Answer
9.4 percent.
Work Step by Step
Let the initial intensity be $I_o$. For the first polarizer, since the incoming light is unpolarized, exactly one-half of the light is transmitted through it.
$$I_1=\frac{1}{2}I_o$$
For the second polarizer, the angle between the light’s polarization axis and the polarizer’s axis is $30^{\circ}$. Use equation 24-5 to find the light transmitted through it.
$$I_2=I_1cos^2\theta_{1,2}=\frac{1}{2}I_o cos^2 30^{\circ}$$
After the light emerges, it is polarized along that axis. We note that the angle given in the problem for the third polarizer is measured relative to the first polarizer, so the third polarizer’s axis is 60 degrees away from the second polarizer’s axis. When light passes through the third polarizer, the angle between the light’s polarization axis and the third polarizer’s axis is $60^{\circ}$. (Had we used 120 degrees instead, the answer would be the same.)
Use Eq. 24–5 again.
$$I_3=I_2cos^2\theta_{2,3}= \frac{1}{2}I_o cos^2 30^{\circ} (cos^2 60^{\circ})$$
$$=\frac{3}{32}I_o\approx0.094I_o$$
About $9.4\%$ of the original light is transmitted through the third polarizer.
