### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2010

An initially parallel cylindrical beam travels in a medium of refractive index $\mu \left( I \right) = {\mu _0} + {\mu _2}\,I,$ where ${\mu _0}$ and ${\mu _2}$ are positive constants and $I$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

The speed of light in the medium is

A
minimum on the axis of the beam
B
the same everywhere in the beam
C
directly proportional to the intensity $I$
D
maximum on the axis of the beam

## Explanation

The speed of light $(c)$ in a medium of refractive index $\left( \mu \right)$ is given by

$\mu = {{{c_0}} \over c},$ where ${c_0}$ is the speed of light in vacuum

$\therefore$ $c = {{{c_0}} \over \mu } = {{{c_0}} \over {{\mu _0} + {\mu _2}\left( I \right)}}$

As $I$ is decreasing with increasing radius, it is maximum

on the axis of the beam. Therefore, $c$ is minimum on the axis of the beam.
2

### AIEEE 2010

An initially parallel cylindrical beam travels in a medium of refractive index $\mu \left( I \right) = {\mu _0} + {\mu _2}\,I,$ where ${\mu _0}$ and ${\mu _2}$ are positive constants and $I$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

As the beam enters the medium, it will

A
diverge
B
converge
C
diverge near the axis and converge near the periphery
D
travel as a cylindrical beam

## Explanation

In the medium, the refractive index will decreases from the axis forwards the periphery of the beam.

Therefore, the beam will move as one move from the axis to the periphery and hence the beam will converge.

3

### AIEEE 2010

An initially parallel cylindrical beam travels in a medium of refractive index $\mu \left( I \right) = {\mu _0} + {\mu _2}\,I,$ where ${\mu _0}$ and ${\mu _2}$ are positive constants and $I$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

The initial shape of the wavefront of the beam is

A
convex
B
concave
C
convex near the axis and concave near the periphery
D
planar

## Explanation

Initially the parallel beam is cylindrical. Therefore, the wave-front will be planar.
4

### AIEEE 2009

A transparent solid cylindrical rod has a refractive index of ${2 \over {\sqrt 3 }}.$ It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure.

The incident angle $\theta$ for which the light ray grazes along the wall of the rod is :

A
${\sin ^{ - 1}}\left( {{\raise0.5ex\hbox{\scriptstyle {\sqrt 3 }} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}} \right)$
B
${\sin ^{ - 1}}\left( {{\raise0.5ex\hbox{\scriptstyle 2} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle {\sqrt 3 }}}} \right)$
C
${\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$
D
${\sin ^{ - 1}}\left( {{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}} \right)$

## Explanation

Applying Snell's law at $Q$

$n = {{\sin {{90}^ \circ }} \over {\sin C}} = {1 \over {\sin C}}$

$\therefore$ $\sin C = {1 \over n} = {{\sqrt 3 } \over 2}$

$\therefore$ $C = {60^ \circ }$

Applying Snell's Law at $P$

$n = {{\sin \theta } \over {\sin \left( {90 - C} \right)}}$

$\Rightarrow \sin \theta = n \times \sin \left( {90 - C} \right);$ from $(1)$

$\Rightarrow \sin \theta = n\,\cos$

$\therefore$ $\theta = {\sin ^{ - 1}}\left[ {{2 \over {\sqrt 3 }} \times \cos {{60}^0}} \right]$

or, $\theta = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$