Answer
We can rank the intensities of light transmitted through the second polarizer, from greatest to smallest:
$a = b \gt e \gt d \gt c$
Work Step by Step
Since the light is horizontally polarized initially, the intensity of the light after passing through the first polarizer is $I_0~cos^2~\theta_1$
We can determine an expression for the intensity after passing through the second polarizer:
$I = (I_0~cos^2~\theta_1)~cos^2(\vert \theta_2-\theta_1\vert)$
For each situation, we can find an expression for the intensity of the light after passing through the two polarizers.
(a) $I = (I_0~cos^2~0^{\circ})~cos^2(30^{\circ}-0^{\circ}) = \frac{3}{4}\times I_0$
(b) $I = (I_0~cos^2~30^{\circ})~cos^2(30^{\circ}-30^{\circ}) = \frac{3}{4}\times I_0$
(c) $I = (I_0~cos^2~0^{\circ})~cos^2(90^{\circ}-0^{\circ}) = 0\times I_0$
(d) $I = (I_0~cos^2~60^{\circ})~cos^2(60^{\circ}-0^{\circ}) = \frac{1}{16}\times I_0$
(e) $I = (I_0~cos^2~30^{\circ})~cos^2(60^{\circ}-30^{\circ}) = \frac{9}{16}\times I_0$
We can rank the intensities of light transmitted through the second polarizer, from greatest to smallest:
$a = b \gt e \gt d \gt c$
