### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

MCQ (Single Correct Answer)
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.
2

### AIEEE 2004

MCQ (Single Correct Answer)
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,

$u = 2F = 20cm$
3

### AIEEE 2003

MCQ (Single Correct Answer)
The image formed by an objective of a compound microscope is
A
virtual and diminished
B
real an diminished
C
real and enlarged
D
virtual and enlarged

## Explanation

A real, inverted and enlarged image of the object is formed by the objective lens of a compound microscope.
4

### AIEEE 2003

MCQ (Single Correct Answer)
To get three images of a single object, one should have two plane mirrors at an angle of
A
${60^ \circ }$
B
${90^ \circ }$
C
${120^ \circ }$
D
${30^ \circ }$

## Explanation

When $\theta = {90^ \circ }$ then ${{360} \over \theta } = {{360} \over {90}} = 4$

is an even number. The number of images formed is given by

$n = {{360} \over \theta } - 1 = {{360} \over {90}} - 1 = 4 - 1 = 3$

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