 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2004

The maximum number of possible interference maximum for slit-separation equal to twice the wavelength in Young's double-slit experiment is
A
three
B
five
C
infinite
D
zero

Explanation

For constructive interference $d\,\sin \theta = n\lambda$

given $d = 2\lambda \Rightarrow \sin \theta = {n \over 2}$

$n = 0,1, - 1,2, - 2$ hence five maxima are possible
2

AIEEE 2004

An electromagnetic wave of frequency $v=3.0$ $MHz$ passes from vacuum into a dielectric medium with permittivity $\in = 4.0.$ Then
A
wave length is halved and frequency remains unchanged
B
wave length is doubled and the frequency becomes half
C
wave length is doubled and the frequency remains unchanged
D
wave length and frequency both remain unchanged. A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is ${4 \over 3}$ and the fish is $12$ $cm$ below the surface, the radius of this circle in $cm$

Explanation

Frequency remains constant during refraction

${v_{med}} = {1 \over {\sqrt {{\mu _0}{ \in _0} \times 4} }} = {c \over 2}$

${{{\lambda _{med}}} \over {{\lambda _{air}}}} = {{{v_{med}}} \over {{v_{air}}}} = {{c/2} \over c} = {1 \over 2}$

$\therefore$ wavelength is halved and frequency remains unchanged
3

AIEEE 2004

The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$) is
A
${\tan ^{ - 1}}\left( {1/n} \right)$
B
${\sin ^{ - 1}}\left( {1/n} \right)$
C
${\sin ^{ - 1}}\left( n \right)$
D
${\tan ^{ - 1}}\left( n \right)$

Explanation

The angle of incidence for total polarization is given by

$\tan \theta = n \Rightarrow \theta = {\tan ^{ - 1}}n$

Where $n$ is the refractive index of the glass.
4

AIEEE 2004

A plano convex lens of refractive index $1.5$ and radius of curvature $30$ $cm$. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object
A
$60$ $cm$
B
$30$ $cm$
C
$20$ $cm$
D
$80$ $cm$

Explanation

KEY CONCEPT : The focal length $\left( F \right)$ of the final mirror

is ${1 \over F} = {2 \over {f\ell }} + {1 \over {{f_m}}}$

Here ${1 \over {{f_\ell }}} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$= \left( {1.5 - 1} \right)\left[ {{1 \over \alpha } - {1 \over { - 30}}} \right] = {1 \over {60}}$

$\therefore$ ${1 \over F} = 2 \times {1 \over {60}} + {1 \over {30/2}} = {1 \over {10}}$

$\therefore$ $F=10cm$

The combination acts as a converging mirror. For the object to be of the same size of mirror,

$u = 2F = 20cm$