Question

The refractive index and the permeability of a medium are $1.5H{m}^{-1}$ and $5\times {10}^{-7}H{m}^{-1}$ respectively. The relative permittivity of the medium is nearly

• A. $25$
• B. $15$
• C. $81$
• D. $10$
• E. $6$

Answer 1

The correct option is E

$6$

Step 1: Given data

Refractive index$\left(n\right)$=$1.5H{m}^{-1}$

Permeability of a medium=$5\times {10}^{-7}H{m}^{-1}$

Step 2: Formula used

The refractive index of the medium is given by,

$n=\frac{c}{v}$where, $n=$refractive index, $c=$speed of light, and $v=$speed of light in vacuum

Step 3: Compute the speed

Substitute the known values in the formula $n=\frac{c}{v}$

$$\begin{gathered} 1·5=\frac{3·0\times {10}^8}{v} \\ v=\frac{3·0\times {10}^8}{1·5} \\ v=2·0\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Step 4: Compute the relative permittivity

It is understood that ,$v=\frac{1}{\sqrt{\mu \epsilon }}=\frac{1}{\sqrt{\mu {\epsilon }_{\circ }{\epsilon }_r}}$

where $\mu =$Permeability of a medium, ${\epsilon }_r=$relative permeability of the medium, ${\epsilon }_0=$permeability of the medium.

Substitute the known values in the formula,

$\begin{array}{rcl}v& =& \frac{1}{\sqrt{\mu \epsilon }}=\frac{1}{\sqrt{\mu {\epsilon }_{\circ }{\epsilon }_r}}\\ \frac{1}{\sqrt{\mu \epsilon }}& =& \frac{1}{\sqrt{\mu {\epsilon }_{\circ }{\epsilon }_r}}\\ \left(2·0\times {10}^8\right)& =& \frac{1}{\sqrt{\left(5\times {10}^{-7}\right)\times {\epsilon }_r\times \left(8·85\times {10}^{-12}\right)}}\\ {\epsilon }_r& =& \frac{1}{{\left(2·0\times {10}^8\right)}^2\times \left(5\times {10}^{-7}\right)\times \left(8·85\times {10}^{-12}\right)}\\ {\epsilon }_r& \simeq & 6\end{array}$

Thus, the relative permittivity of the medium is $6$.

Hence, option E is the correct answer.