 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2013 (Offline)

Two coherent point sources ${S_1}$ and ${S_2}$ are separated by a small distance $'d'$ as shown. The fringes obtained on the screen will be A
points
B
straight lines
C
semi-circles
D
concentric circles

Explanation

It will be concentric circles.
2

JEE Main 2013 (Offline)

A beam of unpolarised light of intensity ${{\rm I}_0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of ${45^ \circ }$ relative to that of $A$. The intensity of the emergent light is
A
${{\rm I}_0}$
B
${{\rm I}_0}/2$
C
${{\rm I}_0}/4$
D
${{\rm I}_0}/8$

Explanation

Relation between intensities ${{\rm I}_r} = \left( {{{{{\rm I}_0}} \over 2}} \right){\cos ^2}\left( {{{45}^ \circ }} \right) = {{{{\rm I}_0}} \over 2} \times {1 \over 2} = {{{{\rm I}_0}} \over 4}$
3

JEE Main 2013 (Offline)

Diameter of a plano-convex lens is $6$ $cm$ and thickness at the center is $3mm$. If speed of light in material of lens is $2 \times {10^8}\,m/s,$ the focal length of the lens is
A
$15$ $cm$
B
$20$ $cm$
C
$30$ $cm$
D
$10$ $cm$

Explanation $\therefore$ $n = {{Velocity\,\,of\,\,light\,\,in\,\,vacuum} \over {Velocity\,\,of\,\,light\,\,in\,\,medium}}$

$\therefore$ $n = {3 \over 2}$

${3^2} + {\left( {R - 3mm} \right)^2} = {R^2}$

$\Rightarrow {3^2} + {R^2} - 2R\left( {3mm} \right) + {\left( {3mm} \right)^2} = {R^2}$

$\Rightarrow R \approx 15\,cm$

${1 \over f} = \left( {{3 \over 2} - 1} \right)\left( {{1 \over {15}}} \right) \Rightarrow f = 30\,cm$
4

AIEEE 2012

An object $2.4$ $m$ in front of a lens forms a sharp image on a film $12$ $cm$ behind the lens. A glass plate $1$ $cm$ thick, of refractive index $1.50$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?
A
$7.2$ $m$
B
$24$ $m$
C
$3.2$ $m$
D
$5.6$ $m$

Explanation

The focal length of the lens

${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$

$= {{20 + 1} \over {240}} = {{21} \over {240}}$

$f = {{240} \over {21}}cm$

Shift$= t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)$

$= 1 \times {1 \over 3}$

Now $v' = 12 - {1 \over 3} = {{35} \over 3}cm$

Now the object distancce $u.$

${1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]$

${1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]$

$u = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m$