Answer
a. 0.
b. 0.11.
c. 0.
Work Step by Step
a. The initial light is unpolarized. Apply equation 24–5.
$$I_1=\frac{1}{2}I_o$$
$$I_2=I_1cos^2 \theta_2= I_1cos^2 90^{\circ}=0 $$
b. Let a new polarizer be inserted. The angle between the first and second polarizers is 56 degrees, and the angle between the second and third polarizers is 34 degrees.
$$I_1=\frac{1}{2}I_o$$
$$I_2=I_1cos^2 \theta_2 $$
$$I_3=I_2cos^2 \theta_3=\frac{1}{2}I_o cos^2 \theta_2 cos^2 \theta_3 $$
$$I_3= \frac{1}{2}I_o cos^2 56^{\circ} cos^2 34^{\circ} =0.11I_o$$
c. If, instead, the new polarizer were to be placed in front of the two crossed polarizers, we would be back to the first case. Two perpendicular polarizers, if they occur one after the other, will allow no light to pass.
