Answer
(a) greater than
(b) III
Work Step by Step
(a) We know that the intensity of the transmitted light after the first polarizer is given as
$I_1=I_{\circ}cos^2(90-\theta)$
and the light transmitted after the second polarizer is given as
$I_2=I_1cos^2(90-\theta)\cdot cos^2\theta$
However, in case (B), the first polarizer has its transmission axis perpendicular to the polarized light; therefore, the intensity of the transmitted light will be zero and we conclude that the transmitted intensity in case A is greater than that in case B.
(b) We know that the best explanation is option (III) -- that is, the transmitted intensity in case B is smaller than in case A; in fact, the transmitted intensity in case B is zero because the first polarizer is oriented vertically.
