The specific rotation of a sugar solution is $0.01 \ SI$ units. $A$ $200 \ kg/m^3$ solution of impure sugar is taken in a polarimeter tube of length $0.25 \ m$,and an optical rotation of $0.4 \ rad$ is observed. The percentage of purity of sugar in the sample is.....$\%$

$A$ 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 $(\cos 45^{\circ} = \frac{1}{\sqrt{2}})$.

The relation $I = I_0 \cos^2 \theta$ is known as (where $I_0$ is the intensity of incident light on the analyser,$I$ is the intensity of emergent light from the analyser,and $\theta$ is the angle between the plane of polarization and the axis of the analyser):

$A$ point source $S$ emits unpolarized light uniformly in all directions. At two points $A$ and $B$,the ratio $r = I_A / I_B$ of the intensities of light is $2$. If a set of two polaroids having $45^{\circ}$ angle between their pass-axes is placed just before point $B$,then the new value of $r$ will be.

Two polaroids $A$ and $B$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now,another polaroid $C$ is placed between $A$ and $B$ bisecting the angle between them. If the intensity of unpolarised light is $I_0$,then the intensity of transmitted light after passing through polaroid $B$ will be:

$A$ polarizer and an analyzer are placed such that the intensity of the transmitted light is maximum. If the analyzer is rotated by $60^{\circ}$,find the intensity of the transmitted light.