Q. Unpolarized light of intensity $I_{0}$ is incident on a series of three polarizing filters. The axis of the second filter is oriented at $45^{\circ}$ to that of the first filter, while the axis of the third filter is oriented at $90^{\circ}$ to that of the first filter. The intensity of the light transmitted through the third filter is $\frac{I_{0}}{n}$. Find $n$.
Wave Optics
Solution:
The first filter always reduces the intensity of the light to half, $I_{1}=1 / 2 I_{0}$
The next filter reduces the intensity by $I_{n}=I_{(n-1)} \cos ^{2} \theta$.
$\theta$ is the angle with respect to the $n^{\text {th }}$ filter.
Thus,
