Question

Prove that $n_{12}\times n_{23}\times n_{31}=1$

where $n_{12}$ is the refractive index of medium-2 with respect to medium-1, $n_{23}$ is the refractive index of medium-3 with respect to medium-2, $n_{31}$ is the refractive index of medium-3 with respect to medium-1.

Answer 1

From the definition the relative refractive index we know, $n_{12}$=$\frac{n_2}{n_1}$, where $n_{12}$ is the refractive index of medium-2 with respect to medium-1, $n_1$ and $n_2$ are absolute refractive indices of medium-1 and 2 respectively.

Similarly, $n_{23}$=$\frac{n_3}{n_2}$ and $n_{31}$=$\frac{n_1}{n_3}$

Now , $n_{12}\times n_{23}\times n_{31}$=$\frac{n_2}{n_1}\times \frac{n_3}{n_2}\times \frac{n_1}{n_3}=1$ (PROVED)