I = 1.6 \(\displaystyle\frac{{W}}{{m}^{{2}}}\)

since the light is unpolarized at first

\(\displaystyle{I}={\left({0.5}\right)}\cdot{\left({1.6}\ \frac{{W}}{{m}^{{2}}}\right)}={0.8}\frac{{W}}{{m}^{{2}}}\)

Then use \(\displaystyle{I}={I}_{{0}}{\cos{{2}}}\theta\)

a.\(\displaystyle{I}={0.8}\cdot{\cos{{2}}}{\left({22.3}^{\circ}\right)}={0.684}\frac{{W}}{{m}^{{2}}}\)

\(\displaystyle{I}={0.684}\cdot{\cos{{2}}}{\left({42.6}^{\circ}-{22.3}^{\circ}\right)}={601.67}{m}\frac{{W}}{{m}^{{2}}}\)

b. \(\displaystyle{I}={0.8}\cdot{\cos{{2}}}{\left({42.6}^{\circ}\right)}={0.433}\frac{{W}}{{m}^{{2}}}\)

\(\displaystyle{I}={0.433}\cdot{\cos{{2}}}{\left({22.3}^{\circ}-{42.6}\right)}={879.6}{m}\frac{{W}}{{m}^{{2}}}\)

since the light is unpolarized at first

\(\displaystyle{I}={\left({0.5}\right)}\cdot{\left({1.6}\ \frac{{W}}{{m}^{{2}}}\right)}={0.8}\frac{{W}}{{m}^{{2}}}\)

Then use \(\displaystyle{I}={I}_{{0}}{\cos{{2}}}\theta\)

a.\(\displaystyle{I}={0.8}\cdot{\cos{{2}}}{\left({22.3}^{\circ}\right)}={0.684}\frac{{W}}{{m}^{{2}}}\)

\(\displaystyle{I}={0.684}\cdot{\cos{{2}}}{\left({42.6}^{\circ}-{22.3}^{\circ}\right)}={601.67}{m}\frac{{W}}{{m}^{{2}}}\)

b. \(\displaystyle{I}={0.8}\cdot{\cos{{2}}}{\left({42.6}^{\circ}\right)}={0.433}\frac{{W}}{{m}^{{2}}}\)

\(\displaystyle{I}={0.433}\cdot{\cos{{2}}}{\left({22.3}^{\circ}-{42.6}\right)}={879.6}{m}\frac{{W}}{{m}^{{2}}}\)