1

### JEE Main 2019 (Online) 9th January Evening Slot

In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength $\lambda$ = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range $-$ 30o$\le $$\theta$$ \le$30o is :
A
640
B
320
C
321
D
641

## Explanation

We know, path difference,

d sin$\theta$ = n$\lambda$

here n = no of bright fringer in the angle

here given

d = 0.32 $\times$ 10-3 m

$\lambda$ = 500 $\times$ 10$-$9 m

$\therefore$  0.32 $\times$ 10$-$3 sin30o = n $\times$ 500 $\times$ 10$-$9

$\Rightarrow$  n = 320

Total number of maxima in the range

$-$ 30o $\theta$ $\le$ 30o is = 320 $\times$ 2 + 1 = 641
2

### JEE Main 2019 (Online) 9th January Evening Slot

Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror (M1) and parallel to the second mirror (M2) is finally reflected from the second mirror (M2) parallel to the first mirror (M1). The angle between the two mirrors will be :
A
45o
B
60o
C
75o
D
90o

## Explanation

Let the angle between two mirrors are $\theta$.

We know sum of angles of triangle = 180o

$\therefore$  3$\theta$ = 180o

$\Rightarrow$  $\theta$ = 60o
3

### JEE Main 2019 (Online) 10th January Morning Slot

In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle ${1 \over {40}}$ by using light of wavelength $\lambda$1. When the light of wavelength $\lambda$2 is used a bright fringe is seen at the same angle in the same set up. Given that $\lambda$1 and $\lambda$2 are in visible range (380 nm to 740 nm), their values are -
A
400 nm, 500 nm
B
625 nm, 500 nm
C
380 nm, 500 nm
D
380 nm, 525 nm

## Explanation

Path difference = d sin$\theta$ $\approx$ d$\theta$

= 0.1 $\times$ ${1 \over {40}}$ mm = 2500nm

or bright fringe, path difference must be integral multiple of $\lambda$.

$\therefore$   2500 = n$\lambda$1 = m$\lambda$2

$\therefore$   $\lambda$1 = 625, $\lambda$2 = 500 (from m = 5)

(for n = 4)
4

### JEE Main 2019 (Online) 10th January Morning Slot

A plano convex lens of refractive index $\mu$1 and focal length ƒ1 is kept in contact with another plano concave lens of refractive index $\mu$2 and focal length ƒ2. If the radius of curvature of their spherical faces is R each and ƒ1 = 2ƒ2, then $\mu$1 and $\mu$2 are related as -
A
$3{\mu _2} - 2{\mu _1}$ = 1
B
${\mu _1} + {\mu _2}$ = 3
C
$2{\mu _1} - {\mu _2}$ = 1
D
$2{\mu _2} - {\mu _1}$ = 1

## Explanation

${1 \over {2{f_2}}} = {1 \over {{f_1}}} = \left( {{\mu _1} - 1} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right)$

${1 \over {{f_2}}} = \left( {{\mu _2} - 1} \right)\left( {{1 \over { - R}} - {1 \over \infty }} \right)$

${{\left( {{\mu _1} - 1} \right)} \over R} = {{\left( {{\mu _2} - 1} \right)} \over {2R}}$

$2{\mu _1} - {\mu _2} = 1$