An unpolarized beam of light is incident on a stack of ideal polarizing filters.

Question

An unpolarized beam of light is incident on a stack of ideal polarizing filters. Find the fraction by which the transmitted beam's intensity is reduced in the following three cases.
(a) Three filters are in the stack, each with its transmission axis at 43.7° relative to the preceding filter.


(b) Four filters are in the stack, each with its transmission axis at 30.4° relative to the preceding filter.


(c) Seven filters are in the stack, each with its axis at 16.6° relative to the proceeding filter.

Explanation / Answer

Let the intensity of unpolarized beam is I
After passing the first filter the intensity becomes I/2
After passing the second filter the intensity becomes I/2 cos243.7
After passing the third filter the intensity becomes I/2 cos243.7 * cos243.7
   i.e. I/2 cos443.7
So the intensity reduced is given by I - I/2 cos443.7
It can be written as, I [1 - (cos443.7)/2 ]
So the required fraction is given by = 1 - (cos443.7)/2 = 0.863
b)
similarly
After passing the first filter the intensity becomes I/2
After passing the second filter the intensity becomes I/2 cos230.6
After passing the third filter the intensity becomes I/2 cos230.6 * cos230.6
   i.e. I/2 cos430.6
So the intensity reduced is given by I - I/2 cos430.6
It can be written as, I [1 - (cos430.6)/2 ]
So the required fraction is given by = 1 - (cos430.6)/2 = 0.726 After passing the first filter the intensity becomes I/2
After passing the second filter the intensity becomes I/2 cos230.6
After passing the third filter the intensity becomes I/2 cos230.6 * cos230.6
   i.e. I/2 cos430.6
So the intensity reduced is given by I - I/2 cos430.6
It can be written as, I [1 - (cos430.6)/2 ]
So the required fraction is given by = 1 - (cos430.6)/2 = 0.726
c)
After passing the first filter the intensity becomes I/2
After passing the second filter the intensity becomes I/2 cos216.6
After passing the third filter the intensity becomes I/2 cos216.6 * cos216.6
   i.e. I/2 cos416.6
So the intensity reduced is given by I - I/2 cos416.6
It can be written as, I [1 - (cos416.6)/2 ]
So the required fraction is given by = 1 - (cos416.6)/2 = 0.578 After passing the first filter the intensity becomes I/2
After passing the second filter the intensity becomes I/2 cos216.6
After passing the third filter the intensity becomes I/2 cos216.6 * cos216.6
   i.e. I/2 cos416.6
So the intensity reduced is given by I - I/2 cos416.6
It can be written as, I [1 - (cos416.6)/2 ]
So the required fraction is given by = 1 - (cos416.6)/2 = 0.578 After passing the first filter the intensity becomes I/2
After passing the second filter the intensity becomes I/2 cos216.6
After passing the third filter the intensity becomes I/2 cos216.6 * cos216.6
   i.e. I/2 cos416.6
So the intensity reduced is given by I - I/2 cos416.6
It can be written as, I [1 - (cos416.6)/2 ]
So the required fraction is given by = 1 - (cos416.6)/2 = 0.578

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