Answer
(a) Less than
(b) $I_u=9.4W/m^2; I_p=12.1W/m^2$
Work Step by Step
(a) We know that $I_u$ is less than $I_p$ -- that is, an unpolarized light intensity will be less than the intensity of the linearly polarized light.
(b) We know that
$I=\frac{I_u}{2}+I_p cos^2 0$
$\implies 16.8=\frac{I_u}{2}+I_p$
This can be rearranged as:
$I_u+2I_p=33.6$......eq(1)
But when the polarized light is at $55^{\circ}$, then
$8.68=\frac{I_u}{2}+I_p cos^2 55^{\circ}$
$\implies I_u+(0.657)I_p=17.36$.....eq(2)
Solving eq(1) and (2), we obtain:
$I_p=12.1W/m^2$
and $I_u+2I_p=33.6$
$\implies I_u=33.6-2I_p$
$\implies I_u=33.6-2(12.1)$
$\implies I_u=9.4W/m^2$
