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Magazine Chapter 5: Q27E (page 1106)
A beam of unpolarized light of intensity passes through a series of ideal polarizing filters with their polarizing axes turned to various angles as shown in Fig. E33.27. (a) What is the light intensity (in terms of) at points A, B, and C? (b) If we remove the middle filter, what will be the light intensity at point C?
Short Answer
(a)The light intensity at point a:
, point b:
, and, point c:
(b) The light intensity at point c after the middle filter is removed is I=0
Step by step solution
Malus’ law
According to Malus' law, the intensity of plane-polarized light that travels through an analyzer varies as the square of the cosine of the angle between the plane of the polarizer and the analyzer鈥檚 transmission axes.
No matter how the Polarizing access is oriented, when polarised light is incident on a perfect polarizer, the intensity of the transmitted light is exactly half that of the incident and price light. Because the incident light is a random mixture of all states of polarisation, the E field of the incident wave is divided into two components, one parallel to the polarising access and one perpendicular to it. Because the incident light is a random mixture of all states of polarisation, these two components are on average equal to the idol's full size.
Light intensity at points a, b and c
(a) 
At point a:
At point b:
At point c;
Hence, the light intensity at point a:
, point b:
, and, point c:
Light intensity at point c middle filter removed
(b) 
- which is consequent on the analyzer and comes from the first polarizer
Hence, the light intensity at point c after the middle filter is removed is I=0.
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