Q.The optical properties of a medium are governed by the relative permitivity ($\varepsilon_r$) and relative permeability ($\mu_r$). The refractive index is defined as $\sqrt{\mu_r \varepsilon_r} = n$. For ordinary material $\varepsilon_r > 0$ 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 $\varepsilon_r < 0$ and $\mu_r < 0$. Since then such 'metamaterials' have been produced in the laboratories and their optical properties studied. For such materials $n = -\sqrt{\mu_r \varepsilon_r}$. As light enters a medium of such refractive index the phases travel away from the direction of propagation.

Negative-index (metamaterial) refraction. For a Veselago medium with $\varepsilon_r<0$ and $\mu_r<0$, the correct root is $n=-\sqrt{\mu_r\varepsilon_r}<0$.

(i) Refracted ray goes to the 3rd quadrant. Take the normal as the vertical axis and the interface as the horizontal axis, with air (quadrants 1 and 2) above and the metamaterial (quadrants 3 and 4) below. An incident ray in the 2nd quadrant strikes the origin at angle $\theta$ from the normal. Snell's law gives

A negative $\sin\theta_r$ places the transmitted ray on the same side of the normal as the incident ray (not the opposite side as in ordinary media). Since it must travel into the lower medium, it lands in the 3rd quadrant. …

The physics of a negative index

The refractive index is defined by $n^2=\mu_r\varepsilon_r$. When both $\mu_r$ and $\varepsilon_r$ are positive we take $n=+\sqrt{\mu_r\varepsilon_r}$. Veselago showed that when both are negative — even though the product $\mu_r\varepsilon_r$ is again positive — the physically consistent choice is the negative root,

because in such a medium the phase advances opposite to the direction of energy flow. It is this sign that reverses the geometry of refraction.

Setting up the coordinate frame

Place the interface along the horizontal axis and the normal along the vertical axis.

• Air ($n_1=1$) occupies the upper half — the 1st and 2nd quadrants.
• The metamaterial ($n_2=-\sqrt{\mu_r\varepsilon_r}<0$) occupies the lower half — the 3rd and 4th quadrants.

An incident ray in the 2nd quadrant comes down onto the origin from the upper-left, making angle $\theta$ with the upward normal. In an ordinary medium the refracted ray would cross to the far side of the normal and continue downward-right into the 4th quadrant.

(i) Locating the refracted ray

Apply Snell's law at the interface:

With $n_1=1$ and $n_2=-\sqrt{\mu_r\varepsilon_r}$,

For an incident ray in the 2nd quadrant, $\sin\theta>0$, so

A negative refraction angle means the transmitted ray does not swing across to the far side of the normal. Instead it stays on the same side of the normal as the incident ray, but travels into the lower medium — i.e. downward and to the left. That direction is the 3rd quadrant.

Equivalently, writing the ray direction vectors: the incident ray points as $(\sin\theta,\,-\cos\theta)$ (down-right of the normal on entry from the upper-left), while the refracted direction $(\sin\theta_r,\,-\cos\theta_r)$ has $\sin\theta_r<0$ and $-\cos\theta_r<0$, i.e. both components negative — squarely in the 3rd quadrant.

(ii) Proving Snell's law holds

Snell's law is a statement of phase matching at the boundary, not a property that depends on the sign of $n$. When a plane wave meets the interface, the component of the wave vector parallel to the surface must be continuous across it (otherwise the wave fronts of the incident and transmitted waves could not stay in step along the boundary): …

Method: Applying Snell's Law When the Refractive Index Is Negative

Use this method for any problem involving an unusual/negative-index medium

where you must (a) determine the geometric direction of the refracted ray, and

(b) show that the ordinary law of refraction still applies in form.

Steps

Step 1: Determine the correct sign of the refractive index

Refractive index is defined via $n^2 = \mu_r\varepsilon_r$. Normally you take

the positive square root. If the medium's electromagnetic parameters are BOTH

negative (a metamaterial), the physically correct choice is the negative

root:

Always check the problem statement (or the sign convention it specifies) rather

than assuming the positive root.

Step 2: Apply Snell's law algebraically, keeping the sign

Substitute the (possibly negative) $n_2$ directly — do not silently drop the

sign. Solving for $\sin\theta_r$ will then come out negative when $n_2<0$ and

the incident angle is positive.

Step 3: Interpret the sign geometrically

A NEGATIVE $\sin\theta_r$ means the refracted ray does not cross to the

opposite side of the normal the way it would in an ordinary medium — it stays

on the same side as the incident ray, while still travelling into the second

medium. Combine "same side of the normal" with "the half-space the ray is

entering" to name the correct quadrant/region.

Step 4 (Applying to this problem): Prove Snell's law holds regardless of sign …