JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2015 (Offline)

On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels, the light beam :
A
bends down wards
B
bends upwards
C
becomes narrower
D
goes horizontally without any deflection

Explanation

2

JEE Main 2014 (Offline)

A green light is incident from the water to the air - water interface at the critical angle $\left( \theta \right)$. Select the correct statement.
A
The entire spectrum of visible light will come out of the water at an angle of ${90^ \circ }$ to the normal.
B
The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.
C
The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.
D
The entire spectrum of visible light will come out of the water at various angles to the normal.

Explanation

For critical angle ${\theta _c},$

$\sin {\theta _c} = {1 \over \mu }$

For greater wavelength or lesser frequency $\mu$ is less.

So, critical angle would be more, So, they will not suffer reflection and come out at angles less then ${90^ \circ }.$
3

JEE Main 2014 (Offline)

Two beams, $A$ and $B$, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam $A$ has maximum intensity (and beam $B$ has zero intensity), a rotation of polaroid through ${30^ \circ }$ makes the two beams appear equally bright. If the initial intensities of the two beams are ${{\rm I}_A}$ and ${{\rm I}_B}$ respectively, then ${{{{\rm I}_A}} \over {{{\rm I}_B}}}$ equals:
A
$3$
B
${3 \over 2}$
C
$1$
D
${1 \over 3}$

Explanation

According to malus law, intensity of emerging beam is given by,

$I = {I_0}{\cos ^2}\theta$

Now, ${I_{A'}} = {I_A}{\cos ^2}{30^ \circ }$

${I_{B'}} = {I_B}{\cos ^2}{60^ \circ }$

As ${I_{A'}} = {I_{B'}}$

$\Rightarrow {I_A} \times {3 \over 4} = {I_B} \times {1 \over 4}$

$\therefore$ ${{{I_A}} \over {{I_B}}} = {1 \over 3}$
4

JEE Main 2014 (Offline)

A thin convex lens made from crown glass $\left( {\mu = {3 \over 2}} \right)$ has focal length $f$. When it is measured in two different liquids having refractive indices ${4 \over 3}$ and ${5 \over 3},$ it has the focal lengths ${f_1}$ and ${f_2}$ respectively. The correct relation between the focal lengths is :
A
${f_1} = {f_2} < f$
B
${f_1} > f$ and ${f_2}$ becomes negative
C
${f_2} > f$ and ${f_1}$ becomes negative
D
${f_1}\,$ and${f_2}\,$ both become negative

Explanation

By Lens maker's formula for convex lens

${1 \over f} = \left( {{\mu \over {{\mu _L}}} - 1} \right)\left( {{2 \over R}} \right)$

for, $\mu {L_1} = {4 \over 3},{f_1} = 4R$

for $\mu {L_2} = {5 \over 3},{f_2} = - 5R$

$\Rightarrow {f_2} = \left( - \right)ve$