Unpolarized light is incident on a set of three polarizing sheets as shown in the figure at right. The polarizing directions of the sheets make angles 10°, 55° and 85° to the vertical.
What fraction of the incident intensity passes through all three sheets?
I am trying to use $I = I_0 \cos^2(\theta)$ and for the first sheet, I am trying to use
$I = 1/2 I_0 \cos^2(\theta)$. But what is $I_0$ for that?
Best Answer
$I_0$ is the intensity of light before it hits a polariser the original intensity of the beam, so called. You need it because you need to compare it to the intensity after it exits the polariser so that you can calculate your fraction of incident intensity.
this fraction requested by the problem is $I\over I_0$, but $I$ refers to intensity of light exiting the third filter (Malus' law only describes intensity in and intensity out for one filter). you would have to use Malus' Law, $I \over I_0$ $= \cos^2(\theta)$ three times. also notice the subtle rearrangement I made to make the left hand side refer to fractional intensity. this will be made clear in abit..
If I was working on the problem I would use $I_0$, $I_1$, $I_2$, and $I_3$ to refer to the original intensity, intensity exiting the first filter, intensity exiting the second filter, and intensity exiting the third filter, respectively. the final answer to the problem would then be $I_3\over I_0$ using my chosen notation...
The solution is easy to obtain once you know that
$I_1 \over I_0$ = $1 \over 2$ , this is a special case to Malus' Law when $I_0$ is unpolarised
$I_2 \over I_1$ = $ \cos^2(55-10)$
$I_3 \over I_2$ = $ \cos^2(85-55)$