}}} \]

So, the correct answer is “Option B”.

Additional Information:
Refractive index: The refractive index is a value determined from the ratio of the velocity of light in the vacuum to that of the higher density of the second medium. The refractive index variable in descriptive text and mathematical equations is most commonly symbolized by the letter \[n\] or \[n'\].

Note:
 A material's refraction value index is a number representing the number of times slower than it is in a vacuum that a light wave will be in that material. So, the optical density increases as the refraction value index increases, and the speed of light in that material decreases.