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

MCQ (Single Correct Answer)
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
2

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

MCQ (Single Correct Answer)
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) 11th January Evening Slot

MCQ (Single Correct Answer)
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
4

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

MCQ (Single Correct Answer)
In a double-slit experiment, green light (5303$$\mathop A\limits^ \circ$$) falls on a double slit having a separation of 19.44 $$\mu$$m and awidht of 4.05 $$\mu$$m. The number of bright fringes between the first and the second diffraction minima is :
A
04
B
05
C
10
D
09

## Explanation

For diffraction

location of 1st minime

y1 = $${{D\lambda } \over a}$$ = 0.2469 D$$\lambda$$

location of 2nd minima

y2 = $${{2D\lambda } \over a}$$ = 0.4938 D$$\lambda$$

Now for interference

Path difference at P.

$${{dy} \over D}$$ = 4.8$$\lambda$$

path difference at Q

$${{dy} \over D}$$ = 9.6$$\lambda$$

So orders of maxima in between P & Q is 5, 6, 7, 8, 9

So 5 bright fringes all present between P & Q.

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