Question

the relative magnetic permeability of a medium is 2.5 and the relative electrical permittivity ofthe medium is 2.25.compute the refractive index of the medium​

Answer 1

Answer:

Explanation:

Relative permeability of the medium,μr = 1 Relative permitivity of the medium, εr = 2.25 Read more on Sarthaks.com - https://www.sarthaks.com/875663/if-the-relative-permeability-and-relative-permittivity-of-the-medium-is-1-0-and-2-25

Answer 2

Given:

The relative magnetic permeability of a medium is 2.5 and the relative electrical permittivity of the medium is 2.25.

To Find:

Refractive index of the medium.

Solution:

We know that speed of light in vacuum $c$, permeability μ$_o$ , and permittivity ε$_o$ of free space are related as

   μ$_o$ε$_o$ $=\frac{1}{c^2}$

For a medium with the speed of light in a vacuum $v$, permeability μ, and permittivity ε,

    με $=\frac{1}{v^2}$

⇒ μ$_r$μ$_o$ε$_r$ε$_o$ $=\frac{1}{v^2}$

⇒ μ$_r$ε$_r$$*\frac{1}{c^2}$ $=\frac{1}{v^2}$

⇒ $\frac{5.625}{c^2} =\frac{1}{v^2}$

⇒ $\frac{c^2}{v^2} =5.625$

We know that the refractive index of a medium can be defined as the ratio of the speed of light in a vacuum to the speed of light in that medium.

   $n=\frac{c}{v}$

⇒ $n=\sqrt{5.625}$

⇒ $n=2.37$

Hence, the refractive index of the medium is equal to 2.37.