Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

$\varepsilon_0$ and $\mu_0$ are the electric permittivity and magnetic permeability of free space respectively. If the corresponding quantities of a medium are $2\varepsilon_0$ and $1.5\mu_0$ respectively, the refractive index of the medium will nearly be:

$\sqrt{2}$

$\sqrt{3}$

Step-by-Step Solution

Speed of Light Relations: The speed of light in vacuum is $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ and in a medium is $v = \frac{1}{\sqrt{\mu \varepsilon}}$ .
Refractive Index: The refractive index $n$ is defined as the ratio $c/v$ . $n = \frac{c}{v} = \sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}}$
Relative Permittivity and Permeability: This can be expressed in terms of relative quantities $\mu_r = \mu/\mu_0$ and $\varepsilon_r = \varepsilon/\varepsilon_0$ as $n = \sqrt{\mu_r \varepsilon_r}$ .
Calculation: Given $\varepsilon = 2\varepsilon_0$ (so $\varepsilon_r = 2$ ) and $\mu = 1.5\mu_0$ (so $\mu_r = 1.5$ ). $n = \sqrt{1.5 \times 2} = \sqrt{3}$

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started