 JEE Mains Previous Years Questions with Solutions

4.5
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1

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

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

AIEEE 2006

The refractive index of a glass is $1.520$ for red light and $1.525$ for blue light. Let ${D_1}$ and ${D_2}$ be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then,
A
${D_1} < {D_2}$
B
${D_1} = {D_2}$
C
${D_1}$ can be less than or greater than ${D_2}$ depending upon the angle of prism
D
${D_1} > {D_2}$

Explanation

For a thin prism, $D = \left( {\mu - 1} \right)A$

Since ${\lambda _b} < {\lambda _r} \Rightarrow {\mu _r} < {\mu _b} \Rightarrow {D_1} < {D_2}$
3

AIEEE 2005

When an unpolarized light of intensity ${{I_0}}$ is incident on a polarizing sheet, the intensity of the light which does not get transmitted is
A
${1 \over 4}\,{I_0}$
B
${1 \over 2}\,{I_0}$
C
${I_0}$
D
zero

Explanation

$I = {I_0}{\cos ^2}\theta$

Intensity of polarized light $= {{{I_0}} \over 2}$

$\Rightarrow$ Intensity of untransmitted light $= {I_0} - {{{I_0}} \over 2} = {{{I_0}} \over 2}$
4

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 $4.$