 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

In Young's double slit experiment , one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If ${{\rm I}_m}$ be the maximum intensity, the resultant intensity ${\rm I}$ when they interfere at phase difference $\phi$ is given by :
A
${{{I_m}} \over 9}\left( {4 + 5\cos \,\phi } \right)$
B
${{{I_m}} \over 3}\left( {1 + 2{{\cos }^2}\,{\phi \over 2}} \right)$
C
${{{I_m}} \over 3}\left( {1 + 4{{\cos }^2}\,{\phi \over 2}} \right)$
D
${{{I_m}} \over 9}\left( {1 + 8{{\cos }^2}\,{\phi \over 2}} \right)$

Explanation

Let ${a_1} = a,\,{I_1} = a_1^2 = {a^2}$

${a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}$

Therefore ${{\rm I}_2} = 4{{\rm I}_1}$

${I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi }$

${I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi$

$\Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Now, ${I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}$

${I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}$

Substituting in equation $\left( 1 \right)$

${I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi$

${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]$

${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]$

${I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]$
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

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

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

$\Rightarrow {{dv} \over {dt}} = {1 \over {15}}m/s$
4

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, $\sqrt 2 \sin i = \sqrt 3 \sin r$

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