The optical properties of a medium are governed by the relative permittivity $(\epsilon_r)$ and relative permeability $(\mu_r)$. The refractive index is defined as $n = \sqrt{\epsilon_r \mu_r}$. For ordinary material $\epsilon_r > 0$ and $\mu_r > 0$ and the positive sign is taken for the square root. In $1964$,a Russian scientist $V$. Veselago postulated the existence of material with $\epsilon_r < 0$ and $\mu_r < 0$. Since then,such 'metamaterials' have been produced in the laboratories and their optical properties studied. For such materials $n = -\sqrt{\epsilon_r \mu_r}$. As light enters a medium of such refractive index,the phases travel away from the direction of propagation.
$(i)$ According to the description above,show that if rays of light enter such a medium from air (refractive index $= 1$) at an angle $\theta_i$ in the $2^{nd}$ quadrant,then the refracted beam is in the $3^{rd}$ quadrant.
$(ii)$ Prove that Snell's law holds for such a medium.

(N/A) $(i)$ Consider the interface between air $(n_1 = 1)$ and the metamaterial $(n_2 = -|n|)$. Let a wavefront $BC$ be incident on the interface at $C$. According to Huygens' principle,the time taken for the wavefront to travel from $B$ to $C$ is $t = \frac{BC}{c}$. In the same time $t$,the secondary wavelet from $A$ must travel a distance $AD = v_2 t = \frac{c}{|n_2|} t = \frac{BC}{|n_2|}$ in the metamaterial.
From the geometry of the incident wavefront,$BC = AC \sin \theta_i$. From the geometry of the refracted wavefront,$AD = AC \sin \theta_r$. Since the phase velocity in a metamaterial is directed towards the interface,the refracted ray must lie on the same side of the normal as the incident ray but in the opposite quadrant relative to the normal,placing it in the $3^{rd}$ quadrant.
$(ii)$ From the triangles $ABC$ and $ADC$,we have $\sin \theta_i = \frac{BC}{AC}$ and $\sin \theta_r = \frac{AD}{AC}$.
Dividing the two,we get $\frac{\sin \theta_i}{\sin \theta_r} = \frac{BC}{AD}$.
Substituting $BC = c t$ and $AD = |v_2| t$,we get $\frac{\sin \theta_i}{\sin \theta_r} = \frac{c}{|v_2|} = |n_2|$.
Since $n_2 = -|n_2|$,we have $\frac{\sin \theta_i}{\sin \theta_r} = n_2$ (taking the magnitude for the ratio of angles),which confirms that Snell's law holds.