Question

Consider three polarizing filters oriented as shown in the figure below. A beam of unpolarized light is directed through the three filters. The first filter has a polarization axis along the vertical, the second has its polarization axis rotated at an angle of 30degree from the vertical, and the third has its axis along the horizontal. (a) What fraction of the light intensity comes out of the filter? I_f/I_0 = (b) What fraction makes it through the system if the middle filter is removed? I_f/I_0 = 0

Answer 1

By malus law from unpolarized to polarized

${I_1}=\frac{1}{2}*{I_o}$

from polarized to polarized

${I_2}=I_1*cos^2(\theta)$

${I_2}=\frac{I_0}{2}*cos^2(30^o)$

${I_2}=\frac{3}{8}I_0$

${I_3}=\frac{3}{8}I_0*cos^2(60^o)$

${I_3}=\frac{3}{32}I_0$

as

$I_3=I_f\Rightarrow \frac{I_f}{I_o}=\frac{3}{32}$