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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

MCQ (Single Correct Answer)
An object $$2.4$$ $$m$$ in front of a lens forms a sharp image on a film $$12$$ $$cm$$ behind the lens. A glass plate $$1$$ $$cm$$ thick, of refractive index $$1.50$$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?
A
$$7.2$$ $$m$$
B
$$24$$ $$m$$
C
$$3.2$$ $$m$$
D
$$5.6$$ $$m$$

Explanation

The focal length of the lens

$${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$$

$$ = {{20 + 1} \over {240}} = {{21} \over {240}}$$

$$f = {{240} \over {21}}cm$$

Shift$$ = t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)$$

$$ = 1 \times {1 \over 3}$$

Now $$v' = 12 - {1 \over 3} = {{35} \over 3}cm$$

Now the object distancce $$u.$$

$${1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]$$

$${1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]$$

$$u = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m$$
2

AIEEE 2012

MCQ (Single Correct Answer)
An electromagnetic wave in vacuum has the electric and magnetic field $$\mathop E\limits^ \to $$and $$\mathop B\limits^ \to $$, which are always perpendicular to each other. The direction of polarization is given by $$\mathop X\limits^ \to $$ and that of wave propagation by $$\mathop k\limits^ \to $$. Then
A
$$\mathop X\limits^ \to ||\mathop B\limits^ \to $$ and $$\mathop X\limits^ \to ||\mathop B\limits^ \to \times \mathop E\limits^ \to $$
B
$$\mathop X\limits^ \to ||\mathop E\limits^ \to $$ and $$\mathop k\limits^ \to ||\mathop E\limits^ \to \times \mathop B\limits^ \to $$
C
$$\mathop X\limits^ \to ||\mathop B\limits^ \to $$ and $$\mathop k\limits^ \to ||\mathop E\limits^ \to \times \mathop B\limits^ \to $$
D
$$\mathop X\limits^ \to ||\mathop E\limits^ \to $$ and $$\mathop k\limits^ \to ||\mathop B\limits^ \to \times \mathop E\limits^ \to $$

Explanation

as The $$E.M.$$ wave are transverse in nature i.e.,

$$ = {{\overrightarrow k \times \overrightarrow E } \over {\mu \omega }} = \overrightarrow H \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

where $$\overrightarrow H = {{\overrightarrow B } \over \mu }$$

and $${{\overrightarrow k \times \overrightarrow H } \over {\omega \varepsilon }} = - \overrightarrow E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$\overrightarrow k $$ is $$ \bot \,\,\overrightarrow H $$ and $$\overrightarrow k $$ is also $$ \bot $$ to $$\overrightarrow E $$

or In other words $$\overrightarrow X ||\overrightarrow E $$ and $$\overrightarrow k ||\overrightarrow E \times \overrightarrow B $$
3

AIEEE 2012

MCQ (Single Correct Answer)
In Young's double slit experiment , one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If $${{\rm I}_m}$$ be the maximum intensity, the resultant intensity $${\rm I}$$ when they interfere at phase difference $$\phi $$ is given by :
A
$${{{I_m}} \over 9}\left( {4 + 5\cos \,\phi } \right)$$
B
$${{{I_m}} \over 3}\left( {1 + 2{{\cos }^2}\,{\phi \over 2}} \right)$$
C
$${{{I_m}} \over 3}\left( {1 + 4{{\cos }^2}\,{\phi \over 2}} \right)$$
D
$${{{I_m}} \over 9}\left( {1 + 8{{\cos }^2}\,{\phi \over 2}} \right)$$

Explanation

Let $${a_1} = a,\,{I_1} = a_1^2 = {a^2}$$

$${a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}$$

Therefore $${{\rm I}_2} = 4{{\rm I}_1}$$

$${I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $$

$${I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi $$

$$ \Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Now, $${I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}$$

$${I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}$$

Substituting in equation $$\left( 1 \right)$$

$${I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi $$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]$$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]$$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]$$
4

AIEEE 2011

MCQ (Single Correct Answer)
A car is fitted with a convex side-view mirror of focal length $$20$$ $$cm$$. A second car $$2.8m$$ behind the first car is overtaking the first car at a relative speed of $$15$$ $$m/s$$. The speed of the image of the second car as seen in the mirror of the first one is :
A
$${1 \over {15}}\,m/s$$
B
$$10\,m/s$$
C
$$15\,m/s$$
D
$${1 \over {10}}\,m/s$$

Explanation

From mirror formula

$${1 \over v} + {1 \over u} = {1 \over f}\,\,\,$$

so, $$\,\,\,{{dv} \over {dt}} = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$$

$$ \Rightarrow {{dv} \over {dt}} = - {\left( {{f \over {u - f}}} \right)^2}{{du} \over {dt}}$$

$$ \Rightarrow {{dv} \over {dt}} = {1 \over {15}}m/s$$

Questions Asked from Ray & Wave Optics

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