Answer
a) $I=\frac{1}{2}I_{\circ}$
b) $0.375I_{\circ}$
c) $0.09375I_{\circ}$
d) $0$
Work Step by Step
(a) We know that when an uplolarized light from the laser passes through the polarizer, then the transmitted intensity is reduced by half and hence the intensity at point A is given as $I=\frac{1}{2}I_{\circ}$
(b) We know that
$I^{\prime}=Icos^2cos30^{\circ}$
We plug in the known values to obtain:
$I^{\prime}=\frac{1}{2}I_{\circ}(0.7499)=0.375I_{\circ}$
(c) The intensity at point C is given as
$I^{\prime \prime}=I^{\prime}cos^2(90-30)$
$I^{\prime \prime}=I^{\prime}cos^2(60^{\circ})$
$I^{\prime \prime}=\frac{1}{2}I_{\circ}cos^2(30^{\circ})cos^2(60^{\circ})$
$I^{\prime \prime}=0.09375I_{\circ}$
(d) We know that if the filter 2 is removed then the intensity at point C is given as
$I=\frac{1}{2}I_{\circ}cos^290^{\circ}$
$I=\frac{1}{2}I_{\circ}(0)=0$
