1

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

The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separateds two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus -
A
2 cm
B
3.1 cm
C
4.0 cm
D
1 cm

## Explanation

${{1.34} \over V} - {1 \over \infty } = {{1.34 - 1} \over {7.8}}$

$\therefore$  V = 30.7 mm
2

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

Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit (S1) ?

A
${\lambda \over {2\left( {5 - \sqrt 2 } \right)}}$
B
${\lambda \over {2\left( {\sqrt 5 - 2} \right)}}$
C
${\lambda \over {\left( {5 - \sqrt 2 } \right)}}$
D
${\lambda \over {\left( {\sqrt 5 - 2} \right)}}$

## Explanation

Path difference, S2P – S1P = ${\lambda \over 2}$

$\Rightarrow$ $\sqrt {4{d^2} + {d^2}} - 2d = {\lambda \over 2}$

$\Rightarrow$ $d\left( {\sqrt 5 - 2} \right) = {\lambda \over 2}$

$\Rightarrow$ d = ${\lambda \over {2\left( {\sqrt 5 - 2} \right)}}$
3

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

In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is ${1 \over 8}$ th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to :
A
0.94
B
0.85
C
0.74
D
0.80

## Explanation

$\Delta$x $=$ ${\lambda \over 8}$

$\Delta$$\phi$ $=$ ${{\left( {2\pi } \right)} \over \lambda }{\lambda \over 8} = {\pi \over 4}$

I $=$ I0cos2$\left( {{\pi \over 8}} \right)$

${{\rm I} \over {{{\rm I}_0}}} =$ cos2$\left( {{\pi \over 8}} \right)$
4

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

An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be :
A
1.16 $\times$ 10–3 m/s towards the lens
B
2.26 $\times$ 10–3 m/s away from the lens
C
3.22 × 10–3 m/s towards the lens
D
0.92 $\times$ 10$-$3 m/s away from the lens

## Explanation

From lens equation

${1 \over v} - {1 \over u} = {1 \over f}$

${1 \over v} - {1 \over {\left( { - 20} \right)}} = {1 \over {\left( {.3} \right)}} = {{10} \over 3}$

${1 \over v} = {{10} \over 3} - {1 \over {20}}$

${1 \over v} = {{197} \over {60}};v = {{60} \over {197}}$

m = $\left( {{v \over u}} \right)$ = ${{\left( {{{60} \over {197}}} \right)} \over {20}}$

velocity of image wrt. to lens is given by

vI/L = m2vO/L

direction of velocity of image is same as that of object

vO/L = 5 m/s