1

### 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

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

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

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

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

### 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
3

### 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 = ${\lambda \over {2\left( {\sqrt 5 - 2} \right)}}$
4

### 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$\left( {{\pi \over 8}} \right)$