### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2005

If ${I_0}$ is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?
A
$4{I_0}$
B
$2{I_0}$
C
${{{I_0}} \over 2}$
D
${I_0}$

## Explanation

$I = {I_0}{\left( {{{\sin \phi } \over \phi }} \right)^2}\,\,$ and $\phi = {\pi \over \lambda }\left( {b\,\sin \theta } \right)$

When the slit width is doubled, the amplitude of the wave at the center of the screen is doubled, so the intensity at the center is increased by a factor $4.$
2

### AIEEE 2005

A thin glass (refractive index $1.5$) lens has optical power of $-5$ $D$ in air. Its optical power in a liquid medium with refractive index $1.6$ will be
A
$-1$ $D$
B
$1$ $D$
C
$-25$ $D$
D
$25$ $D$

## Explanation

${1 \over {{f_a}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

${1 \over {{f_m}}} = \left( {{{{\mu _g}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

${1 \over {{f_m}}} = \left( {{{1.5} \over {1.6}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Dividing $(i)$ by $(ii)$, ${{{f_m}} \over {{f_a}}} = \left( {{{1.5 - 1} \over {{{1.5} \over {1.6}} - 1}}} \right) = - 8$

${P_a} = - 5 = {1 \over {{f_a}}} \Rightarrow {f_a} = - {1 \over 5}$

$\Rightarrow {f_m} = - 8 \times {f_a} = - 8 \times - {1 \over 5} = {8 \over 5}$

${P_m} = {\mu \over {{f_m}}} = {{1.6} \over 8} \times 5 = 1D$
3

### AIEEE 2005

A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is
A
circle
B
hyperbola
C
parabola
D
straight line

## Explanation

The shape of interference fringes formed on a screen in case of a monochromatic source is a straight line. Remember for double hole experiment a hyperbola is generated.
4

### AIEEE 2005

Two point white dots are $1$ $mm$ apart on a black paper. They are viewed by eye of pupil diameter $3$ $mm.$ Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light $=500$ $nm$ ]
A
$1m$
B
$5m$
C
$3m$
D
$6m$

## Explanation

${y \over D} \ge 1.22{\lambda \over d}$

$\Rightarrow D \le {{yd} \over {\left( {1.22} \right)\lambda }}$

$= {{{{10}^{ - 3}} \times 3 \times {{10}^{ - 3}}} \over {\left( {1.22} \right) \times 5 \times {{10}^{ - 7}}}}$

$= {{30} \over {61}} \approx 5m$

$\therefore$ ${D_{\max }} = 5m$

### Graduate Aptitude Test in Engineering

GATE CSE GATE EE GATE ECE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12