Q119 / 180 · NEET 2019 (Odisha) · Physics
For a transparent medium relative permeability and permittivity $\mu_{\mathrm{r}}$ and $\epsilon_{\mathrm{r}}$ are 1.0 and 1.44 respectively. The velocity of light in this medium would be
- Correct answerA$2.5 \times 10^8 \mathrm{~m} / \mathrm{s}$
- B$3 \times 10^8 \mathrm{~m} / \mathrm{s}$
- C$2.08 \times 10^8 \mathrm{~m} / \mathrm{s}$
- D$4.32 \times 10^8 \mathrm{~m} / \mathrm{s}$
Solution
The speed of any electromagnetic wave (EM) like light in any medium of permeability, $\mu$ and permittivity, $\varepsilon$ is given by
$$
\mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}}
$$
Also, $\mu=\mu_0 \mu_r$ and $\varepsilon=\varepsilon_0 \varepsilon_r$ where, $\mu_0$ and $\varepsilon_0$ are permeability and permittivity in free space and $\mu_{\mathrm{r}}$ and $\varepsilon_{\mathrm{r}}$ are relative permeability and relative permittivity of medium respectively.
Here, $\mu_{\mathrm{r}}=1.0$
$$
\varepsilon_{\mathrm{r}}=1.44
$$
$$
\begin{aligned}
\therefore \quad \quad \quad \mathrm{v} & =\frac{1}{\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}} \mu_0 \varepsilon_0}}=\frac{\mathrm{c}}{\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}}}} \quad\left[\because \mathrm{c}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right] \\
& =\frac{3 \times 10^8}{\sqrt{1 \times 1.44}} \\
& =2.5 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.