Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, \sin \theta _1 \over \sin \theta _2 = n_2 \over n_1 , it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation n = ( c \over v ) = \pm \,\sqrt \varepsilon _r\mu _r , where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, \varepsilon r and \mu r, are the relative permittivity and permeability of the medium, respectively. In normal materials, both \varepsilon r and \mu r , are positive, implying positive n for the medium. When both \varepsilon r and \mu r are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating and physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. For light incident from air on a meta-material, the appropriate ray diagram is

Geometrical Optics PYQ — JEE Advanced 2012 · Shift 2 · Q57 with Solution

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Correct answer: C

Key idea: negative refractive index

For ordinary materials, the refractive index is positive: $n=+\sqrt{\varepsilon_r\mu_r}$ and when light goes from air to such a material, the refracted ray bends according to Snell’s law on the other side of the normal.

For a meta-material with $\varepsilon_r<0,\quad \mu_r<0,$ we must take $n=-\sqrt{\varepsilon_r\mu_r}$ so the refractive index is negative.
Apply Snell’s law

Snell’s law is $n_1\sin\theta_1=n_2\sin\theta_2$

Here, light is incident from air, so
$\qquad n_2<0.$$ Therefore, $$\sin\theta_2=\frac{n_1}{n_2}\sin\theta_1$$ Since $n_2$ is negative, $\sin\theta_2$ is negative when $\sin\theta_1$ is positive. This means the refracted angle lies on the **same side of the normal** as the incident ray.$
Physical interpretation

In a negative index medium, the refracted ray bends to the “unusual” side of the normal. So unlike ordinary refraction, the transmitted ray appears on the same side of the normal as the incident ray.
Choosing the correct diagram

Hence the correct ray diagram must show:
- incident ray coming from air,
- refracted ray entering the meta-material,
- refracted ray lying on the same side of the normal as the incident ray.
Among the given options, this corresponds to Option C.
Comparison with stored answer

My derived answer is C, which matches the stored correct answer.

View written solutionFree

Correct answer: C

Key idea: negative refractive index

For ordinary materials, the refractive index is positive: $n=+\sqrt{\varepsilon_r\mu_r}$ and when light goes from air to such a material, the refracted ray bends according to Snell’s law on the other side of the normal.

For a meta-material with $\varepsilon_r<0,\quad \mu_r<0,$ we must take $n=-\sqrt{\varepsilon_r\mu_r}$ so the refractive index is negative.
Apply Snell’s law

Snell’s law is $n_1\sin\theta_1=n_2\sin\theta_2$

Here, light is incident from air, so
$\qquad n_2<0.$$ Therefore, $$\sin\theta_2=\frac{n_1}{n_2}\sin\theta_1$$ Since $n_2$ is negative, $\sin\theta_2$ is negative when $\sin\theta_1$ is positive. This means the refracted angle lies on the **same side of the normal** as the incident ray.$
Physical interpretation

In a negative index medium, the refracted ray bends to the “unusual” side of the normal. So unlike ordinary refraction, the transmitted ray appears on the same side of the normal as the incident ray.
Choosing the correct diagram

Hence the correct ray diagram must show:
- incident ray coming from air,
- refracted ray entering the meta-material,
- refracted ray lying on the same side of the normal as the incident ray.
Among the given options, this corresponds to Option C.
Comparison with stored answer

My derived answer is C, which matches the stored correct answer.