Question

Polarizing filter # $1$ Is oriented so that its polarizing axis is vertical.
Polarizing filter # $2$ is oriented so that its polarizing axis is rotated clockwise ${45}^{\circ }$ from filter # $1$
Polarizing filter # $3$ is oriented so that its polarizing filter is rotated ${90}^{\circ }$ from filter #$1$
Polarizing filter # $4$ oriented so that its polarizing is rotated ${135}^{\circ }$ from filter # $1$
Which sequence of filters-front to back-will block out all light that starts through the front filter?

• A. $1,2,3$
• B. $3,2,1$
• C. $4,1,2$
• D. $1,3,4$
• E. All of the combinations will block our all the light

Answer 1

The correct option is B $3,2,1$

The correct answer is option(B).

The intensity of the light passing through a polarizer is given by Malus' Law, I=I_0 cos(theta)2. When randomly polarized light (unpolarized light) passes through the first polarizer, the intensity is cut in half because there is an even distribution of incident angles, and the mean value of cos(theta)2 is 1/2 from 0 to pi/2. When it passes through the second polarizer, our "new I_0" (for Malus' Law) is 1/2 of the original I_0, and applying Malus' Law, we see that cos(45deg.)2=1/2. In other words, the intensity is halved again and the orientation of the polarization is shifted by 45 degrees. This then happens again on the final polarizer, resulting in a final intensity of 1/8 I_0.

So only case possible for no light is option(B).