Answer
See answers.
Work Step by Step
a. The initial light is unpolarized. Each polarizer is oriented at 30 degrees from the one before. Apply equation 24–5 repeatedly.
$$I_1=\frac{1}{2}I_o$$
$$I_2=I_1cos^2 \theta_2 $$
$$I_3=I_2cos^2 \theta_3=\frac{1}{2}I_o cos^2 \theta_2 cos^2 \theta_3 $$
$$I_4=I_3cos^2 \theta_4=\frac{1}{2}I_o cos^2 \theta_2 cos^2 \theta_3 cos^2 \theta_4$$
$$I_4= \frac{1}{2}I_o cos^230^{\circ} cos^2 30^{\circ}cos^2 30^{\circ}=0.211I_o$$
b. Yes. Suppose the second polarizer were removed. The angle between polarizers 1 and 3 is now 60 degrees.
$$I_1=\frac{1}{2}I_o$$
$$I_3=I_1cos^2 \theta_3=\frac{1}{2}I_o cos^2 \theta_3 $$
$$I_4=I_3cos^2 \theta_4=\frac{1}{2}I_o cos^2 \theta_3 cos^2 \theta_4$$
$$I_4= \frac{1}{2}I_o cos^260^{\circ} cos^2 30^{\circ}=0.0938I_o$$
If the third polarizer were removed, we would calculate the same final intensity. The angle between polarizers 2 and 4 is now 60 degrees.
Therefore, we see that the intensity decreased when either the second or third polarizer was removed.
c. Yes. Remove both the second and third polarizers . The angle between polarizers 1 and 4 is now 90 degrees. No light will be transmitted.
