1

### 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)
2

### 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

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

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

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

$2{\mu _1} - {\mu _2} = 1$
3

### 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
4

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