JEE Main 2019 (Online) 9th January Morning Slot | Ray & Wave Optics Question 195 | Physics

JEE Main 2019 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)

-1

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 :

1.1 cm away from the lens

0.55 cm towards the lens

0.55 cm away from the lens

JEE Main 2019 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)

-1

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 :

JEE Main 2019 (Online) 9th January Morning Slot Physics - Ray & Wave Optics Question 195 English

JEE Main 2019 (Online) 9th January Morning Slot Physics - Ray & Wave Optics Question 195 English

$$\sqrt {{5 \over 3}} $$

$${3 \over {\sqrt 5 }}$$

$${5 \over {\sqrt 3 }}$$

$${4 \over 3}$$

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

A ray of light is incident at an angle of 60^o on one face of a prism of angle 30^o. The emergent ray of light makes an angle of 30^o with incident ray. The angle made by the emergent ray with second face of prism will be :

0^o

90^o

45^o

30^o

JEE Main 2018 (Online) 16th April Morning Slot

MCQ (Single Correct Answer)

-1

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 :

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

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

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

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

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