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. Choose the correct statement.

The speed of light in the meta-material is v = c \over |n| .

The speed of light in the meta-material is v = c|n|.

The speed of light in the meta-material is v = c.

The wavelength of the light in the meta-material ( \lambda m) is given by \lambda _m = \lambda _air|n| , where \lambda _air is wavelength of the light in air.

Geometrical Optics PYQ — JEE Advanced 2012 · Shift 2 · Q58 with Solution

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Correct answer: B

Step-by-step Derivation

Fundamental Definitions The speed of light in a medium (v) is related to its relative permittivity ( $ε_r$ ) and relative permeability ( $μ_r$ ) by the formula: $v = \frac{1}{\sqrt{\varepsilon\mu}} = \frac{1}{\sqrt{\varepsilon_0 \varepsilon_r \mu_0 \mu_r}}$ where ε and μ are the permittivity and permeability of the medium, and $ε_0$ and $μ_0$ are the values in vacuum.

The speed of light in vacuum (c) is given by: $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$

Combining these two equations, we can express v in terms of c: $v = \frac{c}{\sqrt{\varepsilon_r \mu_r}}$
Refractive Index in Meta-materials The problem states that for meta-materials, both $ε_r$ and $μ_r$ are negative. For such materials, the refractive index n is defined by choosing the negative root of the relation $n^2 = ε_r μ_r$ . Therefore, for a meta-material: $n = -\sqrt{\varepsilon_r \mu_r}$ Note that since $ε_r < 0$ and $μ_r < 0$ , their product $ε_r μ_r$ is positive, making n a real, negative number.
Relating Speed v to Refractive Index n The speed of light v in any medium must be a positive scalar quantity. From the equation in Step 1, $v = c / \sqrt{\varepsilon_r \mu_r}$ . Since c is positive, $\sqrt{\varepsilon_r \mu_r}$ must also be positive.

From the definition of n for meta-materials in Step 2, we can find an expression for $\sqrt{\varepsilon_r \mu_r}$ : $n = -\sqrt{\varepsilon_r \mu_r} \implies \sqrt{\varepsilon_r \mu_r} = -n$

Since n is negative for a meta-material, -n is a positive quantity. This positive quantity is the magnitude (or absolute value) of the refractive index, |n|. So, $\sqrt{\varepsilon_r \mu_r} = |n|$ .
Final Expression for Speed v Now, substitute $\sqrt{\varepsilon_r \mu_r} = |n|$ back into the expression for v from Step 1: $v = \frac{c}{\sqrt{\varepsilon_r \mu_r}} = \frac{c}{|n|}$ This shows that the speed of light in the meta-material is v = c / |n|.

Evaluation of Options

A: The speed of light in the meta-material is v = c|n|. This is incorrect. It would imply v > c if |n| > 1, violating the principles of relativity for wave propagation in a medium.
B: The speed of light in the meta-material is v = c / |n|. This matches our derived result. It correctly represents the speed as a positive value, determined by the magnitude of the refractive index.
C: The speed of light in the meta-material is v = c. This is incorrect. It would mean the medium is a vacuum, i.e., |n|=1.
D: The wavelength of the light in the meta-material ( $λ_m$ ) is given by $λ_m = λ_{air}|n|$ . The frequency f of light does not change when it enters a new medium. We have the relations: $λ_{air} = c / f$ $λ_m = v / f$ Dividing the second equation by the first gives: $λ_m / λ_{air} = v / c$ From our result v = c / |n|, we get v / c = 1 / |n|. Therefore, $λ_m = λ_{air} / |n|$ . This option is incorrect.

Thus, the only correct statement is B.

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View written solutionFree

Correct answer: B

Step-by-step Derivation

Fundamental Definitions The speed of light in a medium (v) is related to its relative permittivity ( $ε_r$ ) and relative permeability ( $μ_r$ ) by the formula: $v = \frac{1}{\sqrt{\varepsilon\mu}} = \frac{1}{\sqrt{\varepsilon_0 \varepsilon_r \mu_0 \mu_r}}$ where ε and μ are the permittivity and permeability of the medium, and $ε_0$ and $μ_0$ are the values in vacuum.

The speed of light in vacuum (c) is given by: $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$

Combining these two equations, we can express v in terms of c: $v = \frac{c}{\sqrt{\varepsilon_r \mu_r}}$
Refractive Index in Meta-materials The problem states that for meta-materials, both $ε_r$ and $μ_r$ are negative. For such materials, the refractive index n is defined by choosing the negative root of the relation $n^2 = ε_r μ_r$ . Therefore, for a meta-material: $n = -\sqrt{\varepsilon_r \mu_r}$ Note that since $ε_r < 0$ and $μ_r < 0$ , their product $ε_r μ_r$ is positive, making n a real, negative number.
Relating Speed v to Refractive Index n The speed of light v in any medium must be a positive scalar quantity. From the equation in Step 1, $v = c / \sqrt{\varepsilon_r \mu_r}$ . Since c is positive, $\sqrt{\varepsilon_r \mu_r}$ must also be positive.

From the definition of n for meta-materials in Step 2, we can find an expression for $\sqrt{\varepsilon_r \mu_r}$ : $n = -\sqrt{\varepsilon_r \mu_r} \implies \sqrt{\varepsilon_r \mu_r} = -n$

Since n is negative for a meta-material, -n is a positive quantity. This positive quantity is the magnitude (or absolute value) of the refractive index, |n|. So, $\sqrt{\varepsilon_r \mu_r} = |n|$ .
Final Expression for Speed v Now, substitute $\sqrt{\varepsilon_r \mu_r} = |n|$ back into the expression for v from Step 1: $v = \frac{c}{\sqrt{\varepsilon_r \mu_r}} = \frac{c}{|n|}$ This shows that the speed of light in the meta-material is v = c / |n|.

Evaluation of Options

A: The speed of light in the meta-material is v = c|n|. This is incorrect. It would imply v > c if |n| > 1, violating the principles of relativity for wave propagation in a medium.
B: The speed of light in the meta-material is v = c / |n|. This matches our derived result. It correctly represents the speed as a positive value, determined by the magnitude of the refractive index.
C: The speed of light in the meta-material is v = c. This is incorrect. It would mean the medium is a vacuum, i.e., |n|=1.
D: The wavelength of the light in the meta-material ( $λ_m$ ) is given by $λ_m = λ_{air}|n|$ . The frequency f of light does not change when it enters a new medium. We have the relations: $λ_{air} = c / f$ $λ_m = v / f$ Dividing the second equation by the first gives: $λ_m / λ_{air} = v / c$ From our result v = c / |n|, we get v / c = 1 / |n|. Therefore, $λ_m = λ_{air} / |n|$ . This option is incorrect.

Thus, the only correct statement is B.

Geometrical Optics question

Geometrical Optics question

Step-by-step Derivation

Evaluation of Options

More from Geometrical Optics

Geometrical Optics question

Geometrical Optics question

Step-by-step Derivation

Evaluation of Options

More from Geometrical Optics