$A$ plane polarised light passes through successive polarisers which are rotated by $30^{\circ}$ with respect to each other in the clockwise direction. Neglecting absorption by the polarisers and given that the first polariser's axis is parallel to the plane of polarisation of the incident light,the intensity of light at the exit of the fifth polariser is closest to

The figure shows a glass plate placed vertically on a horizontal table. $A$ beam of unpolarised light falls on its surface at the polarising angle of $57^\circ$ with the normal. The electric vector in the reflected light on screen $S$ will vibrate with respect to the plane of incidence in a:

Three linear polarisers are placed coaxially. The transmission axis of the first polariser is at $30^{\circ}$,the second one is at $60^{\circ}$,and the third at $90^{\circ}$ to the vertical,all in the clockwise sense. Each polariser additionally absorbs $10 \%$ of the light. If a vertically polarised beam of light of intensity $I=100 \, W/m^2$ is incident on this assembly of polarisers,then the final intensity of the transmitted light will be close to ................ $W/m^2$.

$A$ beam of unpolarized light passes through a tourmaline crystal $A$ and then it passes through a second tourmaline crystal $B$ oriented so that its principal plane is parallel to that of $A$. The intensity of emergent light is $I_0$. Now $B$ is rotated by $45^{\circ}$ about the ray. The emergent light will have intensity $(\cos 45^{\circ} = \frac{1}{\sqrt{2}})$.

Which of the following diagrams represents the variation of the electric field vector magnitude with time for a circularly polarized light?

The angle between the polarizer and the analyzer is $45^o$. The percentage of polarized light passing through the analyzer is........$\%$