Answer
9.4% of the initial intensity is transmitted by the system.
Work Step by Step
On the graph, we can see that no light is transmitted when $\theta_2 = 0^{\circ}$
Then either $\theta_1$ or $\theta_3$ is $90^{\circ}$
On the graph, we can see that no light is transmitted when $\theta_2 = 90^{\circ}$
Then either $\theta_1$ or $\theta_3$ is $0^{\circ}$
We can assume that $\theta_1 = 0^{\circ}$ and $\theta_3 = 90^{\circ}$
Note that our final answer will be the same if $\theta_1 = 90^{\circ}$ and $\theta_3 = 0^{\circ}$
Let $I_0$ be the original intensity of the light.
Since the light is initially unpolarized, half the intensity will be transmitted through the first polarizing sheet.
$I_1 = \frac{1}{2}I_0$
Note that the angle between $\theta_1$ and $\theta_2$ is $30^{\circ}$
We can find an expression for $I_2$:
$I_2 = I_1~cos^2~30^{\circ}$
$I_2 = (\frac{1}{2}I_0)~cos^2~30^{\circ}$
$I_2 = 0.375~I_0$
Note that the angle between $\theta_2$ and $\theta_3$ is $60^{\circ}$
We can find an expression for $I_3$:
$I_3 = I_2~cos^2~60^{\circ}$
$I_3 = (0.375~I_0)~cos^2~60^{\circ}$
$I_3 = 0.094~I_0$
We can find the percentage of light that is transmitted:
$0.094 \times 100\% = 9.4\%$
9.4% of the initial intensity is transmitted by the system.
