1

JEE Main 2019 (Online) 9th January Morning Slot

A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is :
A
1.1 cm away from the lens
B
0
C
0.55 cm towards the lens
D
0.55 cm away from the lens

Explanation

Image is formed on the screen.

So, v = 10 cm

and $\mu$ = $-$ 10 cm

Using formula,

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

$\Rightarrow$   ${1 \over {10}} - {1 \over {\left( { - 10} \right)}}$ = ${1 \over f}$

$\Rightarrow$   f = 5 cm

Now a glass block is placed like this,

Because of this glass block source will move t$\left( {1 - {1 \over \mu }} \right)$ in the direction of incident ray.

$\therefore$   S' = t$\left( {1 - {1 \over \mu }} \right)$

= 1.5$\left( {1 - {2 \over 3}} \right)$

= 0.5

$\therefore$   now distance of source from the lens = 10 $-$ 0.5 = 9.5 cm

$\therefore$   $\mu$ = $-$ 9.5 cm

$\therefore$   ${1 \over v} - {1 \over {\left( { - 9.5} \right)}}$ = ${1 \over 5}$

$\Rightarrow$   ${1 \over v}$ = ${1 \over 5} - {2 \over {19}}$

$\Rightarrow$   ${1 \over v}$ = ${9 \over {95}}$

$\Rightarrow$   v = 10.55 cm

So, to get sharp image screen should shift away (10.55 $-$ 10) = 0.55 cm from the lens.
2

JEE Main 2019 (Online) 9th January Morning Slot

Consider a tank made of glass(refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index $\mu$. A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of $\mu$ is :

A
$\sqrt {{5 \over 3}}$
B
${3 \over {\sqrt 5 }}$
C
${5 \over {\sqrt 3 }}$
D
${4 \over 3}$

Explanation

When light is incident on the liquid at 90o, then from ssnells law,

1.sin90o = $\mu$sin$\theta$

$\Rightarrow$   sin$\theta$ = ${1 \over \mu }$ . . . . . . (1)

Between liquid and glass,

$\mu$sin$\theta$ = 1.5 sin r

$\Rightarrow$   $\mu$sin$\theta$ = 1.5 sin (90o $-$ $\theta$)

$\Rightarrow$   $\mu$ tan$\theta$ = 1.5

$\Rightarrow$   tan$\theta$ = ${{1.5} \over \mu }$

$\therefore$   sin$\theta$ = ${{1.5} \over {\sqrt {{\mu ^2} + {{\left( {1.5} \right)}^2}} }}$ . . . . . . (2)

$\therefore$    From (1) and (2) we get,

${1 \over \mu }$ = ${{1.5} \over {\sqrt {{\mu ^2} + {{\left( {1.5} \right)}^2}} }}$

$\Rightarrow$   $\mu$2 + (1.5)2 = $\mu$2 $\times$ (1.5)2

$\Rightarrow$   $\mu$ = ${3 \over {\sqrt 5 }}$
3

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
4

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