### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2015 (Offline)

Monochromatic light is incident on a glass prism of angle $A$. If the refractive index of the material of the prism is $\mu$, a ray, incident at an angle $\theta$. on the face $AB$ would get transmitted through the face $AC$ of the prism provided :
A
$\theta > {\cos ^{ - 1}}\left[ {\mu \,\sin \left( {A + {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$
B
$\theta < {\cos ^{ - 1}}\left[ {\mu \,\sin \left( {A + {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$
C
$\theta > si{n^{ - 1}}\left[ {\mu \,\sin \left( {A - {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$
D
$\theta < si{n^{ - 1}}\left[ {\mu \,\sin \left( {A - {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$

## Explanation

When ${r_2} = C,\,\angle {N_2}Rc = {90^ \circ }$

Where $C =$ critical angle

As $\sin C = {1 \over v} = \sin {r_2}$

Applying snell's law at $'R'$

$\mu \sin {r_2} = 1\sin {90^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

Applying snell's law at $'Q'$

$1 \times \sin \theta = \mu \sin {r_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

But ${r_1} = A - {r_2}$

So, $\sin \theta = \mu \sin \left( {A - {r_2}} \right)$

$\theta = \mu \sin A\cos {r_2} - \cos \,A\,\,\,\,\,...\left( {iii} \right)$ [using $(i)$]

From $(1)$

$\cos \,{r_2} = \sqrt {1 - {{\sin }^2}{r_2}} = \sqrt {1 - {1 \over {{\mu ^2}}}} \,\,\,\,\,\,\,...\left( {iv} \right)$

By eq.$(iii)$ and $(iv)$

$\sin \theta = \mu \sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - \cos A$

on further solving we can show for ray not to transmitted through face $AC$

$\theta = {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right)} \right.} \right]$

So, for transmission through face $AC$

$\theta > {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right.} \right)} \right]$
2

### 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 $= {{1.22\lambda } \over {2\mu \sin \theta }} = 30\,\mu m.$
3

### 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

4

### 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 ${90^ \circ }.$