Answer
(a) $50\%$
(b) $0\%$
(c) $63.4^{\circ}$
(d) $0.138I_{\circ}$
Work Step by Step
(a) We know that for un-polarized light, $I_1=\frac{I_{\circ}}{2}$. We can say that in terms of incident light, it is $50\%$.
(b) We know that if the two fingers are in a crossed position, the transmitted light through the second filter is zero; that is, $I_2=\frac{I_{\circ}}{2}.cos^2(90^{\circ})=0$. Thus, $0\%$ of incident light will pass through the second filter.
(c) We know that
$0.400I_{\circ}=0.500I_{\circ}cos^2(90^{\circ}-\theta)$
This simplifies to:
$\theta=63.4^{\circ}$
(d) We know that the intensity of the transmitted light is given as
$I_2=\frac{1}{2}I_{\circ}cos^2(90^{\circ}-\frac{63.4^{\circ}}{2})$
$\implies I_2=0.138I_{\circ}$
