Answer
7.3% 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 $\theta_2 = 60^{\circ}$
Then either $\theta_1$ or $\theta_3$ is $150^{\circ}$
On the graph, we can see that no light is transmitted when $\theta_2 = 140^{\circ}$
Then either $\theta_1$ or $\theta_3$ is $50^{\circ}$
We can assume that $\theta_1 = 150^{\circ}$ and $\theta_3 = 50^{\circ}$
Note that our final answer will be the same if $\theta_1 = 50^{\circ}$ and $\theta_3 = 150^{\circ}$
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.
$I_1 = \frac{1}{2}I_0$
It is given that $\theta_2 = 90^{\circ}$
Note that the angle between $\theta_1$ and $\theta_2$ is $60^{\circ}$
We can find an expression for $I_2$:
$I_2 = I_1~cos^2~60^{\circ}$
$I_2 = (\frac{1}{2}I_0)~cos^2~60^{\circ}$
$I_2 = 0.125~I_0$
Note that the angle between $\theta_2$ and $\theta_3$ is $40^{\circ}$
We can find an expression for $I_3$:
$I_3 = I_2~cos^2~40^{\circ}$
$I_3 = (0.125~I_0)~cos^2~40^{\circ}$
$I_3 = 0.073~I_0$
We can find the percentage of light that is transmitted:
$0.073 \times 100\% = 7.3\%$
7.3% of the initial intensity is transmitted by the system.
