Two polaroids have their polarizing directions parallel so that the intensity of a transmitted light is maximum. The angle through which either polaroid must be turned if the intensity is to drop by one-half is.....$^o$

$A$ cylindrical tube of length $0.2 \,m$ and radius $R$ with sugar solution of concentration $C$ produces a rotation of $\theta$ in the plane of vibration of a plane-polarized light. The same sugar solution is transferred to another tube of length $0.3 \,m$ of the same radius. The remaining gap is filled by distilled water. Now, the optical rotation produced is:

$A$ beam of natural light falls on a system of $6$ polaroids,which are arranged in succession such that each polaroid is turned through $30^\circ$ with respect to the preceding one. The percentage of incident intensity that passes through the system will be....$\%$

$A$ polariser is used to

$A$ mixture of plane-polarized and unpolarized light falls normally on a polarizing sheet. On rotating the polarizing sheet about the direction of the incident beam,we find that the ratio of maximum to minimum transmitted intensity is $4$. Find the ratio of the intensities $I_P$ and $I_0$,respectively,of the polarized and unpolarized components in the incident beam.

Three polarising sheets are placed co-axially. The pass axes of polaroids $2$ and $3$ make angles of $30^{\circ}$ and $90^{\circ}$ respectively with the pass axis of polaroid $1$. If $I_0$ is the intensity of unpolarised light entering sheet $1$,what is the intensity of the light emerging from sheet $3$? $(\cos 30^{\circ}=\sqrt{3} / 2, \cos 90^{\circ}=0, \cos 60^{\circ}=1 / 2)$