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 1 Offline
MCQ (Single Correct Answer)
+3
-1

A biconvex lens is formed with two planoconvex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both curved surface are of the same radius of curvature R = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be

IIT-JEE 2012 Paper 1 Offline Physics - Geometrical Optics Question 27 English

A
$$-$$280.0 cm
B
40.0 cm
C
21.5 cm
D
13.3 cm
4
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

A
virtual and at a distance of 16 cm from the mirror.
B
real and at a distance of 16 cm from the mirror.
C
virtual and at a distance of 20 cm from the mirror.
D
real and at a distance of 20 cm from the mirror.
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