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प्रश्न
The optical properties of a medium are governed by the relative permittivity (εr) and relative permeability (µr). The refractive index is defined as `sqrt(µ_r ε_r) = n`. For ordinary material εr > 0 and µr > 0 and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with εr < 0 and µr < 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.
- 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, them the refracted beam is in the 3rd quadrant.
- Prove that Snell’s law holds for such a medium.
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उत्तर
i. If given postulate is true, then two parallel rays would proceed as shown in figure (i).

Metamaterials with positive refractive index

Ordinary material with positive refractive index
Again consider figure (i), let AB represent the incident wavefront and DE represent the refracted wavefront. All points on a wavefront must be in the same phase and in turn, must have the same optical path length.
Thus `- sqrt(ε_rµ_r) AE = BC - sqrt(ε_rµ_r) CD`
or `BC = sqrt(ε_rµ_r) (CD - AE)`
`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, figure)
Then, `- sqrt(ε_rµ_r) AE = BC - sqrt(ε_rµ_r) CD`
or `BC = sqrt(ε_rµ_r) (CD - AE)`
If `BC > 0` then `CD > AE`
This is obvious from figure (i). Hence, the postulate is resonable.
However, if the light proceeds in the sense it does for ordinary material (going from 2nd quadrant to 4th quadrant) as shown in figure (i) then proceeding as above
`- sqrt(ε_rµ_r) AE = BC - sqrt(ε_rµ_r) CD`
or `BC = sqrt(ε_rµ_r) (CD - AE)`
As AE > CD, therefore BC < 0 which is not possible. Hence, the given postulate is correct.
ii. From figure (i),
`BC = AC sin theta_i`
And `CD - AE = AC sin theta_r`
As `BC = - sqrt(ε_rµ_r) AC sin theta_r`
∴ `AC sin theta_i = sqrt(ε_rµ_r) AC sin theta_r`
or `sin theta_i/sin theta_r = sqrt(ε_rµ_r) = n`
Which proves Snell's law.
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