A polarizer - analyser set is adjusted such that the intensityof light coming

(2) According to question, the intensity of light coming out of the analyser is just $10 \%$ of the original intensity 

$\left(I_{0}\right)$

Using, $I=I_{0} \cos ^{2} \theta$

$\Rightarrow \frac{I_{0}}{10}=I_{0} \cos ^{2} \theta \Rightarrow \frac{1}{10}=\cos ^{2} \theta$

$\Rightarrow \cos \theta=\frac{1}{\sqrt{10}}=0.316 \Rightarrow \theta \approx 71.6^{\circ}$

Therefore, the angle by which the analyser need to be rotated further to reduced the output intensity to be zero

$\phi=90^{\circ}-\theta=90^{\circ}-71.6^{\circ}=18.4^{\circ}$