CBSE Class 12 · 2017 · Wave OpticsLong Answer

$\text{Polarization by scattering:}$ unpolarised sunlight falling on air molecules sets their electrons oscillating. For light scattered at $90^{\circ }$, only the vibrations perpendicular to the plane containing the incident ray and the line of sight radiate towards the observer, so the scattered light is plane polarized.
$\text{Calculation:}$ A and B are crossed ($90^{\circ }$ apart), and C bisects the angle, so C makes $45^{\circ }$ with both A and B.
After A (unpolarised $I_0$): $I_A=\frac{I_0}{2}$.
Through C ($45^{\circ }$ to A): $I_C=\frac{I_0}{2}{\cos }^{2}45^{\circ }=\frac{I_0}{2}\cdot \frac{1}{2}=\frac{I_0}{4}$.
Through B ($45^{\circ }$ to C): $I_B=\frac{I_0}{4}{\cos }^{2}45^{\circ }=\frac{I_0}{4}\cdot \frac{1}{2}=\frac{I_0}{8}$.

(Without C, B transmits zero; introducing C restores some light through the crossed pair.)