Question

In a medium the speed of light wave decreases to $0.2$ times to its speed in free space The ratio of relative permittivity to the refractive index of the medium is $x: 1$ The value of $x$ is(Given speed of light in free space $=3 \times 10^8 m s ^{-1}$ and for the given medium \(\mu =1\))

Hint:

The refractive index \( n \) is related to the speed of light in a medium, and the relative permittivity \( \epsilon_r \) and permeability \( \mu_r \) are related to the refractive index.

Answer 1

Correct Answer: 5

The correct answer is 5.
$V=\frac{C}{\mu }\Rightarrow \mu =\frac{C}{V}=\frac{C}{0.2C}$
$\mu =5$
$\mu =\sqrt{{ϵ}_r{\mu }_r}$
$\Rightarrow {ϵ}_r=\frac{{\mu }^2}{{\mu }_r}$
$\therefore \frac{{ϵ}_r}{\mu }=\frac{\mu }{{\mu }_r}=5$

Answer 2

The relation between speed, permittivity, and refractive index is given by: \[ V = \frac{C}{\mu} \] where \( C \) is the speed of light in free space, and \( \mu \) is the permeability of the medium. Since the speed of light in the medium is \( 0.2 \) times the speed in free space: \[ \mu = \frac{C}{V} = \frac{C}{0.2C} = 5 \] Thus, \[ \mu = \sqrt{\epsilon_r} \, \mu_0 \] This implies that: \[ \epsilon_r = 25 \] So, the ratio of relative permittivity to the refractive index is: \[ \frac{\epsilon_r}{\mu} = 5 \] Thus, the value of \( x \) is 5.