Statement: It states that the intensity of polarized light transmitted through the analyzer is directly proportional to the square of Cos θ where θ is the angle between the plane of the polarizer and the plane of the analyzer.
|i.e. I ∝ cos2 θ|
Proof: Let ‘a‘ be the amplitude of light transmitted through the polarizer and θ be the angle between the plane of the polarizer and the plane of the analyzer.
Resolve ‘a‘ into two components i.e., a cos θ along OA and a sin θ along OB. Clearly, a cos θ component will pass through the analyzer. We know,
|Intensity ∝ (Amplitude)2|
⇒ I ∝ (a cos θ)2
⇒ I ∝ a2 cos2 θ
⇒ I = k a2 cos2 θ
Put ka2 = Io = constant
⇒ I = Io cos2 θ
⇒ I ∝ cos2 θ
|Sample Problem: Two polaroids are crossed to each other. Now one of them is rotated through 60°. What percentage of incident unpolarized light will pass through the system?|
Solution: Initially the polaroids are at 90°. When one is rotated through 60°, the angle between their transmission axes is 30° i.e., θ = 30°.
Let Io be the intensity of the incident unpolarized beam. The intensity of the beam transmitted by the polarizer will be Io/2 as the other Io/2 is absorbed by the polaroid.
The intensity of light transmitted by the analyzer will, therefore, be given by
Io = Im cos2 θ
where Im = Io/2, θ = 30°, hence
Iθ = (Io/2) cos2 30° = (Io/2) (√3/2)2
∴ (Iθ/Io) x 100 = (3/8) x 100 = 37.5%