A) The intensity of the light passing through thesuccessive polarizers is

\(I_1=(\frac{1}{2})I_o\)

\(\displaystyle{I}_{{2}}={I}_{{1}}{{\cos}^{{2}}\theta_{{2}}}\)

\(\displaystyle={\left({\frac{{{1}}}{{{2}}}}\right)}{I}_{{o}}{{\cos}^{{2}}\theta_{{2}}}\)

\(\displaystyle{I}_{{3}}={I}_{{2}}{{\cos}^{{2}}\theta_{{3}}}\)

\(\displaystyle{\left({\frac{{{1}}}{{{2}}}}\right)}{I}_{{o}}{{\cos}^{{2}}\theta_{{2}}}{{\cos}^{{2}}\theta_{{3}}}\)

\(\displaystyle{I}_{{4}}={I}_{{3}}{{\cos{\theta}}_{{4}}}\)

\(\displaystyle={\left({\frac{{{1}}}{{{2}}}}\right)}{I}_{{o}}{{\cos}^{{2}}\theta_{{2}}}{{\cos}^{{2}}\theta_{{3}}}{{\cos}^{{2}}\theta_{{4}}}\)

\(\displaystyle{\left({\frac{{{1}}}{{{2}}}}\right)}{I}_{{o}}{{\cos}^{{2}}{30}}{{\cos}^{{2}}{30}}{{\cos}^{{2}}{30}}\)

\(\displaystyle={0.21}{I}_{{o}}\)

B) If we remove the second polarizer we get

\(\displaystyle{I}_{{1}}={\left({\frac{{{1}}}{{{2}}}}\right)}{I}_{{o}}\)

\(\displaystyle{I}_{{3}}={I}_{{1}}{{\cos}^{{2}}\theta_{{3}}}'\)

\(\displaystyle={\left({\frac{{{1}}}{{{2}}}}\right)}{I}_{{o}}{{\cos}^{{2}}\theta_{{3}}}'\)

\(\displaystyle{I}-{4}={I}_{{3}}{{\cos}^{{2}}\theta_{{4}}}'\)

\(\displaystyle{\left({\frac{{{1}}}{{{2}}}}\right)}{I}_{{o}}{{\cos}^{{2}}\theta_{{3}}}'{{\cos}^{{2}}\theta_{{4}}}'\)

\(\displaystyle{\left({\frac{{{1}}}{{{2}}}}\right)}{I}_{{o}}{{\cos}^{{2}}{60}}{{\cos}^{{2}}{30}}\)

\(\displaystyle={0.094}{I}_{{o}}\)

Thus we can decrease the intensity byremoving the second or third polarizer.

C) If we remove the second and thirdpolarizers we will have two polarizers with their axesperpendicular to each other and so no light willbe transmitted.