The refractive index of denser medium with respect to rarer medium is 1.125. The difference between the velocities of light in the two media is 0.25 × 10⁸ m/s. Find the velocities of light in the two media and their refractive indices. (c = 3 ×10⁸ m/s)

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$2.0 \times 10^{8} m / s; 2.25 \times 10^{8} m / s; 1.500; 1.333$

$2.5 \times 10^{8} m / s; 2.25 \times 10^{8} m / s; 1.500; 1.333$

$2.0 \times 10^{8} m / s; 2.25 \times 10^{8} m / s; 1.333; 1.500$

$2.25 \times 10^{8} m / s; 2.0 \times 10^{8} m / s; 1.500; 1.333$

answer is A.

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Detailed Solution

$\begin{array}{l} _{r} μ_{d} = 1.125,_{r} μ_{d} = \frac{V_{r}}{V_{d}} \\ V_{r} - V_{d} = 0.25 \times 10^{8} m / s \\ 1.125 V_{d} - V_{d} = 0.25 \times 10^{8} \\ 12.5 V_{d} = 25 \times 10^{8} \times 10 \Rightarrow V_{d} = 2 \times 10^{8} m / s \\ V_{r} = 2.25 \times 10^{8} m / s \\ μ_{d} = \frac{3 \times 10^{8}}{2 \times 10^{8}} = 1.5; μ_{r} = \frac{3 \times 10^{8}}{2.25 \times 10^{8}} = 1.333 \end{array}$