1

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

The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Which of the following graphs is the correct one, if Dm is the angle of minimum deviation?

A
B
C
D

## Explanation

Since Dm = ($\mu$ $-$ 1) A & on increasing the wavelength, $\mu$ decreases & hence Dm decreases.
2

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

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.
3

### 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
4

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