1

### JEE Main 2019 (Online) 11th January Evening Slot

A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $\sqrt 3$, then the angle of incidence is:
A
60o
B
45o
C
90o
D
30o

## Explanation

i = e

r1 = r2 = ${A \over 2}$ = 30o

by Snell's law

1 $\times$ sin i = $\sqrt 3 \times {1 \over 2} = {{\sqrt 3 } \over 2}$

i = 60
2

### JEE Main 2019 (Online) 12th January Morning Slot

A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be :
A
6 V/m
B
10 V/m
C
30 V/m
D
24 V/m

## Explanation

Prefracted = ${{96} \over {100}}Pi$

$\Rightarrow$  K2A$Pi_t^2$ = ${{96} \over {100}}$ K1A$_i^2$

$\Rightarrow$  r2A$_i^2$ = ${{96} \over {100}}$ r1A$_i^2$

$\Rightarrow$  A$_t^2$ = ${{96} \over {100}} \times {1 \over {{3 \over 2}}} \times {\left( {30} \right)^2}$

A1$\sqrt {{{64} \over {100}} \times {{\left( {30} \right)}^2}} = 24$
3

### JEE Main 2019 (Online) 12th January Morning Slot

A point source of light, S is placed at distance L in front of the centre of plane mirror of width d which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror, at a distance 2L as shown below. The distance over which the man can see the image of the light source in the mirror is

A
d
B
d/2
C
3d
D
2d

## Explanation

3d
4

### JEE Main 2019 (Online) 12th January Morning Slot

What is the position and nature of image formed by lens combination shown in figure ? (f1, f2 are focal lengths)

A
${{20} \over 3}$ cm from point B at right, real
B
70 cm from point B at right ; real
C
40 cm from point B at right; real
D
70 cm from point B at left ; virtual

## Explanation

For first lens

${1 \over V} - {1 \over { - 20}} = {1 \over 5}$

V $=$ ${{20} \over 3}$

For second lens

V = ${{20} \over 3}$ $-$ 2 $=$ ${{14} \over 3}$

${1 \over V} - {1 \over {{{14} \over 3}}} = {1 \over { - 5}}$

V $=$ 70cm