1

### JEE Main 2018 (Online) 16th April Morning Slot

Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is $\theta$. Then :
A
cos$\theta$ = ${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$
B
cos$\theta$ = ${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 4}}}}$
C
cos$\theta$ = ${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}$
D
cos$\theta$ = ${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 4}}}}$
2

### 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.
3

### 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

$\Rightarrow$   $\mu$ = ${3 \over {\sqrt 5 }}$
4

### 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