### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2011

A car is fitted with a convex side-view mirror of focal length $20$ $cm$. A second car $2.8m$ behind the first car is overtaking the first car at a relative speed of $15$ $m/s$. The speed of the image of the second car as seen in the mirror of the first one is :
A
${1 \over {15}}\,m/s$
B
$10\,m/s$
C
$15\,m/s$
D
${1 \over {10}}\,m/s$

## Explanation

From mirror formula

${1 \over v} + {1 \over u} = {1 \over f}\,\,\,$

so, $\,\,\,{{dv} \over {dt}} = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$

$\Rightarrow {{dv} \over {dt}} = - {\left( {{f \over {u - f}}} \right)^2}{{du} \over {dt}}$

$\Rightarrow {{dv} \over {dt}} = {1 \over {15}}m/s$
2

### AIEEE 2011

This question has a paragraph followed by two statements, Statement $-1$ and Statement $-2$. Of the given four alternatives after the statements, choose the one that describes the statements.

A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plane. With monochromatic light, this film gives an interference pattern due to light, reflected from the top (convex) surface and the bottom (glass plate) surface of the film.

Statement - $1$: When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of $\pi .$
Statement - $2$: The center of the interference pattern is dark.

A
Statement - $1$ is true, Statement - $2$ is true, Statement - $2$ is the correct explanation of Statement - $1$
B
Statement - $1$ is true, Statement - $2$ is true, Statement - $2$ is not the correct explanation of Statement - $1$
C
Statement - $1$ is false, Statement - $2$ is true
D
Statement - $1$ is true, Statement - $2$ is false.

## Explanation

A phase change of $\pi$ rad appears when the ray reflects at the glass-air interface. Also, the center of the interference pattern is dark.
3

### AIEEE 2011

Let $x$-$z$ plane be the boundary between two transparent media. Medium $1$ in $z \ge 0$ has a refractive index of $\sqrt 2$ and medium $2$ with $z < 0$ has a refractive index of $\sqrt 3 .$ A ray of light in medium $1$ given by the vector $\overrightarrow A = 6\sqrt 3 \widehat i + 8\sqrt 3 \widehat j - 10\widehat k$ is incident on the plane of separation. The angle of refraction in medium $2$ is:
A
${45^ \circ }$
B
${60^ \circ }$
C
${75^ \circ }$
D
${30^ \circ }$

## Explanation

Angle of incidence is given by

$\cos \left( {\pi - i} \right) = {{\left( {6\sqrt 3 \widehat i + 8\sqrt 3 \widehat j - 10\widehat k} \right).\widehat k} \over {20}}$

$- \cos \,i = - {1 \over 2}$

$\angle i = {60^ \circ }$

From Snell's law, $\sqrt 2 \sin i = \sqrt 3 \sin r$

$\Rightarrow$ $\angle r = {45^ \circ }$
4

### 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.