According to question, the intensity of light coming out of the analyser is just 10% of the original intensity (I0) Using, I=I0cos2θ ⇒
I0
10
=I0cos2θ⇒
1
10
=cos2θ ⇒cosθ=
1
√10
=0.316⇒θ≈71.6° Therefore, the angle by which the analyser need to be rotated further to reduced the output intensity to be zero ϕ=90°−θ=90°−71.6°=18.4°