If the absolute refractive indices of water, | vedclass

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(A) The relative refractive index of medium $2$ with respect to medium $1$ is given by the formula: $n_{21} = n_2 / n_1$, where $n_2$ is the refractiv

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Sapphire relative to water

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(A) The relative refractive index of medium $2$ with respect to medium $1$ is given by the formula: $n_{21} = n_2 / n_1$, where $n_2$ is the refractive index of the second medium and $n_1$ is the refractive index of the first medium.Given refractive indices: $n_{	ext{water}} = 1.33$, $n_{	ext{benzene}} = 1.50$, $n_{	ext{sapphire}} = 1.77$.Calculating the values for each option:$A$: Sapphire relative to water = $1.77 / 1.33 \approx 1.3308$$B$: Sapphire relative to benzene = $1.77 / 1.50 = 1.18$$C$: Benzene relative to water = $1.50 / 1.33 \approx 1.1278$$D$: Water relative to benzene = $1.33 / 1.50 \approx 0.8867$Comparing these values, the relative refractive index of sapphire with respect to water $(1.3308)$ is the maximum.

If the absolute refractive indices of water, benzene, and sapphire are $1.33$, $1.50$, and $1.77$ respectively, then which medium has the maximum relative refractive index?

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