Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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:

2.5 × 10⁸ m/s

3 × 10⁸ m/s

2.08 × 10⁸ m/s

4.32 × 10⁸ m/s

Step-by-Step Solution

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) .
Given Data: Relative permeability ( $\mu_r$ ) = $1.0$ . Relative permittivity ( $\varepsilon_r$ ) = $1.44$ .

Speed of light in vacuum ( $c$ ) $\approx 3 \times 10^8 \text{ m/s}$ .

Calculation: Calculate refractive index: $n = \sqrt{1.0 \times 1.44} = \sqrt{1.44} = 1.2$ . Calculate velocity: $v = \frac{3 \times 10^8}{1.2} = \frac{30}{12} \times 10^8 = 2.5 \times 10^8 \text{ m/s}$ .

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