Metamaterials $\rightarrow n = - \sqrt{μ_r ε_r}$

The optical properties of a medium are governed by the relative permitivity $(ε_r)$ and relative permeability ($μ_r)$. The refractive index is defined as $\sqrt{μ_rε_r} = n$. For ordinary material $ε_r > 0$ and $μ_r > 0$ and the positive sign is taken for the square root.

In 1964, a Russian scientist V. Veselago postulated the existence of material with $ε_r < 0$ and $μ_r < 0$. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied.

For such materials $n = - \sqrt{μ_r ε_r}$ . 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 $θ$ in $2nd$ quadrant, then the refracted beam is in the $3rd$ quadrant.

(ii) Prove that Snell’s law holds for such a medium.

$(i)$ Suppose the postulate is true, then two parallel rays would proceed as shown in Figure. Assuming $ED$ shows a wave front then all points on this must have the same phase. All points with the same optical path length must have the same phase. Thus $- \sqrt{ε_rμ_r} AE = BC -\sqrt{ε_rμ_r}CD$ or $BC = \sqrt{ε_rμ_r}(CD - AE)$

$\rightarrow BC > 0, CD > AE$

As showing that the postulate is reasonable. If however, the light proceeded in the sense it does for ordinary material (viz. in the fourth quadrant, Fig. 2) Then $- \sqrt{ε_rμ_r} AE = BC -\sqrt{ε_rμ_r}CD$ or $BC = \sqrt{ε_rμ_r}(CD - AE)$

As $AE > CD, BC < O$, showing that this is not possible. Hence the postulate is correct.

$(ii)$ Prove yourself. $\ddot \smile$ Note by Nishant Rai
4 years, 4 months ago

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Try these - Easy if you know about Metamaterials!

- 4 years, 4 months ago

a comprehension based on this was asked in $JEE$

- 4 years, 4 months ago
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