1

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

A ray of light is incident at an angle of 60o on one face of a prism of angle 30o. The emergent ray of light makes an angle of 30o with incident ray. The angle made by the emergent ray with second face of prism will be :
A
0o
B
90o
C
45o
D
30o
2

### 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$\theta$ = ${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}}$
B
cos$\theta$ = ${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{4}}}}$
C
cos$\theta$ = ${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{2}}}}$
D
cos$\theta$ = ${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{4}}}}$
3

### 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.
4

### 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

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