For a transparent medium, relative permeability and permittivity, $\mu_r$ and $\varepsilon_r$ are 1.0 and 1.44

Question

For a transparent medium, relative permeability and permittivity, $\mu_r$ and $\varepsilon_r$ are 1.0 and 1.44 respectively. The velocity of light in this medium would be:

Accepted Answer

1.  **Formula:** The speed of light ($v$) in a medium is related to the speed of light in a vacuum ($c$) by the refractive index ($n$) of the medium: $v = c/n$. The refractive index is given by $n = \sqrt{\mu_r \varepsilon_r}$, where $\mu_r$ is the relative permeability and $\varepsilon_r$ is the relative permittivity (dielectric constant) .
2.  **Given Data:**
    *   Relative permeability ($\mu_r$) = $1.0$.
    *   Relative permittivity ($\varepsilon_r$) = $1.44$.
    *   Speed of light in vacuum ($c$) $\approx 3 	imes 10^8 	ext{ m/s}$.
3.  **Calculation:**
    *   Calculate refractive index: $n = \sqrt{1.0 	imes 1.44} = \sqrt{1.44} = 1.2$.
    *   Calculate velocity: $v = \frac{3 	imes 10^8}{1.2} = \frac{30}{12} 	imes 10^8 = 2.5 	imes 10^8 	ext{ m/s}$.

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