 JEE Mains Previous Years Questions with Solutions

4.5
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1

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 ${\lambda _b} < {\lambda _r} \Rightarrow {\mu _r} < {\mu _b} \Rightarrow {D_1} < {D_2}$
2

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 $= {I_0} - {{{I_0}} \over 2} = {{{I_0}} \over 2}$
3

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 $4.$
4

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

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

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

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