Calculate the value of n_{32} times n_{21} = | vedclass

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(A) According to the principle of reversibility and the definition of refractive index,the refractive index of medium $2$ with respect to medium $1$ i

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$n_{31}$

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(A) According to the principle of reversibility and the definition of refractive index,the refractive index of medium $2$ with respect to medium $1$ is given by $n_{21} = \frac{n_2}{n_1}$.Similarly,the refractive index of medium $3$ with respect to medium $2$ is $n_{32} = \frac{n_3}{n_2}$.Multiplying these two expressions:$n_{32} 	imes n_{21} = \left( \frac{n_3}{n_2} ight) 	imes \left( \frac{n_2}{n_1} ight) = \frac{n_3}{n_1}$.By definition,the refractive index of medium $3$ with respect to medium $1$ is $n_{31} = \frac{n_3}{n_1}$.Therefore,$n_{32} 	imes n_{21} = n_{31}$.

Calculate the value of $n_{32} \times n_{21} = .... $

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