Answer
a) $I=I_{\circ}$
b) $18.6^{\circ}$
Work Step by Step
(a) We know that the intensity transmitted through eye 1 is given as
$I_1=I_{\circ}cos^2\theta$
and the intensity transmitted through eye 2 is
$I_2=I_{\circ}cos^2(90-\theta)$
$\implies I_2=I_{\circ}sin^2\theta$
Now the sum of intensities is given as
$I=I_1+I_2$
$\implies I=I_{\circ} cos^2\theta+I_{\circ} sin^2\theta$
$\implies I=I_{\circ}$
Thus, we can see that the sum of the intensities of the transmitted lights through theses eyes is equal to the intensities of the incident light.
(b) We know that
$I_{\circ}=I_1+I_2=163+662=825W/m^2$
Now $\theta_1=cos^{-1}(\sqrt{\frac{I}{I_{\circ}}})$
$\implies \theta_1=cos^{-1}(\sqrt{163}{825})=63.6^{\circ}$
and $\theta_2=cos^{-1}(\sqrt{\frac{662}{825}})$
$\theta_2=26.4^{\circ}$
The transmitted intensities of these eyes are equal only when $\theta_1=\theta_2=45^{\circ}$ and this is possible with a rotation of $18.6^{\circ}$ because $63.6^{\circ}-18.6^{\circ}=26.4^{\circ}+18.6^{\circ}=45^{\circ}$.
