With the help of the given ray diagram, calculate the refractive index of Medium $\displaystyle 2$ with respect to Medium 1. (Given that $\displaystyle \sqrt{3}=1 \cdot 732$ )

\[\frac{\operatorname{Sin} \mathrm{i}}{\operatorname{Sin} \mathrm{r}}=n \] \[\angle \mathrm{i}=90^{\circ}-30^{\circ}=60^{\circ} \] \[\frac{\operatorname{Sin} 60^{\circ}}{\operatorname{Sin} 30^{\circ}}=n \] \[\mathrm{n}=\frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} \] \[\mathrm{n}=\sqrt{3} \]