Question

An exotic material having relative permeability and relative permittivity of $\mu_r$ (< 0) and $\epsilon_r$ (< 0) then if ray is incident at angle i then deviation of ray is

• A. $i + \sin^{-1} \frac{\sin i}{\sqrt{\mu_r\epsilon_r}}$
• B. $\pi - i - \sin^{-1} \frac{\sin i}{\sqrt{\mu_r\epsilon_r}}$
• C. $\pi - i + \sin^{-1} \frac{\sin i}{\sqrt{\mu_r\epsilon_r}}$
• D. $\frac{\pi}{2} - i - \sin^{-1} \frac{\sin i}{\sqrt{\mu_r\epsilon_r}}$

Answer 1

Answer: (A)

The deviation of the ray is given by $i + \sin^{-1} \frac{\sin i}{\sqrt{\mu_r\epsilon_r}}$.

Answer 2

For an exotic material with negative relative permeability ($\mu_r < 0$) and negative relative permittivity ($\epsilon_r < 0$), the material is a Left-Handed Material (LHM) or metamaterial.

1. Refractive Index: The refractive index ($n$) for such materials is given by: $n = -\sqrt{\mu_r \epsilon_r}$ Since $\mu_r < 0$ and $\epsilon_r < 0$, their product $\mu_r \epsilon_r$ is positive, so $\sqrt{\mu_r \epsilon_r}$ is a real positive number. Thus, the refractive index $n$ is real and negative.

2. Snell's Law: Let the incident medium be air or vacuum, with refractive index $n_1 = 1$. Let the exotic material be the second medium, with refractive index $n_2 = n = -\sqrt{\mu_r \epsilon_r}$. According to Snell's Law: $n_1 \sin i = n_2 \sin r$ $1 \cdot \sin i = (-\sqrt{\mu_r \epsilon_r}) \sin r$ $\sin r = -\frac{\sin i}{\sqrt{\mu_r \epsilon_r}}$

3. Negative Refraction: The negative sign in the expression for $\sin r$ indicates that the angle of refraction $r$ is negative. This implies that the refracted ray lies on the same side of the normal as the incident ray, unlike conventional refraction where it lies on the opposite side. Let the magnitude of the angle of refraction (the physical angle measured from the normal) be $r_0$. Then, $r_0 = \sin^{-1} \left( \frac{\sin i}{\sqrt{\mu_r \epsilon_r}} \right)$.

4. Deviation: Deviation ($\delta$) is the angle between the incident ray and the refracted ray.

   The incident ray comes from the upper left, making an angle $i$ with the normal. For negative refraction, the refracted ray goes into the lower left, making an angle $r_0$ with the normal.

   To find the deviation, imagine the incident ray continuing in a straight line (undeviated). This undeviated path would make an angle $i$ with the normal on the opposite side of the interface (i.e., in the lower right quadrant). The angle between this undeviated path (lower right) and the actual refracted path (lower left) is the sum of the angles they make with the normal. Angle of undeviated path with normal = $i$. Angle of refracted path with normal = $r_0$. The total angle between the undeviated path and the refracted path is $i + r_0$.

   So, the deviation $\delta = i + r_0$. Substituting the value of $r_0$: $\delta = i + \sin^{-1} \left( \frac{\sin i}{\sqrt{\mu_r \epsilon_r}} \right)$