### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2015 (Offline)

Assuming human pupil to have a radius of $0.25$ $cm$ and a comfortable viewing distance of $25$ $cm$, the minimum separation between two objects that human eye can resolve at $500$ $nm$ wavelength is :
A
$100\,\mu m$
B
$300\,\mu m$
C
$1\,\mu m$
D
$30\,\mu m$

## Explanation

$\sin \theta = {{0.25} \over {25}} = {1 \over {100}}$

Resolving power $= {{1.22\lambda } \over {2\mu \sin \theta }} = 30\,\mu m.$
2

### 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

3

### 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 ${90^ \circ }.$
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 $\mu {L_2} = {5 \over 3},{f_2} = - 5R$

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