Answer
The light that emerges has an intensity of $~~0.50~W/m^2$
Work Step by Step
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 $70^{\circ}$
We can find an expression for $I_2$:
$I_2 = I_1~cos^2~70^{\circ}$
$I_2 = \frac{1}{2}I_0~cos^2~70^{\circ}$
Note that the angle between $\theta_2$ and $\theta_3$ is $40^{\circ}$
We can find an expression for $I_3$:
$I_3 = I_2~cos^2~40^{\circ}$
$I_3 = \frac{1}{2}I_0~cos^2~70^{\circ}~cos^2~40^{\circ}$
Note that the angle between $\theta_3$ and $\theta_4$ is $40^{\circ}$
We can find an expression for $I_4$:
$I_4 = I_3~cos^2~40^{\circ}$
$I_4 = \frac{1}{2}I_0~cos^2~70^{\circ}~cos^4~40^{\circ}$
$I_4 = 0.020~I_0$
We can find the intensity of the light that emerges:
$I = (0.020)(25~W/m^2) = 0.50~W/m^2$
The light that emerges has an intensity of $~~0.50~W/m^2$
