1
IIT-JEE 2012 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

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 $$n = \left( {{c \over v}} \right) = \pm \,\sqrt {{\varepsilon _r}{\mu _r}} $$, 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

A
IIT-JEE 2012 Paper 2 Offline Physics - Geometrical Optics Question 26 English Option 1
B
IIT-JEE 2012 Paper 2 Offline Physics - Geometrical Optics Question 26 English Option 2
C
IIT-JEE 2012 Paper 2 Offline Physics - Geometrical Optics Question 26 English Option 3
D
IIT-JEE 2012 Paper 2 Offline Physics - Geometrical Optics Question 26 English Option 4
2
IIT-JEE 2012 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

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 $$n = \left( {{c \over v}} \right) = \pm \,\sqrt {{\varepsilon _r}{\mu _r}} $$, 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.

A
The speed of light in the meta-material is v = c|n|.
B
The speed of light in the meta-material is $$v = {c \over {|n|}}$$.
C
The speed of light in the meta-material is v = c.
D
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.
3
IIT-JEE 2012 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2

In the given circuit, the AC source has $$\omega$$ = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

IIT-JEE 2012 Paper 2 Offline Physics - Alternating Current Question 7 English

A
The current through the circuit, I is 0.3 A.
B
The current through the circuit, I is 0.3$$\sqrt2$$ A.
C
The voltage across 100 $$\Omega$$ resistor = 10$$\sqrt2$$ V.
D
The voltage across 50 $$\Omega$$ resistor = 10 V.
4
IIT-JEE 2012 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2

A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it, the correct statement(s) is(are)

A
the emf induced in the loop is zero if the current is constant.
B
the emf induced in the loop is finite if the current is constant.
C
the emf induced in the loop is zero if the current decreases at a steady state
D
the emf induced in the loop is finite if the current decreases at a steady state.
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