### JEE Mains Previous Years Questions with Solutions

4.5
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1

### JEE Main 2016 (Offline)

An observer looks at a distant tree of height $10$ $m$ with a telescope of magnifying power of $20.$ To the observer the tree appears:
A
$20$ times taller
B
$20$ times nearer
C
$10$ times taller
D
$10$ times nearer

## Explanation

A telescope magnifies by making the object appearing closer.
2

### JEE Main 2016 (Offline)

The box of a pin hole camera, of length $L,$ has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say ${b_{\min }}$) when :
A
$a = \sqrt {\lambda L} \,$ and ${b_{\min }} = \sqrt {4\lambda L}$
B
$a = {{{\lambda ^2}} \over L}$ and ${b_{\min }} = \sqrt {4\lambda L}$
C
$a = {{{\lambda ^2}} \over L}$ and ${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$
D
$a = \sqrt {\lambda 1}$ and ${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$

## Explanation

Given geometrical spread $=a$

Diffraction spread $= {\lambda \over a} \times L = {{\lambda L} \over a}$

The sum $b = a + {{\lambda L} \over a}$

For $b$ to be minimum ${{db} \over {da}} = 0$ ${d \over {da}}\left( {a + {{\lambda L} \over a}} \right) = 0$

$a = \sqrt {\lambda L}$

$b\min = \sqrt {\lambda L} + \sqrt {\lambda L} = 2\sqrt {\lambda L} = \sqrt {4\lambda L}$
3

### 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 $AC$

$\theta > {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right.} \right)} \right]$
4

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