The optical properties of a medium are governed by the relative permitivity `(epsi_(r))` and relative permeability `(mu_(r))`. The refractive index is defined as `sqrt(mu_(r)epsi_(r))=n`. For ordinary material `epsi_(r)gt0` and `mu_(r)gt0` and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with `epsi_(r)lt0andmu_(r)lt0`. Since then such 'metamaterials' have been produced in the laboratories and their optical properties studied. For such materials `n=-sqrt(mu_(r)epsi_(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.

Explanation

(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=(c)/(v)impliesc=nv" ".......(1)`
If time taken by light ray to travel l distance in above medium is t then,
`v=(l)/(t)impliest=(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=(l_(0))/(t)=(l_(0))/(((l)/(v)))=(vl_(0))/(l)`
`:.(c)/(v)=(l_(0))/(l)`
`:.(c)/(v)=(l_(0))/(l)`
`:.l_(0)=nl" "......(2)`
Above equation is used in the solution present question.

As shown in figure 1, suppose plane wavefront of light (AB) is made incident on the surface MN at time t = 0. Now if prediction made in the statement is correct then at time t, the refracted wavefront ED will be as shown in figure 1 which indicates that when angle of incidence `theta_(i)` is in second quadrant, angle of refraction `theta_(r)` should be obtained in third quadrant.
Here ED is a wavefront. Hence at all the points on it, we should have equal phases of oscillations of light vectors. For this to happen, when light rays emanating from A and B reach respectively at points E and D, their optical path lengths must be obtained same. If these lengths are respectively `r_(1)` and `r_(2)` then using equation (2).
`r_(1)=r_(2)`
`:.` Optical path length equivalent to `vecAE=BC+` optical path length equivalent to `vecCD`
`:.n (AE)=BC+n(CD)`
`:.-sqrt(in_(r)mu_(r))(AE)=BC-sqrt(in_(r)mu_(r))(CD)`
`:.BC=sqrt(in_(r)mu_(r))(CD-AE)" "......(3)`
Since measure of `vecBC` is positive in the figure from above equation it is proved that `CDgtAE`. But this will happen only when if `theta_(i)` is in second quadrant, `theta_(r)` should be in third quadrant. From this fact, prediction regarding existence of metamaterial is proved to be correct. (Because if here lower medium would have been ordinary denser transparent medium instead of metameterial then we should obtain `CDltAE` as shown in figure 2 but in the present case of metamaterial, we obtain `CDgtAE`, which indicates that there must be existence of metamaterial with `nlt0`,
(ii) Proof of Snell.s law,
From equation (3),
`BC=-n(CD-FD)`
`("":.n=-sqrt(in_(r)mu_(r))andAE=FD)`
`:.BC=-n(CF)" "......(4)`
In right angled `DeltaABC`,
`sintheta_(i)=(BC)/(AC)`
`:.BC=AC(sintheta_(i))" "......(5)`
In right angled `DeltaAFC`,
`sintheta_(r)=(CF)/(AC)`
`:.CF=AC(sintheta_(r))" "......(6)`
From equations (4), (5), (6),
`(AC)(sintheta_(i))=-n(AC)sintheta_(r)`
`:.sintheta_(i)=-nsintheta_(r)`
`:.n_(1)sintheta_(i)=-n_(2)sintheta_(r)`
( `:.` Here `n_(1)=1andn_(2)=n` )
Above equation indicates generalised form of Snell.s law.