Question

Determine the fractio of the beam's intensity that is due to the polarised light

Answer 1

0
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 0
I
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, 0
I
I
0
=cos2()
=
cos
2
⁡
(
θ
)
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 0
I
0
, 1
I
1
, 2
I
2
, and 3
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 30
I
3
I
0
using my chosen notation...

The solution is easy to obtain once you know that

10
I
1
I
0
= 12
1
2
, this is a special case to Malus' Law when 0
I
0
is unpolarised

21
I
2
I
1
= cos2(55−10)
cos
2
⁡
(
55
−
10
)

32
I
3
I
2
= cos2(85−55)