Question

Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be $\frac{1}{2}$. Now another identical polarizer C is placed between A and B . The intensity beyond B is now found to be $\frac{I}{8}$. What angle polarizer C makes with A?

Answer 1

Initially polarizer A and B are placed let potarizer B make an angle $\alpha$ with that of A.

Ref. image

after crossing polarizer A, intensity $I_1$ becomes ;

$I_1=\frac{I}{2}$ [unpolarized light intensity reduces to half after crossing polarizer]

Now after passing B, from molus law :

$I_2=I_{1}co{s}^2\alpha$

Given that $I_2=I/2$ so :-

$\frac{I}{2}=\frac{I}{2}co{s}^2\alpha \Rightarrow co{s}^2\alpha =1$

$\Rightarrow cos\alpha =\pm 1\alpha =0^o$ or ${180}^o$

Hence polarizer B is parallel or anti parallel to A

Now when polarizer c is paces between A & B :-

Ref. image

If c makes angle $\theta$ with A , it makes angle O with B also.

when light passes c, intensity :

$I_3=\frac{I}{2}co{s}^2\theta$ [from figure]

again when if passes B,

$I_2=I_{3}co{s}^2\theta$

$\Rightarrow \frac{I}{8}=\frac{I}{2}co{s}^4\theta$ (given $I_2=I/8$)

$\Rightarrow co{s}^4\theta =\frac{1}{4}\Rightarrow cos\theta =\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \theta =\frac{\pi }{4}$ or $\frac{3\pi }{4}$

$\therefore$ polarizer c makes angle $\frac{\pi }{4}$ or $\frac{3\pi }{4}$ with A.