As shown in the figure, after passing through the medium 1. The speed of light v₂ in medium 2 will be :
(Given c = 3 × 10⁸ ms^–1)

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$1.0 \times 10^{8} {ms}^{- 1}$

$0.5 \times 10^{8} {ms}^{- 1}$

$1.5 \times 10^{8} {ms}^{- 1}$

$3.0 \times 10^{8} {ms}^{- 1}$

answer is A.

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Detailed Solution

$\begin{aligned} v = \frac{1}{\sqrt{με}} = \frac{1}{\sqrt{μ_{r} ε_{r} μ_{0} ε_{0}}} = \frac{c}{\sqrt{μ_{r} ε_{r}}} \\ \Rightarrow v_{2} = \frac{c}{\sqrt{9}} = 10^{8} m / s \end{aligned}$

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