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The optical properties of a medium are governed by the relative permitivity $({ \in _r})$ and relative permeability $(\mu _r)$. The refractive index is defined as $n = \sqrt {{ \in _r}{\mu _r}} $. For ordinary material ${ \in _r} > 0$ 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 $\in _r < 0$ and $u_r < 0$. Since then such 'metamaterials' have been produced in the laboratories and their optical properties studied. For such materials $n = - \sqrt {{ \in _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 $ in $2^{nd}$ quadrant, then the refracted beam is in the $3^{rd}$ quadrant.
$(ii)$ Prove that Snell's law holds for such a medium.
Solution
$(i)$ Let us first understand about equivalent optical path length of a given transparent medium.
According to definition, refractive index of a given denser transparent medium is,
$n=\frac{c}{v} \Rightarrow c=n v$
If time taken by light ray to travel $l$ distance in above medium is $t$ then,
$v=\frac{l}{t} \Rightarrow t=\frac{l}{v}$
– Now, the distance that can be travelled by light ray in air or in vacuum in above time is called an equivalent optical path length of a given denser transparent medium. If it is shown by symbol $l_{0}$ then since velocity of light ray in air or vacuum is $c$, we can write,
$c=\frac{l_{0}}{t}=\frac{l_{0}}{\left(\frac{l}{v}\right)}=\frac{v l_{0}}{l}$
$\therefore \frac{c}{v}=\frac{l_{0}}{l}$
$\therefore n=\frac{l_{0}}{l}$
$\therefore l_{0}=n l$$…(2)$
Above equation is used in the solution of present question.
