 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

AIEEE 2009

A mixture of light, consisting of wavelength $590$ $mm$ and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the $4$th bright fringe of the unknown light. From this data, the wavelength of the unknown light is :
A
$885.0$ $nm$
B
$442.5$ $nm$
C
$776.8$ $nm$
D
$393.4$ $nm$

Explanation

Third bright fringe of known light coincides with the 4th bright fringe of the unknown light.

$\therefore$ ${{3\left( {590} \right)D} \over d} = {{4\lambda D} \over d}$

$\Rightarrow \lambda = {3 \over 4} \times 590$

$= 442.5\,nm$
2

AIEEE 2008

A student measures the focal length of a convex lens by putting an object pin at a distance $'u'$ from the lens and measuring the distance $'v'$ of the image pin. The graph between $'u'$ and $'v'$ plotted by the student should look like
A B C D Explanation

This graph obeys the lens equation

${1 \over v} - {1 \over u} = {1 \over f}$

where $f$ is a positive constant for a given convex lens.
3

AIEEE 2007

Two lenses of power $-15$ $D$ and $+5$ $D$ are in contact with each other. The focal length of the combination is
A
$+ 10\,cm$
B
$- 20\,cm$
C
$- 10\,cm$
D
$+ 20\,cm$

Explanation

Power of combination is given by

$P = {P_1} + {P_2} = \left( { - 15 + 5} \right)D$ $\,\,\,\,\,\,\,\,\, = - 10D.$

Now, $P = {1 \over f} \Rightarrow f = {1 \over P} = {1 \over { - 10}}$ metre

$\therefore$ $f = - \left( {{1 \over {10}} \times 100} \right)cm = - 10\,cm.$
4

AIEEE 2007

In a Young's double slit experiment the intensity at a point where the path difference is ${\lambda \over 6}$ ( $\lambda$ being the wavelength of light used ) is $I$. If ${I_0}$ denotes the maximum intensity, ${I \over {{I_0}}}$ is equal to
A
${3 \over 4}$
B
${1 \over {\sqrt 2 }}$
C
${{\sqrt 3 } \over 2}$
D
${1 \over 2}$

Explanation

The intensity of light at any point of the screen where the phase difference due to light coming from the two slits is $\phi$ is given by

$I = {I_0}{\cos ^2}\left( {{\phi \over 2}} \right)\,\,$ where ${I_0}$ is the maximum intensity.

NOTE : This formula is applicable when ${I_1} = {I_2}.$

Here $\phi = {\raise0.5ex\hbox{\pi } \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{3}}$

$\therefore$ ${I \over {{I_0}}} = {\cos ^2}{\pi \over 6} = {\left( {{{\sqrt 3 } \over 2}} \right)^2} = {3 \over 4}$