 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2005

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 $cm$ is
A
${{36} \over {\sqrt 7 }}$
B
${36\sqrt 7 }$
C
${4\sqrt 5 }$
D
${36\sqrt 5 }$

Explanation

$\sin {\theta _c} = {1 \over \mu } = {3 \over 4}$

or $\tan {\theta _c} = {3 \over {\sqrt {16 - 9} }} = {3 \over {\sqrt 7 }} = {R \over {12}}$ $\Rightarrow R = {{36} \over {\sqrt 7 }}\,cm$
2

AIEEE 2004

A light ray is incident perpendicularly to one face of a ${90^ \circ }$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is ${45^ \circ }$, we conclude that the refractive index $n$ A
$n > {1 \over {\sqrt 2 }}$
B
$n > \sqrt 2$
C
$n < {1 \over {\sqrt 2 }}$
D
$n < \sqrt 2$

Explanation

The incident angle is ${45^ \circ }$

Incident angle $>$ critical angle, $i > {i_c}$

$\therefore$ $\sin i > \sin {i_c}$ or $\sin \,45\, > \sin \,{i_c},$ $\sin {i_c} = {1 \over n}$

$\therefore$ $\sin \,{45^ \circ } > {1 \over n}$ or ${1 \over {\sqrt 2 }} > {1 \over n} \Rightarrow n > \sqrt 2$
3

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
4

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