### 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.

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.

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.

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.
3

### 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, $\theta = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$
4

### AIEEE 2009

A mixture of light, consisting of wavelength $590$ $mm$ and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the $4$th bright fringe of the unknown light. From this data, the wavelength of the unknown light is :
A
$885.0$ $nm$
B
$442.5$ $nm$
C
$776.8$ $nm$
D
$393.4$ $nm$

## Explanation

Third bright fringe of known light coincides with the 4th bright fringe of the unknown light.

$\therefore$ ${{3\left( {590} \right)D} \over d} = {{4\lambda D} \over d}$

$\Rightarrow \lambda = {3 \over 4} \times 590$

$= 442.5\,nm$