JEE Challenger
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Refraction of Light at Meta Material Interface with Negative Index

Comprehension Passage

Typically, materials possess a refractive index n>1n > 1. Hence, whenever a light ray travels from air into a naturally occurring substance, according to Snell's law, sin⁡θ1sin⁡θ2=n2n1\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}, the refracted ray bends toward the normal. However, it never emerges on the same side of the normal as the incident ray. From the theory of electromagnetism, the refractive index of a medium is expressed by the relation n=(cv)=±εrμr,n = \left(\frac{c}{v}\right) = \pm \sqrt{\varepsilon_r \mu_r}, in which cc denotes the speed of electromagnetic waves in vacuum, vv is its speed in the medium, and εr,μr\varepsilon_r, \mu_r represent the medium's relative permittivity and relative permeability, respectively.

In conventional substances, both εr\varepsilon_r and μr\mu_r are positive, implying a positive value of nn for the medium. When both εr\varepsilon_r and μr\mu_r are negative, one must choose the negative root of nn. Such negative refractive index materials can now be synthesized artificially and are called meta-materials. They exhibit significantly different optical behavior, without violating any physical laws. Since nn 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 onto a meta-material, identify the appropriate ray diagram:

Options

A
Option A
B
Option B
C
Option C
Correct
D
Option D

Step-by-Step Solution

According to Snell's law of refraction: sin⁡θ1sin⁡θ2=n2n1\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}

where:

  • θ1\theta_1 is the angle of incidence,
  • θ2\theta_2 is the angle of refraction,
  • n1n_1 is the refractive index of the first medium (air, so n1=1>0n_1 = 1 > 0),
  • n2n_2 is the refractive index of the second medium.

For a meta-material, the refractive index is negative, i.e., n2<0n_2 < 0. Therefore, the ratio of the refractive indices is negative: sin⁡θ1sin⁡θ2=n2n1<0\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1} < 0

Since sin⁡θ1>0\sin \theta_1 > 0 for a given positive incident angle with respect to the normal, it follows that: sin⁡θ2<0  ⟹  θ2<0\sin \theta_2 < 0 \implies \theta_2 < 0

A negative angle of refraction θ2\theta_2 implies that the refracted ray is bent to the same side of the normal as the incident ray (negative refraction).

Looking at the given ray diagrams:

  • In diagrams (A), (B), and (D), the refracted ray is on the opposite side of the normal relative to the incident ray, which corresponds to positive refractive indices.
  • In diagram (C), the refracted ray lies on the same side of the normal as the incident ray, correctly representing a negative angle of refraction.

Therefore, the appropriate ray diagram is given in option (C).